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Aortic valve analysis and area prediction using bayesian modeling

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Aortic valve analysis and area prediction using bayesian modeling
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English
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Ghotikar, Miheer S
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Heart valves
Hemodynamics
Valvular stenosis
Porcine valves
Cross-linking agents
Pulse duplicator
Bayesian-learning networks
Dissertations, Academic -- Biomedical Engineering -- Masters -- USF   ( lcsh )
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theses   ( marcgt )
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ABSTRACT: Aortic Valve Analysis and Area Prediction using Bayesian Modeling Miheer S. Ghotikar ABSTRACT Aortic valve stenosis affects approximately 5 out of every 10,000 people in the United States. 3 This disorder causes decrease in the aortic valve opening area increasing resistance to blood flow. Detection of early stages of valve malfunction is an important area of research to enable new treatments and develop strategies in order to delay degenerative progression. Analysis of relationship between valve properties and hemodynamic factors is critical to develop and validate these strategies. Porcine aortic valves are anatomically analogous to human aortic valves. Fixation agents modify the valves in such a manner to mimic increased leaflet stiffness due to early degeneration. In this study, porcine valves treated with glutaraldehyde, a cross-linking agent and ethanol, a dehydrating agent were used to alter leaflet material properties.The hydraulic performance of ethanol and glutaraldehyde treated valves was compared to fresh valves using a programmable pulse duplicator that could simulate physiological conditions. Hydraulic conditions in the pulse duplicator were modified by varying mean flow rate and mean arterial pressure. Pressure drops across the aortic valve, flow rate and back pressure (mean arterial pressure) values were recorded at successive instants of time. Corresponding values of pressure gradient were measured, while aortic valve opening area was obtained from photographic data. Effects of glutaradehyde cross-linking and ethanol dehydration on the aortic valve area for different hydraulic conditions that emulated hemodynamic physiological conditions were analyzed and it was observed that glutaradehyde and ethanol fixation causes changes in aortic valve opening and closing patterns.Next, relations between material properties, experimental conditions, and hydraulic measures of valve performance were studied using a Bayesian model approach. The primary hypothesis tested in this study was that a Bayesian network could be used to predict dynamic changes in the aortic valve area given the hemodynamic conditions. A Bayesian network encodes probabilistic relationships among variables of interest, also representing causal relationships between temporal antecedents and outcomes. A Learning Bayesian Network was constructed; direct acyclic graphs were drawn in GeNIe 2.0ʾ using an information theory dependency algorithm. Mutual Information was calculated between every set of parameters. Conditional probability tables and cut-sets were obtained from the data with the use of Matlabʾ.A Bayesian model was built for predicting dynamic values of opening and closing area for fresh, ethanol fixed and glutaradehyde fixed aortic valves for a set of hemodynamic conditions. Separate models were made for opening and closing cycles. The models predicted aortic valve area for fresh, ethanol fixed and glutaraldehyde fixed valves. As per the results obtained from the model, it can be concluded that the Bayesian network works successfully with the performance of porcine valves in a pulse duplicator. Further work would include building the Bayesian network with additional parameters and patient data for predicting aortic valve area of patients with progressive stenosis. The important feature would be to predict valve degenration based on valve opening or closing pattern.
Thesis:
Thesis (M.S.B.E.)--University of South Florida, 2005.
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by Miheer S. Ghotikar.
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Aortic Valve Analysis and Area Pr ediction using Bayesian Modeling by Miheer S. Ghotikar A thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in Biomedical Engineering Department of Chemical Engineering College of Engineering University of South Florida Major Professor: Michael Vanauker, Ph.D. Joel Strom, M.D. William E. Lee III, Ph.D. Sanjukta Bhanja, Ph.D. Date of Approval: October 28, 2005 Keywords: heart valves, hemodynamics, valvular stenosis, porcine va lves, cross-linking agents, pulse duplicator, Bayesian-Learning networks Copyright 2005, Miheer S. Ghotikar

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Dedication I would like to dedicate my thesis to my late and dear grandfather Mr. Nilkanth Rayarikar. His love and support encouraged me to pursue my dreams and inspired me to accomplish my goals. I hope he is watching and his blessings are always there with me.

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Acknowledgments I take this opportunity to express my grat itude to all the people who have helped me accomplish this work, directly or indirectl y. I would first like to thank my parents for their immense love and motivation. Without th eir support I would not have achieved this. I would like to thank Dr. Michael VanAuker for being my major professor, my guide and giving me the opportunity to work with the Cardiovascular Engineering Research Group, Dr. Sanjukta Bhanja for providi ng me excellent direction in Bayesian Networks and Dr. Joel Strom for providing valu able information and helping me with all of my questions in physiology. I would like to thank Sergio, Rosana and John for their hard work and contribution to the project. I also thank Reetu, Liz, Brian, Monica and everyone else in the cardiovascular team. I would like to acknowledge Dr. William Lee for his being my academic advisor and providi ng his valuable time to discuss and review my work. I thank all of my friends and my roommates Saket, Sam and Pushkar for their help and support. Finally, I thank Carolyn Grai sbery, USF Department of Pediatrics for adjusting my work schedule and providing some va luable suggestions for the project

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i Table of Contents List of Tables iii List of Figures iv Abstract vi Chapter 1 Introduction 1 Chapter 2 Background 3 2.1 Aortic valve 3 2.2 Cusp anatomy 4 2.3 Hydraulic behavior of aortic valve 6 2.4 Aorta 7 2.5 Etiology of aor tic valve stenosis 7 2.6 Different markers to de fine aortic stenosis 9 Chapter 3 Valve Fixation 14 3.1 Porcine tissue valves 14 3.2 Glutaraldehyde fixation 14 3.3 Ethanol fixation 16 Chapter 4 Methods 18 4.1 Valve selection 18 4.2 Method of fixation 20 4.3 Working of pulse duplicator 21 4.4 Calculation of aortic valve areas 26 4.5 Calibration of Image-J 27 Chapter 5 Bayesian Modeling 31 5.1 Bayesian networks 31 5.2 Directed acyclic graphs 32 5.3 Finding cutsets for a pair of nodes 34 5.4 Data for GeNIe 2.0 35 5.4.1 Categorizing data into states 35 5.4.2 Selection of data 36 5.4.3 Order of nodes 36 5.4.4 Bayesian network in GenIe 2.0 36

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ii Chapter 6 Results 39 6.1 Test results 39 6.2 Results for gradient and area relation 43 6.3 Hysteresis loop 45 6.4 Results for Bayesian networks 48 6.4.1 Results for mutual information 48 6.4.2 Cutsets 51 6.4.3 Prediction 53 6.4.4 Hydraulic prediction 58 Chapter 7. Discussion 62 Chapter 8. Conclusions and Recommendations 63 References 64 Appendices 68 Appendix A 69 A1: Program to make Bayesi an file for opening cycle 69 A2: Program to calculate mutual information for opening cycle 78 A3: Program to find cutsets for opening cycle 87 A4: Program for closing cycle 103

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iii List of Tables Table 1. Relation of aort ic valve area (AVA) and se verity of stenosis 11 Table 2. Porcine valve sizes 19 Table 3. Calibration of compliance chamber 24 Table 4. Grouping of va riables in states 35 Table 5. Test results 39 Table 6. Maximum da/dt 48 Table 7. Mutual information for flow rate of 4L/min at 90 bp 49 Table 8. Results for fresh valve 54 Table 9. Results for ethanol fixed valve 55 Table 10. Results for glut araldehyde fixed valve 56

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iv List of Figures Figure 1. Aortic valve 3 Figure 2. Cusp anatomy 4 Figure 3. Cusp cross-section 5 Figure 4. Endothelial disruption 5 Figure 5. Calcific and rheuma tic aortic valve stenosis 9 Figure 6. Representations of monomeric glutaraldehyde 15 Figure 7. Reaction of poly (glutarald ehyde) with amino protein groups 16 Figure 8. Valves used for experiments 20 Figure 9. Pulse duplicator 21 Figure 10. Labview data acquisi tion in pulse duplicator 23 Figure 11. Beat to beat variability for flow 25 Figure 12. Beat to be at variability for gradient 25 Figure 13. Verification of area in pixels to cm2 using a circle 26 Figure 14. Verificat ion of area in pixel to cm2 using a square 27 Figure 15. Fresh valve 1 in pulse duplicator 27 Figure 16. Diameter used for scaling 28 Figure 17. Data in .xls format co llected from pulse duplicator 29 Figure 18. Network given flow as evidence 37 Figure 19. Network given area as evidence 38

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v Figure 20. Maximum area vs. mean flow 42 Figure 21. Peak gradient vs. mean flow 42 Figure 22. Gradient area plot for 3L/min at 90bp 43 Figure 23. Gradient-a rea plot for 4L/min at 90bp 44 Figure 24. Gradient area plot for 5L/min at 90bp 44 Figure 25. Ethanol fixed valve 3 hysteresis loop for 4L/min at 60 bp 45 Figure 26. Ethanol fixed valve 3 hysteresis loop for 4L/min at 90 bp 46 Figure 27. Valve opening 47 Figure 28. Valve closing 47 Figure 29. Back pressure at state 95 to 100 (evi dence) network with and without cutsets 52 Figure 30. Back pressure at state 90 to 95 (evidence) network with and without cutsets 53 Figure 31. Prediction of area in fresh valves 57 Figure 32. Prediction of area in ethanol fixed valve 57 Figure 33. Prediction of area in glutaraldehyde fixed valves 58 Figure 34. Results for fresh vs. fixe d valves for opening cycle given gradient as evidence 59 Figure 35. Results for fresh vs. fixed valves for closing cycle given gradient as evidence 59 Figure 36. Results for fresh valve for high and low back pressure 60 Figure 37. Results for fresh vs. fixed va lves for opening cycle given flow as evidence 61

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vi Aortic Valve Analysis and Area Pr ediction using Bayesian Modeling Miheer S. Ghotikar ABSTRACT Aortic valve stenosis affects approximately 5 out of every 10,000 people in the United States. [3] This disorder causes d ecrease in the aortic valve opening area increasing resistance to blood fl ow. Detection of early stages of valve malfunction is an important area of research to enable new trea tments and develop strategies in order to delay degenerative progression. Analysis of relationship between valve properties and hemodynamic factors is critical to de velop and validate these strategies. Porcine aortic valves are anatomically analogous to human aortic valves. Fixation agents modify the valves in such a manner to mimic increased leaflet stiffness due to early degeneration. In this study, porcine va lves treated with glutaraldehyde, a crosslinking agent and ethanol, a dehydrating agent were used to alter leaflet material properties. The hydraulic performance of ethanol a nd glutaraldehyde tr eated valves was compared to fresh valves using a programm able pulse duplicator that could simulate physiological conditions. Hydraulic conditions in the pulse dup licator were modified by

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vii varying mean flow rate and mean arterial pre ssure. Pressure drops across the aortic valve, flow rate and back pressure (mean arterial pr essure) values were recorded at successive instants of time. Corresponding va lues of pressure gradient were measured, while aortic valve opening area was obtained from photographic data. Effect s of glutaradehyde crosslinking and ethanol dehydration on the aortic valve area for different hydraulic conditions that emulated hemodynamic physiological condi tions were analyzed and it was observed that glutaradehyde and etha nol fixation causes changes in aortic valve opening and closing patterns. Next, relations between material prop erties, experimental conditions, and hydraulic measures of valve performance were studied using a Baye sian model approach. The primary hypothesis tested in this study was that a Bayesian network could be used to predict dynamic changes in th e aortic valve area given the hemodynamic conditions. A Bayesian network encodes proba bilistic relationshi ps among variables of interest, also representing causal relationships between temporal antecedents and outcomes. A Learning Bayesian Network was constructed; di rect acyclic graphs were drawn in GeNIe 2.0 using an information theory depende ncy algorithm. Mutual Information was calculated between every set of parameters. Conditional probability tables and cut-sets were obtained from the data with the use of Matlab. A Bayesian model was built for predicting dynamic values of opening and closing area for fresh, ethanol fixed and glutaradehyde fixed aortic valves for a set of hemodynamic conditions.

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viii Separate models were made for opening and closing cycles. The models predicted aortic valve area for fresh, et hanol fixed and glutaraldehyde fixed valves. As per the results obtained from the model, it can be concluded that the Bayesian network works successfully with the performan ce of porcine valves in a pulse duplicator. Further work would include building the Bayesian network wi th additional parameters and patient data for predicting aortic valve area of patients with progressive stenosis. The important feature would be to predict valve degenera tion based on valve openi ng or closing pattern.

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1 Chapter 1 Introduction Aortic valvular stenosis is defined as an acquired or conge nital narrowing of the aortic valve orifice that obstructs the left ventricular outflow by in creasing resistance to blood flow from left ventricle to the aorta. Th e most common causes of aortic stenosis are leaflet degeneration, congenital valve malf ormations and inflammation, e.g. rheumatic fever. The degenerative form of disease is the prevalent form in Unites States. [36] Analysis of pathophysiology of valvular ste nosis is an important area of research to develop new treatment strategies. Invasi ve (cardiac catheteriza tion, angiography) and non-invasive methods (ultrasound imaging, Do ppler techniques) used currently for diagnosis are based on hydraulic formulae. Th ey combined with patients symptoms are used to determine the timing of aortic valv e replacement, the only available therapeutic option. The progression of aortic stenosis is nonlinear and the patient is asymptomatic for long time. The deterioration of the valve performance can progress rapidly and heart valve surgery is indicated when symptoms develop. Therefore, early detection of the aortic valve disorder is necessary to prevent prog ression of valve disease.

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2 The experiments described in this thesis are related to the assessment and prediction of dynamic values of aortic va lve area based on valve leaflet material properties and other hemodynamic parameters using Bayesian modeling. The data used to build the Bayesian model was collected from fresh (untreated), ethanol treated and glutaraldehyde treated porcine va lves tested in a pulse duplicator under varying conditions of flow and pressure.

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3 Chapter 2 Background 2.1 Aortic valve The aortic valve is situated between th e left ventricular outflow tract and the aorta. It acts as a one-way valve to allow the left ventricle to eject blood into the aorta in systole while preventing regurgitation in to the left ventricle in diastole. Figure 1. Aortic valve [1] The aortic valve is composed of thre e cusps of roughly e qual area. The cusps open against the aortic wall du ring systole to a triangular or ifice (Figure 1). During the diastole, they close rapidly and completely under minimal reverse pressure. The orifice area in a normal size adult is 3.0 to 4.0 cm2. [25] As these cusp s cycle, there are substantial and repetitive changes in size and shape. In particular, the aortic valve cusps

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4 have nearly 50% greater area in diastole than in systole. [5] This requires complex and cyclical structural rearrangements. The ao rtic valve has a highly layered complex structure and highly specialize d, functionally adapted cells and extra cellular matrix. [5] 2.2 Cusp anatomy Figure 2. Cusp anatomy [5] Figure 2 shows a single aortic valve cusp. At the top of the cusp is the free edge, the part of the cusp that is freely movabl e during the blood flow. Just in from the free edge along the upper portion of the cusp is th e coaptation region, which is the portion that joins the neighboring cusps. The curved base portion connects the cusp to the aortic wall. The regions where the free edge meets the ao rta are called the commissures. The corpus arantii (or nodulus of Arantus) is a large collagenous mass in the coaptation region that is believed to aid in valve closur e and reduces regurgitation. [5] A cross-sectional view of a heart valve cusp is shown in Figure 3. The ventricularis, facing the inflow surface is predominantly collagenous with radially

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5 aligned elastic fibers. The centrally located spongiosa is composed of loosely arranged collagen and glycosamaminoglycans (GAGs). Th e fibrosa, facing th e outflow surface is composed predominantly of circumferentiall y aligned, densely packed collagen fibers. They are largely arranged parallel to the cuspal free edge. [5] Figure 3. Cusp cross-section [5] Figure 4. Endothelial disruption [29]

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6 A single layer of endothelia l cells lines the su rfaces of leafle ts separating the fibrosa on the aortic side and the ventricularis on the ventri cular side (Figure 4). When repetitive stresses from blood damage this laye r, small molecules such as cholesterol can invade and alter the structure. Aortic stenos is is thought to start with an endothelial disruption on the aortic side yielding to thickening of the subendothelium and adjacent fibrosa due to accumulation of lipids and inflammatory cells. [37] 2.3 Hydraulic behavior of aortic valve In systole, the pressure in the left ventricular outflow tr act exceeds that of ascending aorta, causing the aortic valve to op en. Valve closure occurs when pressure in aortic root exceeds ventricular pressure. Thus by opening, the valve controls the direction of blood flow, and by closing it allows pressure differentials to exist in a closed system. During the diastole, the initial backflow of blood caused by a drop in ventricular pressure fills the sinuses of Valsalva, this shuts the ao rtic valve and prevents further backflow into the ventricles. The sinuses of Valsalva are three small outpouchings in the most proximal aorta, just above the cusps of the ao rtic valve. They generate systolic blood flow vortexes. Besides avoiding the occlusion of the co ronary ostia, the vortexes push the cusps medially to promote their coaptation as soon as ventricular ejection ceases; thus preventing regurgitation. [30] They also enhance coronary performance.

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7 2.4 Aorta The aorta is the major trunk of a series of vessels that supp ly oxygenated blood to the tissues of the body for nutrition. It comm ences at the outflow tract of the left ventricle, where it is 2.5 to 3.0 cm in di ameter. After ascending for a short distance, arches backward and to the left side, over the ro ot of the left lung; it then descends within the thorax on the left side of the verteb ral column, passes into the abdominal cavity through the aortic hiatus in the diaphragm, and ends, considerably diminished in size (about 1.75 cm in diameter), opposite the lowe r border of the fourth lumbar vertebra, by separating into the right and left common iliac arteries. Thus it is described in several portions, viz., the ascending aorta, the arch of the aorta, and the descending aorta, which last is again divided into the thoracic and abdominal aorta. [7] 2.5 Etiology of aortic valve stenosis In adults, there are three majo r conditions that cause aortic stenosis. They are stated as below. Congenitally malformed valves, e.g. bicuspid valve present from birth. Scarring of the aortic valve caused by infl ammatory diseases e.g. rheumatic fever. Age related degeneration and calci fication of the aortic valve. A congenital presence of bicu spid aortic valve (valve ha ving 2 cusps) is the most common cause of aortic stenosis in adult patients under age 65. About 2% of people are born with bicuspid aortic valves (2 cusp s). Although bicuspid valves usually do not impede blood flow when the patients are young, they do not open as widely as normal

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8 valves with three cusps. Therefore blood flow pattern is distorted, accelerating wear and tear on the valve cusps. Eventually, excessi ve wear and tear l eads to calcification, scarring, and reduced mobility of the valve cu sps. About 10% of bicuspid valves become significantly narrowed to cause the symptoms. [1] Rheumatic fever is an inflammatory c ondition resulting from untreated infection by group A streptococcal bacteria. Cusp dama ge takes the form of cusp thickening, retraction and commissural fusion (Figure 5). Rheumatic aortic stenosis usually occurs with some degree of aortic regurgitation. In aortic regurgitation, the diseased valve allows leakage of blood back into the left ventri cle as the ventricula r muscles relax after pumping. [1] The most common cause of aortic stenosis in patients more than 65 years of age is known as senile calcific aortic stenosis (Figure 5). Along with aging, protein collagen of the valve cusps is destroyed, leading to te ndency of scarring and ultimately calcium deposition. Once when the valve cusp mobility is reduced by fibrosis, the turbulence across the valve increases accelerating this proce ss. In Figure 5 below, calcific stenosis shows deposition of calcium on the valve cu sps while the commissures are relatively unaffected, in contrast rheumatic sten osis results in commissural fusion.

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9 Figure 5. Calcific and rheumatic aortic valve stenosis [15] Other etiologies of aortic valve stenos is include congenital unicuspid valve, postinflammatory (rheumatic) state, which occurs usually in association with mitral valve involvement. Patients with aortic stenosis may develop concentric left ventricular hypertrophy, pulmonary hypertension, infective endocarditis, and systemic embolization. Currently, there is no known therapy that ca n slow or reverse disease progression in patients with calcific aortic stenosis However some studies indicate that, statins may significantly delay hemodynamic progression both in mild-to-moderate and in severe aortic stenosis. [31] Current management includes monitoring disease progression, and ensuring patient awareness of the need fo r antibiotic prophylaxis against infective endocarditis. For those patients with severe symptomatic disease, the only therapeutic option is aortic valve replacement. [7] 2.6 Different markers to quantify severity of aortic stenosis 1. Pressure drop: The fundamental hydrauli c effect of narrowed aortic valve is development of increased pressure drop acros s the valve. Use of Doppler ultrasound to estimate the severity of a valve stenosis is based principally on the fact that such obstructions result in an increa se in the velocity of flow. For any given pressure gradient

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10 there is a corresponding increase in veloci ty, as predicted by the simplified Bernoulli equation: p1 p2 = 4V 2 Where, p1 p2 = pressure drop across the valve and V = peak velocity in the aortic jet. General assumptions leading to simplified Bernoulli equation are: 1. Fluid is incompressible 2. Steady state of flow 3. Fully developed flow (There is negligible viscous loss) 4. Pressure recovery is small co mpared to transvalvular losses When the blood flows from aortic valve it is spatially accelerated from the left ventricular outflow tract to th e vena contracta. During the acceleration a part of static pressure (potential energy) is converted in to dynamic pressure (kinetic energy). As the blood flows to the ascending aorta, a certain amount of dynamic pressure is converted to static pressure. The phenomenon of pressure recovery may be clin ically relevant in patients with moderate or severe stenosis. [35] It should be recognized th at knowledge of the gradient across a stenotic valve does not provide all the information necessary to assess the severi ty of obstruction. The gradient varies with flow rate across the stenotic valve orifice. 2. Aortic valve orifice area (AVA) : It has been reported that th e rate of change in aortic valve area in the cardiac cycle can predic t the rate of hemodynamic progression. [4] The

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11 degree of aortic stenosis depends on the decreas e in aortic valve orifice area. [25] On the basis of hemodynamic and natural history, the disease is graded as shown in Table 1. Table 1. Relation of aortic valve area ( AVA) and severity of stenosis [25] Aortic StenosisAVA (cm2) Mild >1.5 Moderate 1.0 to 1.5 Severe <=1.0 There are several ways to estimate orifice area. a) Continuity equation: This employs Doppler echocardiographic data to calculate AVA. [14] AVA = LVOT velocity x LVOT area / Aortic Valve Velocity b) Gorlins equation: Gorlin formula is used with invasive measurement of cardiac output and transvalvular pr essure drop. [18] AVA = cardiac output (mLmin-1) / {heart rate (min-1) x systolic ejection period (secs) x 51.6 x Cd x P} In the above equation, Cd (coefficient of di scharge) is the rati o of effective area vs. anatomic area. Valve areas derived by the Gorlin formula have been observed to vary with transvalvular flow rate. [26] The con tinuity equation measures the area of vena contracta, gives underestimates compared with the Go rlin formula and it is not clear which is the more accurate. [27] Es timates of orifice area in an individual valve as judged by any of the equations tested should be seen as a guide to rather than as a precise measure of actual orifice area. [27] c) The rate of change in AVA is an add itional measure of disease severity and may be used to predict an indivi duals risk for subsequent rapid disease progression. [4]

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12 Rate of change in area = AVA [1/2D]/ AVA [1/2A] Where, AVA [1/2D]: aortic valve area during half deceleration AVA [1/2A]: aortic valve area du ring half acceleration d) Aortic valve area can be measured direc tly by planimetry, which is tracing out of aortic valve opening in a still image obtai ned during echocardiogr aphic acquisition in systole. 3. Valve resistance: Doppler-echo estimates of aortic valve resistance (AVR) may be used as an alternative index of aortic st enosis severity. AVR equation does not use a constant and treats the pressure gradient a nd the cardiac output without favoring either, it has been proposed as a more accurate index of the severity of aortic stenosis. [18] AVR = 1.333 4Vmax2 / area LVOT velocity LVOT (Doppler) AVR = (1.333 P) / (CO/HR SEP) (Catheterization) 4. Percent stroke work: Because aortic stenosis results in the loss of le ft ventricular stroke work (due to resistance to flow through the valve and turbulen ce in the aorta), the percentage of stroke work that is lost can reflect the sever ity of stenosis. This index can be calculated from pressure data alone. The relation between percent stroke work loss and anatomic aortic valve orifice area can be investigated. [18] Percent stroke work loss = (mean systolic pr essure gradient / mean ventricular systolic pressure) x 100%

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13 5. Aortic jet velocity: Aort ic jet velocity measured by Doppler echocardiographic methods as maximum velocity across the ao rtic valve has been shown as a predictive symptom onset and clinical result in asymptom atic patients and in patients who have a symptomatic disease. [18] 6. Ejection fraction velocity ratio: The EFVR is a simple noninvasive method for screening patients with aortic valve area of 1.0 cm2. It could be used as a screening test or in lieu of the continuity equation particular ly when there is problematic measurement of either the LVOT diameter or velocity. [18] EFVR = ejection fraction (%) / ma ximal aortic velocity (m/sec) Aortic valve area is the most important a nd affirmative indication of stenosis. The actual valve opening area can be visualized using transesophageal echo (TEE). Valve area can be calculated from Doppl er data using continuity eq uation is effective orifice area that account for flow convergence in stenot ic jet. The relationship of anatomic valve area and effective valve area is therefore complex. Clinical estimates of AVA do not reflect how it varies through out the cy cle, but reflect maximum valve opening.

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14 Chapter 3 Valve Fixation 3.1 Porcine tissue valves Aortic porcine valves have close rese mblance to human aortic valves. They exhibit similar hemodynamic performance and also have the advant age of availability. [32] After the introduction by Carpentier of glutaraldehyde fixation of the valve tissue porcine valves proved to be more durable [7] but hydraulic performance is compromised because the leaflets are stiffer and more st enotic. In this study, porcine valves were treated with glutaraldehyde and et hanol to emulate early stenosis. 3.2 Glutaradehyde fixation Most of the aortic tissue valves used fo r implants are either glutaraldehyde fixed porcine aortic valves or valves made of glutaraldehyde fixed bovi ne pericardium. [8] Glutaraldehyde (HCO-CH2-CH2-CH2-CHO) acts through formation of cross-links between protein end groups. An aqueous soluti on of glutaraldehyde (glutaric dialdehyde) is a complex mixture at room temperature, consisting of approximately 4% free aldehyde, 16% monohydrate, 9% dihydrate and 70% hemiacetal [8] It is also suggested that pure, monomeric glutaraldehyde (Figure 6) is the best fixative and much less inhibitory to enzymes than is the mixed polymeric product. [8] The success of glutaraldehyde as a

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15 cross-linking agent also depends on the la rge range of different molecules present simultaneously in the fixation solution. [8] There are many variations in the prepar ation of this fixative, including the percentage of glutaraldehyde, additives, and buffers. Because of its low penetration, only small blocks of tissues (1-2 mm3) fix well at temperat ures of 1-4C. [8] Glutaraldehyde is comprised of two aldehyde groups, separated by a flexible chain of 3 methylene bridges. In aqueous solu tions, glutaraldehyde is present largely as polymers of variable size. Fr ee aldehyde groups combine with any protein nitrogens with which they come into contact forming cro ss-linked protein molecules (Figure 7). [12] The number of cross-linked molecules crea ted depends on the number of available primary amines coupled with their interm olecular distance. [8] There are also many leftover aldehyde groups (not bound to anything) that cannot be washed out of the tissue. Figure 6. Representations of monomeric glutaraldehyde [12]

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16 Figure 7. Reaction of poly (glutaralde hyde) with amino protein groups [12] Bioprosthetic heart valves fabricated from glutaraldeyde fixed porcine valves fail frequently due to calcification. [21] They also develop fraying of collagen and over the time they can become perforated and torn. Glutaraldehyde fixed ti ssues exhibit altered mechanical properties compared to fresh tissues. Porcine aortic valve fixed with glutaraldehyde tend to be stiffer than fresh valves and have stress relaxation rates about 60% of that of fresh valves. [33] 3.3 Ethanol fixation Ethanol, C2H5OH, (also called Ethyl Alcohol) is the second member of the aliphatic alcohol series. [8] Ethanol fixation mainly causes dehydration. Alteration of the structure of proteins brought about by ethanol is primar ily due to disruption of the hydrophobic bonds that contribute to the maintenan ce of the tertiary structure of proteins. Hydrogen bonds appear to be more stable in me thanol and ethanol than in water so that

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17 while affecting the tertiary structure of proteins, these alcohols may preserve their secondary structure. [8] The primary protein structure is relative ly unaltered by ethanol fixation. [22] Ethanol is closely related in structure to water and replaces water molecules in the tissues, unbound as well as bound, during fixati on. While absolute ethanol preserves glycogen, it can cause distortion of nuclear detail and shrinkage of cytoplasm. If fixation is prolonged, the alcohols remove histones from the nuclei and later extract RNA and DNA. [8] Ethanol-water system can have two hydrogen bonding stru ctures corresponding to the two possible heterodimers (or isom ers) where alcohol is a proton acceptor, RHOH2O or a proton donor, ROHOH2. [16] Ethanol treatment of glutar aldehyde fixed tissue is show n to be highly effective anti-calcification treatment by subcutaneous implantation in rats and implantation in mitral position in sheep. [21] Absolute ethanol is most commonly used coagulant fixative. It appears to afford greater pr eservation of immuno-reactivity of filament proteins. [22] Ethanol treatment in porcine valves has shown to cause permanent alteration in collagen conformation leading significant change in material properties. [34]

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18 Chapter 4 Methods 4.1 Valve selection During this study, the fresh, ethanol fixed and glutaraldehyde fixed valves were tested using a programmable pulse duplicator Data was collected from nine porcine valves, which was used to construct a Bayesi an model that can lear n causal relationships and predict area for given hemodynamic conditions. The porcine hearts were obt ained from La Casa Sierra Pig Slaughter House, Land O Lakes, Florida. Aortic valves were excise d from the hearts. They were divided in 3 groups: 1) fresh 2) ethanol fixed 3) glut araldehyde fixed. (Figure 8). Each group had three valves. For each valve 6 different condi tions were implemented Flow rates of 3, 4 and 5 L/min were considered at the back pressures of 60 mmHg (dynamic range of 5080) and 90 mmHg (dynamic range of 80-110 mmH g). In adults 90 to 100 mmHg is the Mean Arterial Pressure (ra nge being 70-110 mmHg); while in children of age less than 12 years it is 50 to 75 mmHg. [23]

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19 Table 2. Porcine valve sizes Valve A ortic root diameter (cm) Annulus diameter (cm) Fresh 1 1.81.4 Fresh 2 1.71.1 Fresh 3 2.11.6 Ethanol 1 1.81.4 Ethanol 2 1.851.4 Ethanol 3 1.861.6 Glutaraldehyde 1 1.81.5 Glutaraldehyde 2 1.71.2 Glutaraldehyde 3 1.871.6 A rubber ring of circumference equal to that of the inner circumference of the aortic valve testing chamber of the pulse duplicator was suture d to every valve. Surgical sutures used were 4-0 Dermalon (1.5 Metr ic) Clear-Monofilament Nylon PRE-2 13mm. For fresh and ethanol fixed valves, due to delicacy a collapse of the part of aortic root was observed during closure of the valv e cusps. Thus a nylon fabric was carefully stuck with epoxy around the aortic root to avoid its collapse and get clear images of cusps opening and closing actions.

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20 Figure 8. Valves used for experiments 4.2 Method of fixation Tissue fixation is dependent of reagent penetration and rate of reaction. Fixation for 96 hours has no harmful effect on the tissu e morphology, however it allows increased cross-linking. [22] The fixati on of valves by glutaradehyde was carried on for time of 96 hours to allow maximum cross-lin king. Ethanol fixation was also carried out for the time of 96 hours to allow si gnificant dehydration. The pH of blood is maintained at close to 7.4 and thus acts as a buffer to organs. Hence, the valves were stored in 10 X Phos phate Buffer Saline (PBS) solution, which has a pH of 7.4. A 10X concentrated solution was preferred to allow minimum fluctuation of the pH. [24]

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21 4.3 Working of pulse duplicator Figure 9. Pulse duplicator Experiments were performed in a pulse duplicator interfaced with a computer controller (Figure 9) to simula te blood flow from left vent ricle to the aorta using a 40% solution of glycerol in distilled water. The solution has viscosity (0.04 poise) and density (1.056 gm/cc) similar to that of blood but is optically clear. The pulse duplicator consisted of two separate chambers to mimi c aortic compliance and resistance. The mean arterial pressure was simulated by appl ying pressure in the compliance chamber downstream of the valve and th e peripheral resistance of the arteries was simulated by clamping the flow tube downstream. The pulse duplicator was interfaced with a computer with Labview software, board, and a custom-bui lt control box for pulse control and data acquisition (National Instruments, Houston TX). This system controlle d the heart rate and systolic ejection time produced by the pulse duplicator; it acquired a nd displayed pressure and flow data, and contained subroutines fo r data analysis. Flow measurements were

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22 made with a Transonic Systems T110R ultras ound flow meter with cannulating probes (Transonic Systems, Inc., Ithaca NY). The flow transducer was factory calibrated for the working fluid in Tygon tubing. Three pressure transducers (Kalvico P 155 15G) were placed as first (P1) 3 cm proximal, second (P2) 3cm distal and third (P3) 10cm distal (P3) to the aortic valve. The pressure transducer s were calibrated by the laboratory. A vacuum pump (GE 0522P177 G-180DX) wa s connected with an adju stable regulator to the solenoid valve of the ventricular chamber to allow full relaxation of the ventricular bulb. A fixed alignment was selected for the positi oning of the camera and strobe light. Strobe, aortic valve testing chamber and camera were mounted in fixed positions over a wooden base. An electronic circuit board assembly consisted of an asynchronous counter circuit that would trigger the camera and strobe simultaneously to capture an image. Pressure (P1, P2, P3, pressure in compliance chamber) a nd flow rate values were recorded at the same instant of time. Measurement of aortic valve area is done using a photographic technique. A strobe light is triggered ever y fixed millisecond (n) in a card iac cycle (c), after which the camera shutter opens to capture an image. Af ter four cardiac cycles (c + 4), the strobe then fires for next millisecond (n+1) and a picture is captured, subsequently for 300 milliseconds in separate cardiac cycles, but comp leting a series of pictures for one cardiac cycle. A gap of four cardiac cycles is used between two pictures as the camera shutter opening time is close to 2 seconds. The cam era used for capturing photographic data was Canon EOS 10D. Compact Flash Type II card was used for memory storage. A stroboscopic flash tube (DS303 22-95) was used instead of the camera flash. The heart

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23 rate for the system was set to 70 beats/ min. Systolic ejection period was set to 300 milliseconds, thus the diastolic period lasted for 557 milliseconds. Figure 10. Labview data acqui sition in pulse duplicator Due to high pressure in the compliance chamber, air tried to escape out of the chamber through the pressure bulb (used in back pressure control, Figure 9). This effect was prevented by attaching a stop valve at th e end of tubing connected to the compliance chamber. There was an error in back pressure recording as the back pressure value was influenced by the pressure induced by the tubing. Hence the back pr essure indicator was connected directly to the compliance chambe r and voltage values were recorded for accurate back pressures. Calibration of b ack pressure reading was done as follows:

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24 Table 3. Calibration of compliance chamber Run Voltage 1 Voltage 2 Voltage 3 Pressure (mmHg) 13.9033.9033.90420 3.8473.8463.84340 3.7823.7823.78960 3.6623.6613.668100 3.63.6063.608120 23.9013.9013.89920 3.8453.8483.84340 3.7893.7843.78560 3.7263.7233.73180 3.6613.6663.662100 3.5973.6023.6120 33.93.8973.90120 3.8443.8393.83940 3.7893.7873.78960 3.723.7223.72480 3.6653.6653.663100 3.5983.63.604120 The calibration equation (Pressu re = -333.48 (Voltage) + 1321.7, R2=0.9994) derived from linear curve fitting obtained from the plot of mean voltage vs. pressure (values in Table 3) was impleme nted in the Labview routine. Before the experiments, the tubing wa s replaced with a new one. Tubing was cut short to minimize resistance. Leaks were ch ecked, and system was cleaned using bleach solution then was rinsed using lab algicide and distilled water to remove and prevent fungi. The alignment of whole assembly was arranged to allow minimum twists in the tubing.

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25 During collection of data it was assumed th at all cardiac cycles for a valve from which the data was collected were similar. The following illustration shows variability between four consecutive b eats for flow (Figure 11, ma ximum variance: 1.35, average variance: 0.14) and pressure gradient (Figure 12, maxi mum variance: 8.28, average variance: 0.22). Beat to beat variability for flow-6 -4 -2 0 2 4 6 8 10 12 14 0200400600800 Time (ms)Flow (L/min) beat1 beat 2 beat 3 beat 4 variance Figure 11. Beat to beat variability for flow Beat to beat variability for gradient-120 -100 -80 -60 -40 -20 0 20 40 0200400600800 Time (ms)Gradient (mmHg ) beat1 beat 2 beat 3 beat 4 variance Figure 12. Beat to beat variability for gradient

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26 4.4 Calculation of valve areas Image-J is a public domain Java imageprocessing program inspired by NIH Image for Windows and Macintosh. It runs, either as an online applet or as a downloadable application, on any PC with a Ja va 1.1 or later virtual machine. Image-J measures the number of pixels and has a facility of converting pixels to defined scale. As cm2 is the measure of area, we need to convert number of pixels to cm2. To check if the results of Image-J were right, two basic Figures were drawn and their areas were verified by Image-J calculated area. Figure 13. Verification of area in pixels to cm2 using a circle The radius of circle was marked 1 cm and area was calculated as the area of brighter pixels. Result given by Image-J: To tal Area: 3.136 cm^2. (0.19 % error)

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27 Figure 14. Verification of area in pixel to cm2 using a square For the square the length f sides was marked as 2cm, and area was calculated. Result given by Image-J: Total Ar ea: 3.917 cm^2. (2.075 % error) 4.5 Calibration of Image -J The cross sectional diameter of the in ner tubing 3.16 cm (which is prominently visible) was used for scaling of data from pixel2 to cm2. Figure 15. Fresh valve 1 in pulse duplicator

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28 Figure 16. Diameter used for scaling For every experiment, data was collec ted as 150-200 datasets in Excel file. Photographs obtained were matched with th e datasets, and corres ponding instantaneous area values (obtained from ImageJ) were included in the file Gradient was calculated as the instantaneous pressure difference P1-P2 in mmHg. Figure 17 shows the excel file that was used as a base for constructing Bayesian Network. Bayesian networks are described in the next chapter.

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29 Figure 17. Data in .xls format collected from pulse duplicator In this Figure, A: Instantaneous time (seconds) B: Instantaneous pressure (mmHg) by transducer P1 C: Instantaneous pressure (mmHg) by transducer P2 D: Instantaneous pressure (mmHg) by transducer P3 E: Instantaneous flow (L/min) F: Back Pressure (mmHg) G: Status of solenoid valve 1 (0: closed, 5: open) H: Status of solenoid valve 2 (0: closed, 10: open) I: Camera status (15: shutter open)

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30 J: Strobe light status (0: ne gative edge of strobe signal) K: Picture number L: Gradient (P1-P2 in mmHg) M: Area calculated by Image-J (cm2)

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31 Chapter 5 Bayesian Modeling 5.1 Bayesian networks Bayesian Networks are used to determine causal relationships and can approximate unknown parameters, provided prior knowledge is available. [19] They are mathematically defined in terms of probabi lity and conditional independence statements. As such, they are useful for causality anal ysis and through statistical induction they support a form of automatic learning. [20] Bayesian networks are derived from Bayes formula. Bayes formula provides the mathema tical tool that combines prior knowledge with current data to produce a posterior di stribution. [19] Bayes fo rmula states that Posterior = (Likeli hood Prior) / Evidence p(H|E,c) = {p(E|H,c) x p(H|c)} / p(E|c) p(H|E,c): posterior or probabil ity of parameter H for after considering effect of E on c p(E|H,c): likelihood or probability of E assuming H and background information c P(H|c): prior or probability of H given c alone p(E|c): evidence or normalizing constant or scaling factor independent of H In Bayesian modeling the infe rence of all parameters is derived from the posterior distribution.

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32 The joint probability distribution of a Bayesian Netw ork is given by Where a (v): parents or direct ancestors of vertex v having directed edges connected to v. 5.2 Directed acycl ic graphs [20] A Bayesian network consists of di rected acyclic graphs (DAGs) of nodes representing variables and arcs signifying conditional dependencies between a pair of nodes. In a DAG there is no path that starts and ends at the same node, for example if there is an outgoing arc from A to B, there could not exist an incoming arc from B to A. Associated with each node is a conditional pr obability of the variable given its parents. Hemodynamic parameters considered were ba ck pressure (mmHg), pressure gradient (mmHg), flow (L/min) and valve area (cm2). On constructing Bayesian networks from databases, we make use of nodes to represent database attributes (back pressure, gradient, flow and area in our case). If two node s are dependent, knowing the value of one node will give some information about the value of the other node. Knowledge of mutual information betw een two variables can tell us about dependency relation between them. The mutual information between two nodes xi and xj is defined as follows. [13] I(xi,xj) = P(xi,xj). Log10 (P(xi,xj) / P(xi)P(xj)) Pxi Pxj = probabilities of xi and xj respectively P(xi,xj) = joint probab ility of xi and xj

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33 A learning dependency algorithm was im plemented. Phases of algorithm to construct the DAG are as follows: [13] Phase I: (Drafting) 1. Initiate a graph, E V G where V = {all the attr ibutes of a data set }, E = {}. Initiate an empty list L. 2. For all the pairs of nodes sort them base d on their mutual information values and put these pairs of nodes into list L from large to small. Create a pointer p that points to the first pair of nodes in L. 3. Get the first two pairs of nodes of lis t L and remove them from it. Add the corresponding arcs to E. Shift the pointer p to the next pair of nodes. (directions of the arcs are decided by the node ordering.) 4. Get the pair of nodes from L pointed by the pointer p. If there is no open path between the two nodes, add the correspond ing arc to E and remove this pair of nodes from L. 5. Move the pointer p to the ne xt pair of nodes and go back to step 4 unless p is pointing to the end of L. Phase II: (Thickening) 6. Move the pointer p to the first pair of nodes in L. 7. Get the pair of nodes (node1, node2) from L at the position of the pointer p. Call (Current graph, node1, node2) to find a cutset which can separate node1 and node2 in the current graph. Use a conditional indepe ndence test to see if node1 and node2 are

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34 conditionally independent given th e cut-set. If so, go to next step; otherwise, connect the pair of nodes by addition of corresponding arc to E. 8. Move the pointer p to the ne xt pair of nodes and go back to step 7 unless p is pointing to the end of L. Phase III: (Thinning) 9. For each arc (node1, node2) in E, if there are other paths besides this arc between the two nodes then remove this arc from E tempor arily and find a cut-set that can separate node1 and node2 in the current graph. Use a cond itional independence test to see if node1 and node2 are conditionally independent given the cut-set. If so, eliminate the arc permanently; or else add this arc back to E. 5.3 Finding cutsets for a pair of nodes Considering node A and node B, if P(A, B/C) = P(A/C) then nodes A and B are conditionally independent given C. Thus we can say that C is the cutest of A and B. C is a path (consisting of one or more nodes). Thus the algorithm for finding out the cutsets between node A and node B goes as: 1. Determine all possible pathways between node A and B and add them to C. 2. Now consider first path C1 (consisting of one or more nodes) in C, if P(A,B/C1) = P(A/C1), then C1 is the cutest of A and C. 3. If not, Repeat step 2 for next pathway C2 in C.

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35 5.4 Data for GeNIe 2.0 Matlab was used as a tool for converti ng the data into a GeNIe 2.0 specific format. The code is included in the Appendices section. 5.4.1 Categorizing data into states If the data is discrete and has no repeat ed values, then the result of conditional probability of any parameter given that data is 1. Hence due to discrete nature of the data, it was categorized into states for obtaining better estimates of c onditional probabilities. One more reason for dividing it into states wa s to decrease the data as the final product (area) has its outcome as exponential functi on of the parents (back pressure, flow and gradient). Hence each parameter was categorized into 10 states. The states are listed in Table 4 below. Table 4. Grouping of variables in states Back pressure Flow Gradient Area State 1 <=55 <=0 <= -50 0 State 2 55 to 60 0 to 1 -50 to 0 to 0.1 State 3 60 to 65 1 to 2 -30 to 0.1 to 0.2 State 4 65 to 70 2 to 3 -10 to 0 0.2 to 0.3 State 5 70 to 80 3 to 4 0 to 5 0.3 to 0.4 State 6 80 to 85 4 to 5 5 to 10 0.4 to 0.5 State 7 85 to 90 5 to 6 10 to 15 0.5 to 0.6 State 8 90 to 95 6 to 7 15 to 20 0.6 to 0.7 State 9 95 to 100 7 to 8 20 to 25 0.7 to 0.8 State 10 > 100 > 8 More than 25 > 0.8

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36 5.4.2 Selection of data A variable p was marked when the valve st arts to open (first value of area more than 0). Selection of data was done using this variable. 50 values before and after p were selected. A limitation of 100 values is due to the range of data available from the experiments. For closing cycle p was marked when the valve closed (first value of area equal to 0). 50 values before and after p were selected. 5.4.3 Order of nodes Order of nodes for the network is 1. back pressure, 2. flow, 3. gradient and 4. area, since it is known that ar ea is the final outcome, while back pressure and flow are user controlled parameters in the pulse dupl icator. Node ordering is responsible for making topological order in the DAGs, thus re presents causal relationship from parent nodes to children nodes. We can determine node ordering if we have prior knowledge of independent and dependent parameters. Node ordering approach makes Bayesian models preferable than multivariate correlation models. 5.4.4 Bayesian network in GeNIe 2.0 A text file was created from the excel f ile using Matlab (program in appendix A1), which was saved as hugin.NET format for GeNIe 2.0. Evidence can be set for one or more parameters to obtain predicted values of other parameters. Figure 18 shows evidence of flow set to 4 to 5 L/min, while the other values are pr edicted probabilities.

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37 Figure 18. Network given flow as evidence

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38 Figure 19. Network given area as evidence

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39 Chapter 6 Results 6.1 Test results Experiments were carried out at the mean flow rates of 3, 4 and 5 L/min for back pressure values of 60 and 90 mmHg. Heart rate was set to 70 beats/min while systolic ejection period was set to 300 ms. Table 5 su mmarizes the systolic flow rate, peak pressure gradient, maximum opening area (anatomic area calculated from photography, effective area calculated from Gorlin equatio n) and discharge coefficient Cd (anatomic area/effective area). We would expect discharge coe fficient between 0.8 and 1. It has been observed that, valve areas derived by the Gorlin formula have been observed to vary with transvalvular volume flow rate. [26] For small size valves (16-21mm annulus) the Gorlin formula shows increased area by 10 to 15% [18], and this may explain some of the errors in the dataset (Table 5). Table 5. Test results Valve Mean back pressure (mmHg) Systolic flow rate (L/min) Peak gradient (P1-P2) Maximum opening area (cm2) Gorlin equation area Discharge coefficient Cd Fresh 1 607.514.410.640.55 0.85 Fresh 2 607.9119.780.570.49 0.86 Fresh 3 608.88.10.750.85 1.14 Ethanol 1 607.4732.260.420.36 0.87

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40 Table 5. Continued Ethanol 2 608.0116.580.480.54 1.13 Ethanol 3 608.2424.620.610.46 0.75 Glutaraldehyde 1 607.0130.780.410.35 0.85 Glutaraldehyde 2 608.5138.70.210.38 1.8 Glutaraldehyde 3 6010.9530.120.610.55 0.9 Fresh 1 609.434.90.550.44 0.8 Fresh 2 6010.4427.070.70.55 0.79 Fresh 3 6010.313.540.930.77 0.83 Ethanol 1 609.3351.250.460.36 0.78 Ethanol 2 6010.21529.860.570.52 0.91 Ethanol 3 609.7831.60.660.48 0.73 Glutaraldehyde 1 6010.08440.710.42 0.59 Glutaraldehyde 2 6010.855.010.260.4 1.55 Glutaraldehyde 3 6013.47380.650.6 0.93 Fresh 1 6011.0825.580.90.61 0.67 Fresh 2 6012.7837.930.590.57 0.97 Fresh 3 6014.1925.131.190.78 0.66 Ethanol 1 6011.7364.150.540.4 0.75 Ethanol 2 6012.238.350.480.54 1.13 Ethanol 3 6011.76440.660.49 0.74 Glutaraldehyde 1 6013.1851.430.760.51 0.67 Glutaraldehyde 2 6011.8363.30.340.41 1.21 Glutaraldehyde 3 6013.72381.10.61 0.47 Fresh 1 907.6915.410.670.54 0.81 Fresh 2 907.2927.440.420.38 0.92 Fresh 3 909.512.260.810.75 0.93 Ethanol 1 907.58934.840.420.36 0.85 Ethanol 2 908.0221.820.660.47 0.72 Ethanol 3 908.44250.670.47 0.7 Glutaraldehyde 1 907.9144.710.340.33 0.96 Glutaraldehyde 2 906.9742.90.180.29 1.63 Glutaraldehyde 3 9012.3836.490.840.57 0.67 Fresh 1 9010.4825.180.540.58 1.07 Fresh 2 909.9843.840.470.42 0.89 Fresh 3 9012.5319.211.120.79 0.71 Ethanol 1 9010.2563.440.360.36 0.99 Ethanol 2 9010.53360.570.48 0.85 Ethanol 3 9010.86380.680.49 0.72

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41 Table 5. Continued Glutaraldehyde 1 9010.359.050.40.37 0.93 Glutaraldehyde 2 9010.6455.080.270.4 1.47 Glutaraldehyde 3 9013.743.210.540.58 1.07 Fresh 1 9011.8633.540.760.57 0.74 Fresh 2 9011.7746.210.50.48 0.96 Fresh 3 9014.088290.570.72 1.27 Ethanol 1 9011.33630.490.39 0.8 Ethanol 2 9011.5249.140.670.45 0.68 Ethanol 3 9011.7747.80.750.47 0.63 Glutaraldehyde 1 9012.460.480.390.44 1.13 Glutaraldehyde 2 9012.2567.210.260.41 1.59 Glutaraldehyde 3 9014.9547.10.460.6 1.31 The illustrations below (Figure 20, Figur e 21) indicate maximum area and peak flow changing with mean flow rates of 3, 4 and 5 L/min at 60 mmHg back pressure. The maximum area and peak gradient in following plots are the average values for all three valves of each type. It has been seen that maximum area and peak gradient increase with increasing mean flow. Also, ma ximum area is more from fresh valves than for fixed valves while peak gradient is lower fo r fresh valves than for fixed valves.

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42 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 3L/min4L/min5L/minMean flow (L/min)Maximum area (cm2) Fresh valves Ethanol fixed valves Glutaraldehyde fixed valves Figure 20. Maximum area vs. mean flow 0 10 20 30 40 50 60 3L/min4L/min5L/minMean flow (L/min)Peak gradient (mmHg) Fresh valves Ethanol fixed valves Glutaraldehyde fixed valves Figure 21. Peak gradient vs. mean flow

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43 6.2 Results for gradient and area relation It was observed that fresh valves required minimum pressure gradient to open and a fully open state was achieved quicker. In case of higher back pressure (90 mmHg), values of pressure gradient required for the valve opening (instant that which valve opened) were almost similar to that of lo wer back pressure (6 0 mmHg). While for the closing cycle (instant at which valve closed), the pressure gradient for high back pressure (90 mmHg) was 25 to 50 mmHg lesser than that for lower back pressure (60 mmHg). The illustrations below compare fresh valve 1, etha nol fixed valve 1 and glutaradehyde fixed valve 1. Since the valves are not anatomically similar, valves with similar size are chosen for comparison. Gradient-Area plot for opening cycle, flow of 3lpm at 90backpressure 0.0 0.2 0.4 0.6 0.8 1.0 -20.000.0020.0040.0060.00 Gradient (mmHg)Area (cm2) Fresh Valve Ethanol fixed valve Glutaraldehyde fixed valve Figure 22. Gradient area plot for 3L/min at 90bp

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44 Gradient-Area plot for opening cycle, flow of 4lpm at 90 backpressure 0.0 0.2 0.4 0.6 0.8 1.0 -20.000.0020.0040.0060.00 Gradient (mmHg)Area (cm2) Fresh valve Ethanol exposed valve Glutaraldehyde fixed valve Figure 23. Gradient-area plot for 4L/min at 90bp Gradient-Area plot for opening cycle, flow of 5lpm at 90 backpressure0.0 0.2 0.4 0.6 0.8 1.0 -20.000.0020.0040.0060.00 Gradient (mmHg)Area (cm2) Fresh valve Ethanol exposed valve Glutaraldehyde fixed valve Figure 24. Gradient area plot for 5L/min at 90bp

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45 6.3 Hysteresis loop The instantaneous relationship between pressure gradient and opening area is different for valve opening and closing cycles. We define the plot of this as hysteresis loop. It was observed that the area of hysteresis loop in case of higher back pressure (90 mmHg) was higher than that of the loop are in case of lower back pressure (60 mmHg). Narrow hysteresis loop implies small amount of dissipated energy, wh ile a broader loop implies greater dissipated energy. etoh3 hysteresis, 4lpm at 60 back pressure 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 -50-30-10103050 Gradient (mmHg)Area (cm2) closing area opening area Figure 25. Ethanol fixed valve 3 hyst eresis loop for 4L/min at 60 bp

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46 etoh3 Hysteresis, 4lpm at 90 back pressure 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 -50-30-10103050 Gradient (mmHg)Area (cm2) closing area opening area Figure 26. Ethanol fixed valve 3 hyst eresis loop for 4L/min at 90 bp A consistent behavior was observed for fresh and fixed valves in opening and closing cycles. It is observed that fo r first 40 milliseconds of the opening phase, the fresh valve and ethanol fixed valve fo llowed rapid opening as compared to glutaraldehyde fixed valve. The following illu strations show fresh valve 1, ethanol fixed valve 1 and glutaraldehyde fixed valve 1 in ope ning and closing cycles at mean flow rate of 5 L/min for back pressure of 60 mmHg.

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47 Valve opening0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0204060Picture numberArea (cm2) Fresh valve Ethanol fixed valve Glutaraldehyde fixed valve Figure 27. Valve opening Figure 28. Valve closing Valve closing0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400204060 Picture numberArea (cm2) Fresh valve Ethanol fixed valve Glutaraldehyde fixed valve

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48 The illustrations (Table 6) below summarizes maximum rate of change in area vs. time (da/dt) while opening for the mean flow rate of 5L/min at 90 mmHg back pressure. The figures show fresh and ethanol fi xed valves follow rapid opening while gluraraldehyde fixed valves follow slow opening. Table 6. Maximum da/dt Type of valve Maximum da/dt Fresh valve 1 0.13 Fresh valve 2 0.13 Fresh valve 3 0.15 Ethanol valve 1 0.1 Ethanol valve 2 0.16 Ethanol valve 3 0.15 Glutaraldehyde valve 1 0.04 Glutaraldehyde valve 2 0.05 Glutaraldehyde valve 3 0.076 6.4 Results for Bayesian networks 6.4.1 Results for mutual information It was observed that mutual information mainly depends on the variability of the parameters. If both of the parameters show si milar variability the mutual information is the highest. But even if one of the parameter is less diverse i.e. the range of values of original data set is small, it contributes less to the mu tual information. Hence back pressure was observed to share least mutu al information as in a cardiac cycle instantaneous back pressure only varied +/15% from the mean valu e. Flow contributed to the highest mutual information with area. The mutual information is proportional to the correlation coefficient of the two parameters Univariate correlation coefficients were

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49 calculated from Matlab using corrcoef functi on where, R = corrcoef(X) returns a matrix R of correlation coefficients calcul ated from an input matrix X. The matrix R = corrcoef (X) is related to the covariance matrix C = cov(X) by R(j,k) = c(j,k)/ {(C(j,j)C(k,k)}. Illustration below (Table 7) summarizes mutual information and correlation coefficient for all valves for mean fl ow rate of 4L/min at back pressure of 90 mmHg. Table 7. Mutual information for flow rate of 4L/min at 90 bp Valve ParametersMutual Information Regression coefficient Fresh1BP-G 1.96 0.371 FL-G 2.4636 0.5978 BP-FL 2.8619 0.7836 BP-A 5.4927 0.8125 FL-A 4.3678 0.9356 G-A 2.36 0.5956 Fresh2BP-G 0.57 0.3369 FL-G 4.27 0.8981 BP-FL 0.14 0.334 BP-A 0.032 0.2862 FL-A 4.15 0.9446 G-A 3.43 0.8022 Fresh3BP-G 0.13 Na FL-G 3.04 0.6106 BP-FL 0.052 Na BP-A 0.14 Na FL-A 4.07 0.8601 G-A 1.39 0.4099 Etoh1 BP-G 0.8364 0.1569 FL-G 2.0357 0.7797 BP-FL -0.108 0.1147 BP-A 0.5887 0.061 FL-A 2.9455 0.7806 G-A 8.3453 0.9427 Etoh2 BP-G 0.1055 Na

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50 Table 7. Continued FL-G 3.3283 0.8086 BP-FL 0.0555 Na BP-A 0.1546 Na FL-A 4.1279 0.893 G-A 3.2475 0.6962 Etoh3 BP-G 0.1425 Na FL-G 2.7474 0.7695 BP-FL 0.0587 Na BP-A 0.1669 Na FL-A 4.7095 0.9359 G-A 1.1962 0.6311 Glut1 BP-G 0.2076 -0.0212 FL-G 4.0307 0.8076 BP-FL 0.2949 -0.0499 BP-A 0.0987 -0.0749 FL-A 7.1686 0.9078 G-A 5.1665 0.9285 Glut2 BP-G 0.0717 Na FL-G 4.0045 0.8318 BP-FL 0.1473 Na BP-A 0.18 Na FL-A 7.4643 0.8175 G-A 5.8385 0.9599 Glut3 BP-G 0.1403 Na FL-G 1.7328 0.6374 BP-FL 0.1334 Na BP-A 0.1414 Na FL-A 8.4245 0.9457 G-A 2.9953 0.7012

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51 6.4.2 Cutsets Cut-sets are used to shorten the data and remove unnecessary arcs between the nodes. A Matlab program was written to dete rmine cutsets between any two parameters (nodes). The program is included in appendi ces section A3. The program chooses all possible paths (nodes or set of nodes connect ing the two nodes) be tween the two nodes and checks if the two nodes are conditionally in dependent given the probability of path. If they are independent, the path is declared as a cutest between them. It was observed that in the presence or absence of the path between the conditionally independent nodes, the probabilit ies of the child node was same. The data in this case does not need to be reduced, as there are only four nodes in this Bayesian network, hence even if we do not shorten the ne twork with cutsets, the result is fast and accurate. The following example shows the results for Fresh valve 3 at 4L/min opening cycle at 60 mmHg back pressure. Path cont aining flow and area was confirmed as a cutset between back pressure and gradient Thus the arc between back pressure and gradient was removed and results were obtai ned. Back pressure was set as evidence in both available states and it was verified that the result was same in case of network with cutsets and network without cutsets.

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52 Figure 29. Back pressure at state 95 to 100 (e vidence) network with and without cutsets

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53 Figure 30. Back pressure at state 90 to 95 (e vidence) network with and without cutsets 6.4.3 Prediction Table 8, Table 9 and Table 10 below s how the results obtained from modeling compared with the actual results. As there we re three valves available for each type, for comparing every experimental result, the modeling results are obtained from the models made from other two valves of the same t ype (fresh, ethanol fixed or glutaraldehyde fixed). A random value was chosen from the data excel file, and it was compared to the results predicted by the models built from ot her two valves. The second column in the tables shows the condition of the experiment The state of evidence for conditions (back pressure, flow and gradient) was chosen to the closest available value. The predicted value stated in the table is the value of hi ghest probability of area given by the model, probability stated in the last column.

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54 Table 8. Results for fresh valve Condition back pressure (evidence) flow (evidence) gradient (evidence) area (predicted) Prediction probability Dataset Fresh1 4L/min 60BP closing 76.17 5.41 -1.98 0.65 Model 1 Fresh2 4L/min 60BP closing 80 to 85 5 to 6 0 to -10 0.5 to 0.6 1 Model 2 Fresh3 4L/min 60BP closing 95 to 100 4 to 5 0 to -10 0.7 to 0.8 1 Dataset Fresh1 5L/min 60BP opening 51.52 8.91 20.9 0.82 Model 1 Fresh2 5L/min 60BP opening 70 to 80 >8 15 to 20 0.5 to 0.6 1 Model 2 Fresh3 5L/min 60BP opening 90 to 95 >8 10 to 15 0.7 to 0.8 0.261 Dataset Fresh2 4L/min 90BP closing 102.91 0.33 -5.43 0.22 Model 1 Fresh1 4L/min 90BP closing >100 0 to 1 0 to -10 0.2 to 0.3 1 Model 2 Fresh3 4L/min 90B closing >100 0 to 1 0 to -10 0.4 to 0.5 / 0.6 to 0.7 0.5/0.5 Dataset Fresh2 4L/min 90BP opening 97.42 6.55 17.5 0.41 Model 1 Fresh1 4L/min 90BP opening 95-100 4 to 5 10 to 15 0.4 to 0.5 / 0.5 to 0.6 0.5/0.5 Model 2 Fresh3 4L/min 90BP opening >100 7 to 8 10 to 15 0.3 to 0.4 1 Dataset Fresh3 5L/min 60BP closing 99.62 3.03 -4.32 0.686 Model 1 Fresh1 5L/min 90BP closing 95 to 100 3 to 4 0 to -10 0.3 to 0.4 / 0.4 to 0.5 0.5/0.5 Model 2 Fresh2 5L/min 60BP closing 80 to 85 3 to 4 0 to -10 0.4 to 0.5 0.889 Dataset Fresh3 3L/min 60BP opening 81.5 7.41 4.36 0.608 Model 1 Fresh1 3L/min 60BP opening 60 to 65 6 to 7 5 to 10 0.6 to 0.7 1 Model 2 Fresh2 3L/min 60BP opening 70 to 80 6 to 7 5 to 10 0.5 to 0.6 1

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55 Table 9. Results for ethanol fixed valve Condition back pressure (evidence) flow (evidence) gradient (evidence) area (predicted) Prediction probability Dataset Etoh1 4L/min 90BP closing 101.86 7.01 6.36 0.32 Model 1 Etoh2 4L/min 90BP closing >100 6 to 7 0 to 5 0.5 to 0.6 1 Model 2 Etoh3 4L/min 60BP closing >100 4 to 5 0 to -10 0.5 to 0.6 1 Dataset Etoh1 5L/min 90BP opening 101.39 10.69 39.42 0.47 Model 1 Etoh2 5L/min 90BP opening >100 >8 20 to 25 0.4 to 0.5 1 Model 2 Etoh3 5L/min 90BPopening > 100 >8 20 to 25 0.6 to 0.7 1 Dataset Etoh2 4L/min 90BP closing 125.92 -0.27 -8.78 0.16 Model 1 Etoh1 4L/min 90BP closing >100 0 to 1 0 to 0 to 0.1 1 Model 2 Etoh3 4L/min 90BP closing >100 0 to 1 0 to 0 1 Dataset Etoh2 4L/min 90BPopening 120.96 10.11 26.03 0.5 Model 1 Etoh1 4L/min 90BPopening 95-100 3 to 4 20 to 25 0.2 to 0.3 1 Model 2 Etoh3 4L/min 90BPopening >100 >8 15 to 20 0.6 to 0.7 1 Dataset Etoh3 5L/min 90BPclosing 115.5 -4.41 -60.77 0 Model 1 Etoh1 5L/min 90BPclosing >100 <0 <-50 0 1 Model 2 Etoh2 5L/min 90BPclosing >100 <0 <-50 0 1 Dataset Etoh3 3L/min 90BPopening 116.63 9.07 19.1 0.634 Model 1 Etoh1 3L/min 90BPopening 90 to 95 7 to 8 15 to 20 0.3 to 0.4 1 Model 2 Etoh2 3L/min 90BPopening >100 6 to 7 10 to 15 0.6 to 0.7 1

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56 Table 10. Results for glutaraldehyde fixed valve Condition back pressure (evidence) flow (evidence) gradient (evidence) area (predicted) Prediction probability Dataset Glut1 3L/min 60BP closing 91 -0.46 -52.77 0.19 Model 1 Glut2 3L/min 60BP closing 70 to 80 <0 <50 0 1 Model 2 Glut3 3L/min 60BP closing 85 to 90 1 to 2 -10 to -30 0 1 Dataset Glut1 4L/min 90BP opening 85.57 1.59 23.6 0.16 Model 1 Glut2 4L/min 90BP opening >100 1 to 2 15 to 20 0.1 to 0.2 1 Model 2 Glut3 4L/min 90BP opening >100 1 to 2 10 to 15 0.1 to 0.2 1 Dataset Glut2 3L/min 60BP closing 79.07 1.7 -13.51 0.008 Model 1 Glut1 3L/min 60BP closing 65 to 70 0 to 1 -10 to -30 0 1 Model 2 Glut3 3L/min 60BP closing 85 to 90 1 to 2 0 to -10 0 1 Dataset Glut2 4L/min 60BP opening 70.97 6.09 30.93 0.261 Model 1 Glut1 4L/min 60BP opening 55 to 60 6 to 7 >25 0.2 to 0.3 0.429 Model 2 Glut3 4L/min 60BP opening >100 5 to 6 10 to 15 0.4 to 0.5 0.429 Dataset Glut3 5L/min 90BP closing 119.52 11.9 2.96 0.392 Model 1 Glut1 5L/min 90BP closing >100 >8 0 to 5 0.3 to 0.4 0.8 Model 2 Glut2 5L/min 90BP closing >100 >8 0 to 5 0.2 to 0.3 1 Dataset Glut3 3L/min 90BP opening 111.39 1.53 10.79 0.28 Model 1 Glut1 3L/min 90BP opening 85 to 90 1 to 2 15 to 20 0.1 to 0.2 1 Model 2 Glut2 3L/min 90BP opening >100 1 to 2 15 to 20 0.1 to 0.2 0.75

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57 The following figures show that most of the estimates of area given the closest possible evidence match the actual area. A vari ation is seen in th e results due to the difference in the size of the valves and experimental dissimilarities. Prediction in fresh valves 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Area (cm2) Dataset Model 1 Model 2 Figure 31. Prediction of area in fresh valves Prediction in ethanol fixed valves0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5Area (cm2) Dataset Model 1 Model 2 Figure 32. Prediction of area in ethanol fixed valves

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58 Prediction in glutaraldehyde fixed valves0 0.1 0.2 0.3 0.4 0.5 0.6 0.7Area (cm2) Dataset Model 1 Model 2 Figure 33. Prediction of area in glutaraldehyde fixed valves 6.4.4 Hydraulic prediction Hydraulic predictions for area were obtain ed for fresh valves compared to fixed valves to observe their opening pattern. Illustration below (Figure 34) shows opening cycle for fresh valve 1, ethanol fixed valve 1, and glutaraldehyde fixed valve 1 for the flow rate of 5L/min at 90 mmHg back pressu re. The evidence of gradient was varied from lower state to higher state and corre sponding states of ar eas were plotted.

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59 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 to 55 to 1010 to 1515 to 2020 to 25>25Gradient (mmHg)Area (cm2) Fresh valve Ethanol fixed valve Glutaraldehyde fixed valve Figure 34. Results for fresh vs. fixed valves fo r opening cycle given gradient as evidence The following illustration (Figure 35) shows fresh valve 1, ethanol fixed valve 1 and glutaraldehyde fixed valve 1 predicted area vs. gradient (eviden ce) while closing for the flow rate of 5 L/min at 60 mmHg back pressure. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0 to -50 to -10-10 to -30-30 to -50<-50Gradient (mmHg)Area (cm2) Fresh valve Ethanol fixed valve Glutaraldehyde fixed valve Figure 35. Results for fresh vs. fixed valves fo r closing cycle given gr adient as evidence

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60 A distinct opening pattern is observed for the valves working at lower back pressure (60 mmHg) vs. higher back pressure (90 mmHg). Il lustration below (Figure 36) shows opening phase of fresh valve 1 predicte d area vs. gradient (e vidence) at 60 mmHg and 90 mmHg backpressure. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 00-55-1010-1515-2020-25>25 Gradient (mmHg)Area (cm2) Fresh valve at 60 back pressure Fresh valve at 90 back pressure Figure 36. Results for fresh valve for high and low back pressure The following illustration (Figure 37) shows fresh valve 1, ethanol fixed valve 1 and glutaraldehyde fixed valve 1 predicted ar ea vs. instantaneous flow (evidence).

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61 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9<00 to 11 to 22 to 33 to 44 to 55 to 66 to 77 to 8> 8Flow (L/min)Area (cm2) Fresh valve Ethanol fixed valve Glutaraldehyde fixed valve Figure 37. Results for fresh vs. fixed valves for opening cycle given flow as evidence

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62 Chapter 7 Discussion As observed from the experiments, fixed valves compared to fresh valves were more stenotic, determined by opening area and peak pressure drops. Also the fixed valves, compared to fresh valves had different opening and closing patterns. For all the valves instantaneous relations hip of pressure drop and ar ea was different for opening and closing cycles. The Bayesian model was successful in capturing the hydraulic behavior of fresh and fixed valves. There were certain limitations in this study. The main limitation of the study was small size of the valves, and thus they did not follow physics applicable to normal size valves. (e.g. Gorlin equation failed, there wa s no pressure recovery, unlike what is observed in clinical situations). The data set was small for constructing fully functional Bayesian network, and it was not validated on a totally independent data set. The Bayesian model used in this project consis ted of basic set of parameters; it could be improved by including more parameters like type of fixation treatment, valve size (annulus, aortic root diameter), time of testi ng, pressure reading P3 and valve resistance.

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63 Chapter 8 Conclusions and Recommendations Analysis of probabilistic relations hips between aortic valve area and hemodynamic factors should give a better estimate to understand valve performance. As the Bayesian networks work for porcine valves tested in pulse duplicator, if used with available patient data, they should predict values of aortic area for patients with progressive stenosis. More number of nodes wo uld be considered for patients and they would differ from the ones used in the network for pulse duplicator. Further step would be to construct a Ba yesian network that can develop functional relationship between the parameters and pred ict for unknown datasets. More research can be carried out in studying the hysteresi s patterns between valve opening and valve closing and understanding th eir significance with aort ic valve hemodynamics.

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64 References [1] www.MedicineNet.co m. A network of U.S. Board Certified Physicians and Allied Health Professionals [2] S. Joanna Cowell, David E. Newby, Nicholas A. Boon, Andrew T. Elder. Calcific aortic stenosis: same old story? Oxford Journals Age and Aging, Volume 33, August 2004, 538-544 [3] http://www.nlm.nih.gov/medlineplus. A se rvice of the U.S. National Library of medicine and the National Institutes of Hea lth and the National Institutes of Health [4] Steven J. Lester, Doff B. McElhinney, Joseph P. Miller, Juergen T. Lutz, Catherine M. Otto, Rita F. Redberg Rate of Change in Aortic Valve Area duri ng a Cardiac Cycle Can Predict the Rate of Hemodynamic Progression of Aortic Stenosis AHA Circulation. 2000, 101:1947 [5] http://heartlab.robarts.ca. The John P. R obarts Research Institute Heart Valve Lab [6] Anatomy of Human Body, Henry Gray, 1995, 15th Ed [7] C. H. Peels, L. H. B. Baur, 2004 Valve Surgery at Turn of Millennium [8] Anthony S-Y Leong Fixation and fixatives Extract from Woods and Ellis, Laboratory Histopathology: A Complete Reference, 1994 Churchill Livingstone [9] Michael D. VanAuker, Katherin e Rhodes, Alok Singh, Joel A. Strom Development of Markers of Valve Stiffne ss for Prediction of Hemodynamic Progression of Aortic Stenosis Proceedings of 26th Annual Internation Conference of IEEE EMBS, September 1-5, 2004 [10] Arthur E. Weyman, Marielle Scherrer-Crosbie Aortic Stenosis: Physics and Physiol ogyWhat Do the Numbers Really Mean? Rev Cardiovasc Med. 2005 Winter 6(1), 23-32

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65 [11] Lian Cheng, David G. Bostwick Essentials of anatomic pathology, September 2005 [12] John A. Kiernan, Formaldehyde, formalin, paraformaldehyde and glutaraldehyde: What they are and what they do Article published in Microscopy Today 00-1, 2000, 8-12 [13] Jie Cheng, David Bell, Weiru Liu Learning Bayesian Networks from Data: An Efficient Approach Based on Information Theory Conference on Information and Knowledge Management, Proceedings of the sixth international conference on Information and knowledge management, 1997, 325 331 [14] Otto C.M., Valvular Heart diseas e, W.B. Saunders Company, Pennsylvania 1999 [15] Edmund Kenneth Kerut, Elizabet h F. Mcllwain, Gary D. Plotnick Handbook of Echo Doppler Interpretation, 2nd Edition [16] Eudes E. Fileti, Puspitapa llab Chaudhuri, Sylvio Canuto Relative strength of hydrogen bond intera ction in alcohol-water complexes Chem. Phys. Lett. 2004, 400 494 [17] Rahimtoola,S.H.,Aortic Valve Disease,Hu rst's Diseases of The Heart,10th Edition, Vol. 2, 1682-1695 [18] Yannick Dumont and Marie Arsenault An Alternative to Standard Continuity Equa tion for the Calculation of Aortic Valve Area by Echocardiography J Am Soc Echocardiogr. Dec 2003, 16(12): 1309-15 [19] http://en.wikipedia.org/wiki/Bayesia n_network. Free online encyclopedia [20] Judea Pearl, Stuart Russell Bayesian Networks. UCLA Cognitive Systems Laboratory, Technical Report (R-277), Nov 2000 [21] Anil Madhav Patwardhan, Pradeep Vaideeswar Stress strain characteristics of glutaraldehyde treated porcine aortic valve tissue following ethanol treatment IJTCVS 2004, 20, 67-71 [22] Timothy J. O'Leary Advanced Diagnostic Methods in Pathology: Pr inciples, Practice, a nd Protocols, Chapter II Immunohistochemical and In Situ Hybridization Techniques, 95-123

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66 [23] Blood Pressure Tables fo r Children and Adolescents Fourth Report on the Diagnosis, Evaluation, a nd Treatment of High Blood Pressure in Children and Adolescents, National Heart Lung and Blood Institute Pediatrics, 2004; 114:555-576. NIH Publication No. 05-5267 [24] http://www.answers.com. Online encyclopedia [25] ACC/AHA Guidelines for the Management of Patients with Va lvular Heart Disease III, Specific Valve Lesions, 1998 [26] IG Burwash, DD Thomas, M Sadahiro, AS Pearlman, ED Verrier, R Thomas, CD Kraft and CM Otto Dependence of Gorlin formula and continui ty equation valve areas on transvalvular volume flow rate in valv ular aortic stenosis AHA Journals Circulation, Vol 89, 827-835 [27] JB Chambers, DC Sprigings, T Coch rane, J Allen, R Morris, MM Black and G Jackson Continuity equation and Gorlin formula compar ed with directly obs erved orifice area in native and prosthetic aortic valves Br. Heart J. 1992; 67: 193-199 [28] Joseph A. Kisslo, MD, David B. Adams, RDCS Doppler Evaluation of Valvular Stenosis #3 [29] Peter F. Davies, Anthony G. Passerini, Craig A. Simmons Aortic Valve Turning Over a New Leafle t in Endothelial Phenotypic Heterogeneity Arteriosclerosis, Thrombosis, and Vascular Biology. 2004, 24:1331 [30] Paulo Jos F. Ribeiro, Paulo Roberto B. vora, Walter V. A. Vicente, Antonio Carlos Menardi Ribeiro Preto Reconstructive Surgery fo r Aortic Valve Disease Arq. Bras. Cardiol. Vol.74 n.5 So Paulo [31] Raphael Rosenhek, Florian Rader, Nico le Loho, Harald Gabriel, Maria Heger, Ursula Klaar, Michael Schemper, Thomas Binder, Gerald Maurer, Helmut Baumgartner Statins but Not Angiotensin-Converting Enzyme Inhibitors Delay Progression of Aortic Stenosis AHA Journal, Circulation. Sep 7 2004, 110 (10):1291-5 [32] Ali A, Lim E, Halstead J, Ashrafian H, Ali Z, Khalpey Z, Theodorou P, Chamageorgakis T, Kumar P, Jackson C, Pepper J. Porcine or human stentless valves for aortic valve replacement? Results of a 10-year comparative study J Heart Valve Dis. July 2003; 12 (4), 430-5; discussion 435

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67 [33] Talman EA, Boughner DR Glutaraldehyde fixation alters the internal shear properties of porcine aortic heart valve tissue Ann Thorac Surg. 60 (2 Suppl), Aug 1995, S369-73 [34] Narendra R. Vyavahare Danielle Hirsch Eyal Lerner Jonathan Z. Baskin Robert Zand Frederick J. Schoen, Robert J. Levy Prevention of calcification of glutaraldehyde-c rosslinked porcine aor tic cusps by ethanol preincubation: Mech anistic studies of protein st ructure and waterbiomaterial relationships J Biomed Mater Res. 40 (4), June 15 1998, 577-85 [35] Niederberger J, Schima H, Maurer G, Baumgartner H. Importance of pressure recovery for assessmen t of aortic stenosis by Doppler ultrasound: role of aortic size, aortic valve area, and direction of ste notic jet in vitro Circulation 1996; 94, 1934-1940 [36] http://www.cardiologychannel.com/aorticstenosis/. Cardiology community [37] Muiesan Maria Loren za, Losi Maria Angela Aortic valve sclerosis: new help from echocardiography in the assessment of cardiovascular risk Journal of Hypertension: Volume 23(4) April 2005, 721-723 [38] VanAuker MD, Cape EG, Sigf usson G, Tacy TA, del Nido PJ Potential role of mechanical stress in the etio logy of pediatric heart disease: septal shear stress in subaortic stenosis Journal of the American Colle ge of Cardiology, 1997, 30:247-254

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68 Appendices

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69 Appendix A A1: Program to make a Bayesian file for opening cycle % This program creates a GeNIe compatible .t xt file to construct a Bayesian network. % This file needs to be exported to the GeNIe so ftware and saved as hugin.NET file to make it % executable. %----------User selection of .xls file---------[filename, pathname] = uigetfile(' *.xls', 'Pick the xls-file'); if isequal(filename,0) | isequal(pathname,0) disp('User pressed cancel') else disp(['User selected ', fullfile(pathname, filename)]) end % ----------Reading .xls file---------Database = xlsread(fullfile(pathname, filename)); % Reading variables Gr = Database(:,12); ar = Database(:,13); Fl = Database(:,5); BPR = Database(:,6); %----------Selection of 100 points in the database---------p=0; for a = 1:length(ar) if ar(a)==0 p=p+1; end; end; x=(p-50); for l=1:100 %----------Assigning 100 points to a new database---------Gradient(l)= Gr(x); Flow(l)= Fl(x); BP(l) = BPR(x); area(l) = ar(x); x=x+1; end;

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70Appendix A: (Continued) %----------Assigning States to Area---------for k = 1:100 if area(k) == 0 sta(k) = 1; elseif area(k) <= 0.1 sta(k) = 2; elseif area(k) <= 0.2 sta(k) = 3; elseif area(k) <= 0.3 sta(k) = 4; elseif area(k) <= 0.4 sta(k) = 5; elseif area(k) <= 0.5 sta(k) = 6; elseif area(k) <= 0.6 sta(k) = 7; elseif area(k) <= 0.7 sta(k) = 8; elseif area(k) <= 0.8 sta(k) = 9; else sta(k) = 10; end; end; %----------Assigning States to BP---------for k = 1:100 if BP(k) <= 55 stBP(k) = 1; elseif BP(k) <= 60 stBP(k) = 2; elseif BP(k) <= 65 stBP(k) = 3; elseif BP(k) <= 70

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71Appendix A: (Continued) stBP(k) = 4; elseif BP(k) <= 80 stBP(k) = 5; elseif BP(k) <= 85 stBP(k) = 6; elseif BP(k) <= 90 stBP(k) = 7; elseif BP(k) <= 95 stBP(k) = 8; elseif BP(k) <= 100 stBP(k) = 9; else stBP(k) = 10; end; end; %----------Assigning States to Flow---------for k = 1:100 if Flow(k) <= 0 stf(k) = 1; elseif Flow(k) <= 1 stf(k) = 2; elseif Flow(k) <= 2 stf(k) = 3; elseif Flow(k) <= 3 stf(k) = 4; elseif Flow(k) <= 4 stf(k) = 5; elseif Flow(k) <= 5 stf(k) = 6; elseif Flow(k) <= 6 stf(k) = 7; elseif Flow(k) <= 7 stf(k) = 8;

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72Appendix A: (Continued) elseif Flow(k) <= 8 stf(k) = 9; else stf(k) = 10; end; end; %----------Assigning States to Gradient---------for k = 1:100 if Gradient(k) <= -50 stGr(k) = 1; elseif Gradient(k) <= -30 stGr(k) = 2; elseif Gradient(k) <= -10 stGr(k) = 3; elseif Gradient(k) <= 0 stGr(k) = 4; elseif Gradient(k) <= 5 stGr(k) = 5; elseif Gradient(k) <= 10 stGr(k) = 6; elseif Gradient(k)<= 15 stGr(k) = 7; elseif Gradient(k) <= 20 stGr(k) = 8; elseif Gradient(k) <= 25 stGr(k) = 9; else stGr(k) = 10; end; end; %-----------------------------------------------------------fid = fopen('model.txt', 'W'); % create file for Bayesian Network

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73Appendix A: (Continued) %----------Counting states area, BP, flow and gradient---------for k = 1:100 countarea(k)=0; temp(k)=sta(k); for l = 1:100 tem(l)=sta(l); if temp(k)== tem(l) countarea(k) = countarea(k)+1; % count area end; end; end; for k = 1:100 countBP(k)=0; temp(k)=stBP(k); for l = 1:100 tem(l)=stBP(l); if temp(k)== tem(l) countBP(k) = countBP(k)+1; %count back pressure end; end; end; for k = 1:100 countFlow(k)=0; temp(k)=stf(k); for l = 1:100 tem(l)=stf(l); if temp(k)== tem(l) countFlow(k) = countFlow(k)+1; %count flow end; end; end; for k = 1:100 countgr(k)=0; temp(k)=stGr(k); for l = 1:100 tem(l)=stGr(l); if temp(k)== tem(l) countgr(k) = countgr(k)+1; % count gradient end; end; end;

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74Appendix A: (Continued) %----------Probabilities calculation---------for k = 1:100 ufBP(k) = stf(k)*10000 + stBP(k); ufBPGr(k)=stf(k)*10000 + stBP(k)*100 + stGr(k); ufBPGrA(k) = stf(k)*1000000 + stBP(k)*10000 + stGr(k)*100 + sta(k); end; for k = 1:100 countufBP(k)=0; temp(k)=ufBP(k); for l = 1:100 tem(l)=ufBP(l); if temp(k)== tem(l) countufBP(k) = countufBP(k)+1; end; end; end; for k = 1:100 countufBPGr(k)=0; temp(k)=ufBPGr(k); for l = 1:100 tem(l)=ufBPGr(l); if temp(k)== tem(l) countufBPGr(k) = countufBPGr(k)+1; end; end; end; for k = 1:100 countufBPGrA(k)=0; temp(k)=ufBPGrA(k); for l = 1:100 tem(l)=ufBPGrA(l); if temp(k)== tem(l) countufBPGrA(k) = countufBPGrA(k)+1; end; end; end; for k = 1:100 PflowgivenBP(k) = countufBP(k)/c ountBP(k); %P(flow/back pressure) PgrgivenBPandflow(k) = countufBPGr(k)/countufBP(k); %P(gradient/ bp, flow) PareagivenBPflowGr(k) = countufBPGrA(k)/countufBPGr(k); %P(area/ bp, flow, gradient) end;

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75Appendix A: (Continued) %------------------------------------------------------------------------BackPressure=zeros(10,1); for k=1:100 for i=1:10 if stBP(k) ==i BackPressure(i) = countBP(k)/100; end; end; end; %------------------------------------------------------------------------FlowgivenBP=zeros(100,1); for i=1:10 for k=1:100 if stBP(k)== i t = (10*(i-1) + stf(k)); FlowgivenBP(t) = PflowgivenBP(k); end; end; end; %------------------------------------------------------------------------GrgivenFLBP = zeros(1000,1); for k = 1:100 for i=1:10 if stBP(k) == i for j=1:10 if stf(k)==j d = (10*(j-1) + 100*(i-1) + stGr(k)); GrgivenFLBP(d) = PgrgivenBPandflow(k); end; end; end; end; end; %------------------------------------------------------------------------AreagivenGrFLBP = zeros(10000,1); for k=1:100 for i=1:10 if stBP(k)==i for j=1:10

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76Appendix A: (Continued) if stf(k)==j for m=1:10 if stGr(k)==m e = (10*(m-1) + 100*(j-1) + 1000*(i-1) + sta(k)); AreagivenGrFLBP(e) = PareagivenBPflowGr(k); end; end; end; end; end; end; end; %----------File for Bayesian Net for syntax compatible with GeNIe 2.0---------fprintf (fid,'net \n{ \n node_size = (30 30);\n}\nnode BP\n{\nlabel = "BP";\nposition = (212 163);\nstates = ("less than equal to 55" "55 to 60" "6 0 to 65" "65 to 70"\n "70 to 80" "80 to 85" "85 to 90" "90 to 95" "95 to 100" "more than 100");\n}\n'); fprintf (fid,'\nnode F \n{\nlabel = "Flow";\nposition = (135 79);\nstates = ("less than equal to 0" "0 to 1" "1 to 2" "2 to 3"\n "3 to 4" "4 to 5" "5 to 6" "6 to 7" "7 to 8" "more than 8");\n}\n'); fprintf (fid,'\nnode G \n{\nlabel = "Gradient";\npos ition = (305 75);\nstates = ("less than -50" "50 to -30" "-30 to -10" "-10 to 0"\n "0 to 5" "5 to 10" "10 to 15" "15 to 20" "20 to 25" "more than 25");\n}\n'); fprintf (fid,'\nnode A \n{\nlabel = "Area";\nposition = (210 -15);\nstates = ("0" "0 to 0.1" "0.1 to 0.2" "0.2 to 0.3" "0.3 to 0.4"\n "0.4 to 0.5" "0.5 to 0.6" "0.6 to 0.7" "0.7 to 0.8" more than 0.8");\n}\n'); %-------------------------------------------------------------------------fprintf(fid, '\npotential (BP |)'); % Probability for all states of back pressure fprintf(fid, '\n{\n\tdata = ('); for i=1:length(BackPressure) fprintf(fid, ['%9.8f'], BackPressure(i)); if i ~= length(BackPressure) fprintf(fid, '); if mod(i, 5) == 0 fprintf(fid, '\n\t\t'); end; end; end; fprintf(fid, ');\n}'); %-------------------------------------------------------------------------fprintf(fid, '\n\npotential (F | BP)'); % Probability for all states of flow given back pressure fprintf(fid, '\n{\n\tdata = (('); for i=1:length(FlowgivenBP) fprintf(fid, ['%9.8f'], FlowgivenBP(i));

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77Appendix A: (Continued) if i ~= length(FlowgivenBP) if mod(i, 10) ~= 0 fprintf(fid, '); end; if mod(i, 5) == 0 if mod(i, 10) == 0 fprintf(fid, ')\n\t\t('); else fprintf(fid, '\n\t\t'); end; end; end; end; fprintf(fid, '));\n}'); %-------------------------------------------------------------------------fprintf(fid, '\n\npotential (G | BP F)'); % Probability for all states of gradient given bp and flow fprintf(fid, '\n{\n\tdata = ((('); for i=1:length(GrgivenFLBP) fprintf(fid, ['%9.8f'], GrgivenFLBP(i)); if i ~= length(GrgivenFLBP) if mod(i, 10) ~= 0 fprintf(fid, '); end; if mod(i, 5) == 0 if mod(i, 10) == 0 && mod(i,100) ~=0 fprintf(fid, ')\n\t\t('); elseif mod(i, 100) == 0 fprintf(fid, '))\n\t\t(('); else fprintf(fid, '\n\t\t'); end; end; end; end; end; fprintf(fid, ')));\n}'); %-------------------------------------------------------------------------fprintf(fid, '\n\npotential (A | BP F G)'); % Probability for all states of area given bp, flow, gr fprintf(fid, '\n{\n\tdata = (((('); for i=1:length(AreagivenGrFLBP) fprintf(fid, ['%9.8f'], AreagivenGrFLBP(i)); if i ~= length(AreagivenGrFLBP) if mod(i, 10) ~= 0

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78Appendix A: (Continued) fprintf(fid, '); end; if mod(i, 5) == 0 if mod(i, 10) == 0 && mod(i,100) ~=0 && mod(i,1000) ~= 0 fprintf(fid, ')\n\t\t('); elseif mod(i, 100) == 0 && mod(i,1000) ~=0 fprintf(fid, '))\n\t\t(('); elseif mod(i,1000) == 0 fprintf(fid, ')))\n\t\t((('); else fprintf(fid, '\n\t\t'); end; end; end; end; fprintf(fid, '))));\n}'); fclose(fid); A2: Program to calculate Mutual information for opening cycle % This program calculates mutual information betw een every pair of parameters for opening % cycle % -----------User selection of xls file------------------[filename, pathname] = uigetfile(' *.xls', 'Pick the xls-file'); if isequal(filename,0) | isequal(pathname,0) disp('User pressed cancel') else disp(['User selected ', fullfile(pathname, filename)]) end % -----------Reading xls file------------------Database = xlsread(fullfile(pathname, filename)); % Reading variables Gr = Database(:,12); ar = Database(:,13); Fl = Database(:,5); BPR = Database(:,6); %----------Selection of 100 points in the database---------p=0; for a = 1:length(ar) if ar(a)==0 p=p+1; end;

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79Appendix A: (Continued) end; x=(p-50); for l=1:100 %----------Assigning 100 points to a new database---------Gradient(l)= Gr(x); Flow(l)= Fl(x); BP(l) = BPR(x); area(l) = ar(x); x=x+1; end; %----------Assigning States to Area---------for k = 1:100 if are(k) == 0 area(k) = 1; elseif are(k) <= 0.1 area(k) = 2; elseif are(k) <= 0.2 area(k) = 3; elseif are(k) <= 0.3 area(k) = 4; elseif are(k) <= 0.4 area(k) = 5; elseif are(k) <= 0.5 area(k) = 6; elseif are(k) <= 0.6 area(k) = 7; elseif are(k) <= 0.7 area(k) = 8; elseif are(k) <= 0.8 area(k) = 9; else area(k) = 10; end; end;

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80Appendix A: (Continued) %----------Assigning States to BP---------for k = 1:100 if BPr(k) <= 55 BP(k) = 1; elseif BPr(k) <= 60 BP(k) = 2; elseif BPr(k) <= 65 BP(k) = 3; elseif BPr(k) <= 70 BP(k) = 4; elseif BPr(k) <= 80 BP(k) = 5; elseif BPr(k) <= 85 BP(k) = 6; elseif BPr(k) <= 90 BP(k) = 7; elseif BPr(k) <= 95 BP(k) = 8; elseif BPr(k) <= 100 BP(k) = 9; else BP(k) = 10; end; end; %----------Assigning States to Flow---------for k = 1:100 if Flo(k) <= 0 Flow(k) = 1; elseif Flo(k) <= 1 Flow(k) = 2; elseif Flo(k) <= 2 Flow(k) = 3; elseif Flo(k) <= 3

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81Appendix A: (Continued) Flow(k) = 4; elseif Flo(k) <= 4 Flow(k) = 5; elseif Flo(k) <= 5 Flow(k) = 6; elseif Flo(k) <= 6 Flow(k) = 7; elseif Flo(k) <= 7 Flow(k) = 8; elseif Flo(k) <= 8 Flow(k) = 9; else Flow(k) = 10; end; end; %----------Assigning States to Gradient---------for k = 1:100 if Gra(k) <= -50 Gradient(k) = 1; elseif Gra(k) <= -30 Gradient(k) = 2; elseif Gra(k) <= -10 Gradient(k) = 3; elseif Gra(k) <= 0 Gradient(k) = 4; elseif Gra(k) <= 5 Gradient(k) = 5; elseif Gra(k) <= 10 Gradient(k) = 6; elseif Gra(k)<= 15 Gradient(k) = 7; elseif Gra(k) <= 20 Gradient(k) = 8;

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82Appendix A: (Continued) elseif Gra(k) <= 25 Gradient(k) = 9; else Gradient(k) = 10; end;end; %----------Count Area & calculate Pxj (probability of area)---------for k = 1:100 countarea(k)=0; temp(k)=area(k); for l =1:100 tem(l)=area(l); if temp(k)== tem(l) countarea(k) = countarea(k)+1; end; end; end; for k =1:100 Pxj(k)= countarea(k)/100; end; %----------GRADIENT---------%----------Count Gradient---------for j =1:100 countgradient(j)=0; temp(j)= Gradient(j); for i =1:100 tem(i)=Gradient(i); if temp(j)== tem(i) countgradient(j) = countgradient(j)+1; end; end; end; %----------Calculate Probability of Gradient---------for k = 1:100 Pgrxi(k) = countgradient(k)/100; end; %define a unique term for calculating P(grxi,xj) for k = 1:100 Uniquegr(k) = Gradient(k)*1000 + (area(k)); end;

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83Appendix A: (Continued) %count Unique Term for k =1:100 countuniquegr(k)=0; temp(k)=Uniquegr(k); for l =1:100 tem(l)=Uniquegr(l); if temp(k)== tem(l) countuniquegr(k) = countuniquegr(k)+1; end; end; end; for k =1:100 Pgrxixj(k)= (countuniquegr(k)/countarea(k))*Pxj(k); % P(gradient,area) end; % Calculation of Mutual Information between area and gradient for k =1:100 FinalAnswergr(k) = (Pgrxixj(k))*(Log10 (Pgrxixj(k)/(Pgrxi(k)*Pxj(k)))); end; Relationship_Gradient_Area = sum(FinalAnswergr) %----------FLOW---------%----------Count Flow---------for j =1:100 countflow(j)=0; tempfl(j)= Flow(j); for i =1:100 temfl(i)=Flow(i); if tempfl(j)== temfl(i) countflow(j) = countflow(j)+1; end; end; end; %----------Calculate Probability of Flow---------for k =1:100 Pflxi(k) = countflow(k)/100; end; %define a unique term for calculating P(grxi,xj) for k =1:100 Uniquefl(k) = Flow(k)*1000 + (area(k)); end;

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84Appendix A: (Continued) %count Unique Term for k =1:100 countuniquefl(k)=0; temp(k)=Uniquefl(k); for l =1:100 tem(l)=Uniquefl(l); if temp(k)== tem(l) countuniquefl(k) = countuniquefl(k)+1; end; end; end; for k =1:100 Pflxixj(k)= (countuniquefl(k)/countarea(k))*Pxj(k); end; % Calculation of Mutual Information between flow and area for k =1:100 FinalAnswerfl(k) = (Pflxixj(k))*(Log10 (Pflxixj(k)/(Pflxi(k)*Pxj(k)))); end; Relationship_Flow_Area = sum(FinalAnswerfl) %----------BACK PRESSURE---------%----------Count back pressure---------for j =1:100 countBP(j)=0; temp(j)= BP(j); for i =1:100 tem(i)=BP(i); if temp(j)== tem(i) countBP(j) = countBP(j)+1; end; end; end; %----------Calculate Probability of BP---------for k =1:100 PBPxi(k) = countBP(k)/101; end; %define a unique term for calculating P(grxi,xj) for k =1:100 UniqueBP(k) = BP(k)*1000 + (area(k)); end;

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85Appendix A: (Continued) %count Unique Term for k = 1:100 countuniqueBP(k)=0; temp(k)=UniqueBP(k); for l =1:100 tem(l)=UniqueBP(l); if temp(k)== tem(l) countuniqueBP(k) = countuniqueBP(k)+1; end; end; end; for k =1:100 PBPxixj(k)= (countuniqueBP(k)/countarea(k))*Pxj(k); end; % Calculation of Mutual Information between back pressure and area for k =1:100 FinalAnswerBP(k) = (PBPxixj(k))*(Log10 (PBPxixj(k)/(PBPxi(k)*Pxj(k)))); end; Relationship_BP_Area = sum(FinalAnswerBP) %----------Flow & Gradient---------%define a unique term for k =1:100 Uniqueflgr(k) = Flow(k)*1000 + (Gradient(k)); end; %count Unique Term for k = 1:100 countuniqueflgr(k)=0; temp(k)=Uniqueflgr(k); for l =1:100 tem(l)=Uniqueflgr(l); if temp(k)== tem(l) countuniqueflgr(k) = countuniqueflgr(k)+1; end; end; end; for k =1:100 Pflgrxixj(k)= (countuniqueflgr(k)/countgradient(k))*Pgrxi(k); end;

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86Appendix A: (Continued) % Calculation of Mutual Informa tion between flow and gradient for k =1:100 FinalAnswerflgr(k) = (Pflgrxixj(k))*(Lo g10 (Pflgrxixj(k)/(Pflxi(k)*Pgrxi(k)))); end; Relationship_Flow_Gradient = sum(FinalAnswerflgr) %----------BP & Gradient---------for k =1:100 UniqueBPgr(k) = BP(k)*1000 + (Gradient(k)); end; %count Unique Term for k =1:100 countuniqueBPgr(k)=0; temp(k)=UniqueBPgr(k); for l =1:100 tem(l)=UniqueBPgr(l); if temp(k)== tem(l) countuniqueBPgr(k) = countuniqueBPgr(k)+1; end; end; end; for k =1:100 PBPgrxixj(k)= (countuniqueBPgr(k)/countgradient(k))*Pgrxi(k); end; % Calculation of Mutual Information between back pressure and gradient for k = 1:100 FinalAnswerBPgr(k) = (PBPgrxixj(k))*(Log1 0 (PBPgrxixj(k)/(PBPxi(k)*Pgrxi(k)))); end; Relationship_BP_Gradient = sum(FinalAnswerBPgr) %----------BP & Flow---------%define a unique term for calculating P(grxi,xj) for k = 1:100 UniqueBPfl(k) = BP(k)*1000 + (Flow(k)); end; %count Unique Term for k = 1:100 countuniqueBPfl(k)=0; temp(k)=UniqueBPfl(k); for l = 1:100 tem(l)=UniqueBPfl(l);

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87Appendix A: (Continued) if temp(k)== tem(l) countuniqueBPfl(k) = countuniqueBPfl(k)+1; end; end; end; for k = 1:100 PBPflxixj(k)= (countuniqueBPfl(k)/countflow(k))*Pflxi(k); end; % Calculation of Mutual Information between flow and back pressure for k = 1:100 FinalAnswerBPfl(k) = (PBPflxixj(k))*(Log10 (PBPflxixj(k)/(PBPxi(k)*Pflxi(k)))); end; Relationship_BP_Flow = sum(FinalAnswerBPfl) A3: Program to find cutsets for opening cycle % This program determines cutsets between every two parameters for opening cycle % -----------User selection of xls file------------------[filename, pathname] = uigetfile(' *.xls', 'Pick the xls-file'); if isequal(filename,0) | isequal(pathname,0) disp('User pressed cancel') else disp(['User selected ', fullfile(pathname, filename)]) end % Reading xls file Database = xlsread(fullfile(pathname, filename)); % Reading variables Gr = Database(:,12); ar = Database(:,13); Fl = Database(:,5); BPR = Database(:,6); %----------Selection of 100 points in the database---------p=0; for a = 1:length(ar) if ar(a)==0 p=p+1; end; end;

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88Appendix A: (Continued) x=(p-50); for l=1:100 %----------Assigning 100 points to a new database---------Gradient(l)= Gr(x); Flow(l)= Fl(x); BP(l) = BPR(x); area(l) = ar(x); x=x+1; end; %----------Assigning States to Area---------for k = 1:100 if are(k) == 0 area(k) = 1; elseif are(k) <= 0.1 area(k) = 2; elseif are(k) <= 0.2 area(k) = 3; elseif are(k) <= 0.3 area(k) = 4; elseif are(k) <= 0.4 area(k) = 5; elseif are(k) <= 0.5 area(k) = 6; elseif are(k) <= 0.6 area(k) = 7; elseif are(k) <= 0.7 area(k) = 8; elseif are(k) <= 0.8 area(k) = 9; else area(k) = 10; end; end;

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89Appendix A: (Continued) %----------Assigning States to BP---------for k = 1:100 if BPr(k) <= 55 BP(k) = 1; elseif BPr(k) <= 60 BP(k) = 2; elseif BPr(k) <= 65 BP(k) = 3; elseif BPr(k) <= 70 BP(k) = 4; elseif BPr(k) <= 80 BP(k) = 5; elseif BPr(k) <= 85 BP(k) = 6; elseif BPr(k) <= 90 BP(k) = 7; elseif BPr(k) <= 95 BP(k) = 8; elseif BPr(k) <= 100 BP(k) = 9; else BP(k) = 10; end; end; %----------Assigning States to Flow---------for k = 1:100 if Flo(k) <= 0 Flow(k) = 1; elseif Flo(k) <= 1 Flow(k) = 2; elseif Flo(k) <= 2 Flow(k) = 3; elseif Flo(k) <= 3

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90Appendix A: (Continued) Flow(k) = 4; elseif Flo(k) <= 4 Flow(k) = 5; elseif Flo(k) <= 5 Flow(k) = 6; elseif Flo(k) <= 6 Flow(k) = 7; elseif Flo(k) <= 7 Flow(k) = 8; elseif Flo(k) <= 8 Flow(k) = 9; else Flow(k) = 10; end; end; %----------Assigning States to Gradient---------for k = 1:100 if Gra(k) <= -50 Gradient(k) = 1; elseif Gra(k) <= -30 Gradient(k) = 2; elseif Gra(k) <= -10 Gradient(k) = 3; elseif Gra(k) <= 0 Gradient(k) = 4; elseif Gra(k) <= 5 Gradient(k) = 5; elseif Gra(k) <= 10 Gradient(k) = 6; elseif Gra(k)<= 15 Gradient(k) = 7; elseif Gra(k) <= 20 Gradient(k) = 8;

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91Appendix A: (Continued) elseif Gra(k) <= 25 Gradient(k) = 9; else Gradient(k) = 10; end; end; %----------Counting area & calculate Pxj (probability of area)---------for k = 1:100 countarea(k)=0; temp(k)=area(k); for l = 1:100 tem(l)=area(l); if temp(k)== tem(l) countarea(k) = countarea(k)+1; end; end; end; for k = 1:100 Pxj(k)= countarea(k)/100; end; %----------GRADIENT---------%----------Count Gradient---------for j = 1:100 countgradient(j)=0; temp(j)= Gradient(j); for i = 1:100 tem(i)=Gradient(i); if temp(j)== tem(i) countgradient(j) = countgradient(j)+1; end; end; end; %------------FLOW-----------%----------Count Flow--------for j = 1:100 countflow(j)=0; tempfl(j)= Flow(j); for i = 1:100 temfl(i)=Flow(i);

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92Appendix A: (Continued) if tempfl(j)== temfl(i) countflow(j) = countflow(j)+1; end; end; end; %------------BACK PRESSURE-----------%----------Count Back pressure---------for j = 1:100 countBP(j)=0; temp(j)= BP(j); for i = 1:100 tem(i)=BP(i); if temp(j)== tem(i) countBP(j) = countBP(j)+1; end; end; end; %--------------------------------------------% Calculating P(Flow/area) & P(area/Flow) for k = 1:100 Uniquefl(k) = (Flow(k)*1000 + area(k)); end; %count Unique Term for k = 1:100 countuniquefl(k)=0; temp(k)=Uniquefl(k); for l = 1:100 tem(l)=Uniquefl(l); if temp(k)== tem(l) countuniquefl(k) = countuniquefl(k)+1; end; end; end; for k = 1:100 PAreagivenflow(k) = (countuniquefl(k)/countflow(k)); PflowgivenArea(k) = (countuniquefl(k)/countarea(k)); end; %--------------------------------------------% Calculating P(BP/area) & P(area/BP) for k = 1:100 PBPxi(k) = countBP(k)/100;

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93Appendix A: (Continued) end; %define a unique term for k = 1:100 UniqueBP(k) = BP(k)*10000 + (area(k)); end; %count Unique Term for k = 1:100 countuniqueBP(k)=0; temp(k)=UniqueBP(k); for l = 1:100 tem(l)=UniqueBP(l); if temp(k)== tem(l) countuniqueBP(k) = countuniqueBP(k)+1; end; end; end; for k = 1:100 PAreagivenBP(k) = (countuniqueBP(k)/countBP(k)); PBPgivenArea(k) = (countuniqueBP(k)/countarea(k)); end; %------------Probability of Gradient---------% Calculating P(Gradient/area) & P(area/Gradient) for k = 1:100 Pgrxi(k) = countgradient(k)/100; end; %define a unique term for k = 1:100 Uniquegr(k) = (Gradient(k)*10000 + area(k)); end; %count Unique Term for k = 1:100 countuniquegr(k)=0; temp(k)=Uniquegr(k); for l = 1:100 tem(l)=Uniquegr(l); if temp(k)== tem(l) countuniquegr(k) = countuniquegr(k)+1; end; end; end;

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94Appendix A: (Continued) for k = 1:100 PAreagivengradient(k) = (countuniquegr(k)/countgradient(k)); Pgradientgivenarea(k) = (countuniquegr(k)/countarea(k)); end; %------------Flow & Gradient-----------% Calculating P(Gradient/Flow) & P(Flow/Gradient) %define a unique term for k = 1:100 Uniqueflgr(k) = (Flow(k)*10000 + Gradient(k)); end; %count Unique Term for k = 1:100 countuniqueflgr(k)=0; temp(k)=Uniqueflgr(k); for l = 1:100 tem(l)=Uniqueflgr(l); if temp(k)== tem(l) countuniqueflgr(k) = countuniqueflgr(k)+1; end; end; end; for k = 1:100 Pgradientgivenflow(k) = countuniqueflgr(k)/countflow(k); Pflowgivengradient(k) = countuniqueflgr(k)/countgradient(k); end; %------------BP & Flow-----------% Calculating P(BP/Flow) & P(Flow/BP) %define a unique term for k = 1:100 UniqueBPfl(k) = (BP(k)*10000 + Flow(k)); end; %count Unique Term for k = 1:100 countuniqueBPfl(k)=0; temp(k)=UniqueBPfl(k); for l = 1:100 tem(l)=UniqueBPfl(l); if temp(k)== tem(l) countuniqueBPfl(k) = countuniqueBPfl(k)+1; end; end; end; for k = 1:100 PflowgivenBP(k) = countuniqueBPfl(k)/countBP(k);

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95Appendix A: (Continued) PBPgivenflow(k) = countuniqueBPfl(k)/countflow(k); end; %------------BP & Gradient-----------%define a unique term for calculating for k = 1:100 UniqueBPgr(k) = (BP(k)*10000 + Gradient(k)); end; %count Unique Term for k = 1:100 countuniqueBPgr(k)=0; temp(k)=UniqueBPgr(k); for l = 1:100 tem(l)=UniqueBPgr(l); if temp(k)== tem(l) countuniqueBPgr(k) = countuniqueBPgr(k)+1; end; end; end; for k = 1:100 PgrgivenBP(k) = (countuniqueBPgr(k)/countBP(k)); PBPgivengr(k) = (countuniqueBPgr(k)/countgradient(k)); end; %************************************************* % Determining Cutsets %-------------------------------------------------------------------------% Determine cutsets between Flow and Area %Gr as cutset for k = 1:100 Uniquegrflarea(k) = (Gradient(k)*10000 + Flow(k)*10 + area(k)); end; %count Unique Term for k = 1:100 countuniquegrflarea(k)=0; temp(k)=Uniquegrflarea(k); for l = 1:100 tem(l)=Uniquegrflarea(l); if temp(k)== tem(l) countuniquegrflarea(k) = countuniquegrflarea(k)+1; end; end;

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96Appendix A: (Continued) end; for k = 1:100 PAreagivenflowandgradient(k) = (count uniquegrflarea(k)/countuniqueflgr(k)); end; %Display Result c=0; for k =1:100 if PAreagivenflowandgradient(k )== PAreagivengradient(k) c= c; else c=c+1; end; end; if c==0answer = 'Gradient is the cutset of area and Flow'; end; %BP as cutset for k = 1:100 UniqueBPflarea(k) = (BP(k)*100000 + Flow(k)*10 + area(k)); end; %count Unique Term for k = 1:100 countuniqueBPflarea(k)=0; temp(k)=UniqueBPflarea(k); for l = 1:100 tem(l)=UniqueBPflarea(l); if temp(k)== tem(l) countuniqueBPflarea(k) = countuniqueBPflarea(k)+1; end; end; end; for k = 1:100 PAreagivenflowandBP(k) = (countuniqueBPflarea(k)/countuniqueBPfl(k)); end; %Display Result c=0; for k =1:100 if PAreagivenflowandBP(k)== PAreagivenBP(k) c= c; else c=c+1; end; end; if c==0 answer = 'BP is the cutset of area and Flow' end;

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97Appendix A: (Continued) %----------------------------------------------------------------------% Determine cutsets between Area and BP %Flow as cutset %Display Result c=0; for k =1:100 if PAreagivenflowandBP(k)== PAreagivenflow(k) c= c; else c=c+1; end; end; if c==0 answer = 'Flow is the cutset of area and BP' end; %Gradient as cutset for k = 1:100 UniqueBPgrarea(k) = (BP(k)*100000 + Gradient(k)*10 + area(k)); end; %count Unique Term for k = 1:100 countuniqueBPgrarea(k)=0; temp(k)=UniqueBPgrarea(k); for l = 1:100 tem(l)=UniqueBPgrarea(l); if temp(k)== tem(l) countuniqueBPgrarea(k) = countuniqueBPgrarea(k)+1; end; end; end; for k = 1:100 PAreagivengradientandBP(k) = (countuni queBPgrarea(k)/countuniqueBPgr(k)); PBPgivenAreaandgradient(k) = (countuniqueBPgrarea(k)/countuniquegr(k)); end; %Display Result e=0; for k = 1:100 if PAreagivengradientandBP(k)== PAreagivengradient(k) e= e; else e=e+1;

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98Appendix A: (Continued) end; end; if e==0 answer = 'Gradient is the cutset of area and BP' end; %----------------------------------------------------------------------% Determine cutsets between Area and Gradient %Flow as cutset %Display Result d=0; for k = 1:100 if PAreagivenflowandgradie nt(k)== PAreagivenflow(k) d= d; else d=d+1; end; end; if d==0answer = 'Flow is the cutset of area and gradient' end; %BP as cutset %Display Result e=0; for k = 1:100 if PAreagivengradientandBP(k)== PAreagivenBP(k) e= e; else e=e+1; end; end; if e==0 answer = 'BP is the cutset of area and gradient' end; %----------------------------------------------------------------------% Determine cutsets between BP and Gradient for k = 1:100 UniqueBPgrfl(k) = (BP(k)*100000 + gradient(k) + Flow(k)*0.01); end; %count Unique Term for k = 1:100 countuniqueBPgrfl(k)=0; temp(k)=UniqueBPgrfl(k); for l = 1:100 tem(l)=UniqueBPgrfl(l);

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99Appendix A: (Continued) if temp(k)== tem(l) countuniqueBPgrfl(k) = countuniqueBPgrfl(k)+1; end; end; end; for k = 1:100 PBPgivenflowandgradient(k) = countuniqueBPgrfl(k)/countuniqueflgr(k); end; %Flow as cutset %Display Result d=0; for k = 1:100 if PBPgivenflowandgradient(k)== PBPgivenflow(k) d = d; else d=d+1; end; end; if d==0 answer = 'Flow is the cutset of BP and gradient' end; %Area as cutset %Display Result e=0; for k = 1:100 if PBPgivenAreaandgradient(k)== PBPgivenArea(k) e= e; else e=e+1; end; end; if e==0 answer = 'Area is the cutset of BP and gradient' end; %----------------------------------------------------------------------% Determine cutsets between BP and Flow for k=1:100 PflowgivengradientandBP(k) = count uniqueBPgrfl(k)/countuniqueBPgr(k); PBPgivenflowandArea(k)= countuni queBPflarea(k)/countuniquefl(k); PflowgivenBPandArea(k) = countuniqueBPflarea(k)/countuniqueBP(k); PflowgivengradientandArea(k) = count uniquegrflarea(k)/countuniquegr(k); end;

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100Appendix A: (Continued) %Gradient as cutset %Display Result d=0; for k = 1:100 if PBPgivenflowandgradient(k)== PBPgivengr(k) d= d; else d=d+1; end; end; if d==0 answer = 'Gradient is the cutset of BP and Flow' end; %Area as cutset %Display Result e=0; for k = 1:100 if PBPgivenflowandArea(k)== PBPgivenArea(k) e= e; else e=e+1; end; end; if e==0 answer = 'Area is the cutset of BP and flow' end; %----------------------------------------------------------------------% Determine cutsets between Flow and Gradient %BP as cutset %Display Result d=0; for k = 1:100 if PflowgivengradientandBP(k)== PflowgivenBP(k) d= d; else d=d+1; end; end; if d==0 answer = 'BP is the cutset of Flow and gradient' end; %Area as cutset %Display Result e=0; for k = 1:100

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101Appendix A: (Continued) if PflowgivengradientandArea(k)== PflowgivenArea(k) e= e; else e=e+1; end; end; if e==0; answer = 'Area is the cutset of Flow and gradient' end; % DUAL CUTSETS (cutsets comprising of path having 2 nodes)----------------for k=1:100 Uniqueall(k) = BP(k)*10000000 + Gradient(k)*10000 + Flow(k) + area(k)*0.1; end; for k = 1:100 countuniqueall(k)=0; temp(k)=Uniqueall(k); for l = 1:100 tem(l)=Uniqueall(l); if temp(k)== tem(l) countuniqueall(k) = countuniqueall(k)+1; end; end; end; for k = 1:100 PBPgivenflowgradientarea(k) = countuniqueall(k)/countuniquegrflarea(k); PFlgivenBPgradientarea(k) = count uniqueall(k)/countuniqueBPgrarea(k); PareagivenBPflowgradient(k) = countuniqueall(k)/countuniqueBPgrfl(k); end; % Determine cutsets between BP and Gradient c=0; for k=1:100 if PBPgivenflowgradientarea(k) == PBPgivenflowandArea(k) c=c; else c=c+1; end; end; if c==0 cutset = 'Area,Flow is the cutset of BP and gradient' end; % Determine cutsets between BP and Flow c=0; for k=1:100

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102Appendix A: (Continued) if PBPgivenflowgradientarea(k) ~= PBPgivenAreaandgradient(k) c = c+1; end; end; if c==0 cutset = 'Gradient,Area is the cutset of BP and flow' end; % Determine cutsets between BP and Area c=0; for k=1:100 if PBPgivenflowgradientarea(k) == PBPgivenflowandgradient(k) c=c; else c=c+1; end; end; if c==0 cutset = 'Flow,gradient is the cutset of BP and area' end; % Determine cutsets between Flow and area c=0; for k=1:100 if PFlgivenBPgradientarea(k) == PflowgivengradientandBP(k) c=c; else c=c+1; end; end; if c==0 cutset= 'BP,gradient is the cutset of Flow and Area' end; % Flow and Gr c=0; for k=1:100 if PFlgivenBPgradientarea(k) == PflowgivenBPandArea(k) c=c; else c=c+1; end; end; if c==0 cutset = 'BP,area is the cutset of Flow and gradient' end;

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103Appendix A: (Continued) % Determine cutsets between Area and Gradient c=0; for k=1:100 if PareagivenBPflowgradient(k)==PAreagivenflowandBP(k) c=c; else c=c+1; end; end; if c==0 cutset= 'Flow,BP is the cutset of Area and gradient' end; A4: Program for closing cycle The following section is the only change in all of the programs for closing cycle, rest all of the sections are same. %----------Selecting database for closing cycle---------p=0; for a = 1:length(ar) if ar(a)~=0 p=p+1; end; end;


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Aortic valve analysis and area prediction using bayesian modeling
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[Tampa, Fla.] :
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2005.
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Thesis (M.S.B.E.)--University of South Florida, 2005.
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Includes bibliographical references.
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Text (Electronic thesis) in PDF format.
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ABSTRACT: Aortic Valve Analysis and Area Prediction using Bayesian Modeling Miheer S. Ghotikar ABSTRACT Aortic valve stenosis affects approximately 5 out of every 10,000 people in the United States. [3] This disorder causes decrease in the aortic valve opening area increasing resistance to blood flow. Detection of early stages of valve malfunction is an important area of research to enable new treatments and develop strategies in order to delay degenerative progression. Analysis of relationship between valve properties and hemodynamic factors is critical to develop and validate these strategies. Porcine aortic valves are anatomically analogous to human aortic valves. Fixation agents modify the valves in such a manner to mimic increased leaflet stiffness due to early degeneration. In this study, porcine valves treated with glutaraldehyde, a cross-linking agent and ethanol, a dehydrating agent were used to alter leaflet material properties.The hydraulic performance of ethanol and glutaraldehyde treated valves was compared to fresh valves using a programmable pulse duplicator that could simulate physiological conditions. Hydraulic conditions in the pulse duplicator were modified by varying mean flow rate and mean arterial pressure. Pressure drops across the aortic valve, flow rate and back pressure (mean arterial pressure) values were recorded at successive instants of time. Corresponding values of pressure gradient were measured, while aortic valve opening area was obtained from photographic data. Effects of glutaradehyde cross-linking and ethanol dehydration on the aortic valve area for different hydraulic conditions that emulated hemodynamic physiological conditions were analyzed and it was observed that glutaradehyde and ethanol fixation causes changes in aortic valve opening and closing patterns.Next, relations between material properties, experimental conditions, and hydraulic measures of valve performance were studied using a Bayesian model approach. The primary hypothesis tested in this study was that a Bayesian network could be used to predict dynamic changes in the aortic valve area given the hemodynamic conditions. A Bayesian network encodes probabilistic relationships among variables of interest, also representing causal relationships between temporal antecedents and outcomes. A Learning Bayesian Network was constructed; direct acyclic graphs were drawn in GeNIe 2.0 using an information theory dependency algorithm. Mutual Information was calculated between every set of parameters. Conditional probability tables and cut-sets were obtained from the data with the use of Matlab.A Bayesian model was built for predicting dynamic values of opening and closing area for fresh, ethanol fixed and glutaradehyde fixed aortic valves for a set of hemodynamic conditions. Separate models were made for opening and closing cycles. The models predicted aortic valve area for fresh, ethanol fixed and glutaraldehyde fixed valves. As per the results obtained from the model, it can be concluded that the Bayesian network works successfully with the performance of porcine valves in a pulse duplicator. Further work would include building the Bayesian network with additional parameters and patient data for predicting aortic valve area of patients with progressive stenosis. The important feature would be to predict valve degenration based on valve opening or closing pattern.
590
Adviser: Michael VanAuker, Ph.D.
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Heart valves.
Hemodynamics.
Valvular stenosis.
Porcine valves.
Cross-linking agents.
Pulse duplicator.
Bayesian-learning networks.
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Dissertations, Academic
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Masters.
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t USF Electronic Theses and Dissertations.
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