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Case study of Global Terrain Method

Case study of Global Terrain Method. Jeonghwa Moon and Andreas A. Linninger 02/01/2006 Laboratory for Product and Process Design , Department of Chemical Engineering, University of Illinois, Chicago, IL 60607, U.S.A. INTRODUCTION.

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Case study of Global Terrain Method

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  1. Case studyof Global Terrain Method Jeonghwa Moon and Andreas A. Linninger 02/01/2006 Laboratory for Product and Process Design, Department of Chemical Engineering, University of Illinois, Chicago, IL 60607, U.S.A.

  2. INTRODUCTION • Basic concept of Global Terrain Method (A Lucia and Y.Feng,2002) • Intelligent, reliable and efficient method for finding all physically meaningful solutions and singular points • Based on intelligent movement along the valleys and ridges • The task : tracing out ‘connected’ lines in a multi dimensional space • Mathematical background • The backbone function of Global terrain method • Initial movement • It can be calculated from M or H using Lanzcos or some other eigenvalue-eigenvector technique (Sridhar and Lucia) • Direction • Downhill: Eigendirection of negative Eigenvalue • Uphill: Eigendirection of positive Eigenvalue V = {opt gTg such that FTF = L, for all L єL}----(Eq.1) F: a vector function , g:2JTF , J: Jacobian matrix

  3. Case study 1

  4. Problem definition • Equations • Feasible region • Starting point • (1,2.3) 3D space of case 1

  5. Previous Result (12/01/2005) Result of previous version S4 S3 S1 We couldn’t find S2 and S5 S5 S2

  6. Current Result (02/01/2006) S3 S4 2 S1 4 3 S5 1 S2

  7. v1 λ1 -v1 v2 λ2 -v2 no Newton-like Method yes What is updated?(02/01/06) • Exploration direction is expanded • This expansion increase possibility to find solution! • But it also increase unnecessary exploration! • The magnitude of initial uphill movement is changed! Modified algorithm for initial uphill movement If initial step is too small it cannot be escaped from saddle point area and newton-like method is not performed properly!!! Saddle point Alpha=0.005(too small) Alpha=5(too big)

  8. Result of Case1 ############################################################ Start new :k=4, starting point is (-0.8181,2.1590) eigenvalue is (3.8710,30.4831) 1. Direction is (-0.8399,-0.5427)meet boundary 2. Direction is (0.8399,0.5427) point found (0.5295,2.1891) duplicated with S3! 3. Direction is (-0.5427,0.8399) point found (0.5320,2.1880) duplicated with S3! 4. Direction is (0.5427,-0.8399) point found (0.5319,2.1880) duplicated with S3! ############################################################ Start new :k=5, starting point is (-0.2369,-0.8328) eigenvalue is (-19.6756,-2.8165) 1. Direction is (-0.9761,0.2175)found point (negative eig) is (-0.8181,2.1590) duplicated with S4! 2. Direction is (0.9761,-0.2175)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 3. Direction is (-0.2175,-0.9761)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 4. Direction is (0.2175,0.9761)found point (negative eig) is (-0.8181,2.1590) duplicated with S4! ############################################################ Start new :k=1, starting point is (1.5959,0.9521) eigenvalue is (2.5184,46.8559) 1. Direction is (0.3394,-0.9406) , point found (-0.5734,-2.2449) (S2) 2. Direction is (-0.3394,0.9406) , point found (0.5610,2.1741) (S3) 3. Direction is (-0.9406,-0.3394) , point found (0.5607,2.1742) duplicated with S3! 4. Direction is (0.9406,0.3394) , point found (0.5605,2.1743) duplicated with S3! ############################################################ Start new :k=2, starting point is (-0.5734,-2.2449) eigenvalue is (-1.0893,29.6883) 1. Direction is (-0.8948,0.4465)found point (negative eig) is (-0.8181,2.1590)(S4) 2. Direction is (0.8948,-0.4465)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 3. Direction is (0.4465,0.8948)near saddle, point found (-0.5734,-2.2448) duplicated with S2! 4. Direction is (-0.4465,-0.8948)meet boundary ############################################################ Start new :k=3, starting point is (0.5610,2.1741) eigenvalue is (-2.3162,27.3930) 1. Direction is (-0.8976,0.4409) found point (negative eig) is (-0.8181,2.1590) duplicated with S4! 2. Direction is (0.8976,-0.4409 )found point (negative eig) is (1.5959,0.9521) duplicated with S1! 3. Direction is (0.4409,0.8976) point found (0.5746,2.2789) duplicated with S3! 4. Direction is (-0.4409,-0.8976) point found (-0.2369,-0.8328)(S5) MIN MIN SAD MAX SAD

  9. Maximum point Saddle point Minimum point Connection diagram S4 S3 S1 S5 S2 Connection Diagram of case 1

  10. Case study 2

  11. Problem definition • Equations • Feasible region • Starting point • (1.1,1.12)

  12. Result of case2

  13. Result of case2 3 1 S4 S2 S1 5 2 S3 4 S6 S8 S5 7 6 S7

  14. Eigendirection of case2

  15. Result of Case2 Result table

  16. Maximum point Saddle point Minimum point Connection Diagram • From maximum point • It cannot find saddle point. • From some minimum points • It cannot find saddle point S2 S1 S4 S3 ? ? S5 S6 S8 ? S7

  17. Future work • Study background knowledge concerned with GTM • condition of saddle point • Newton like method • Etc. • Try another way to find maximum point • Finding maximum point of f(x) is same as finding minimum point of –f(x) • Solve the case which has more than 2 variables • The algorithm for selecting good eigenvalues is needed! • Use numerical method to Get Jaccobian and Hessian matrix • Consider about performance (DIFFICULT) • method for reducing steps • How to avoid the solution duplication. • Case study 3 • Catalytic Pellet Reactor (Celia Xue ,2005 REU)

  18. Case Study3 –Future WorkCatalytic Pellet Reactor3

  19. Overview • Packed bed configuration • Spherical Pellets as Catalysts • 2nd Order reaction • A2B • Concentration Profile The mission is finding all of solutions of Concentration Profile using Global Terrain Method!

  20. Problem Definition • Collocation method (i=1..n) …(2) …(1) X1 X2 … Xn …(3) Insert (1) & (2) into (3) Collocation method is used for approximation N variables, n equations!!! (eqn 5) In this case, n=5(multivariable) will be used …(4) …(5)

  21. END!!!

  22. S1:1.5959 0.9521 S2:0.5601 2.1745 S3:-0.2370 -0.8332 S4:-0.8181 2.1590 S2-1 S4-2 S1-4 S4 Initial point S2-3 S4-3 S2 S4-1 S1-2 S2-2 S1 S3-4 S4-4 S2-4 S3-1 S1-3 S3-2 S3-3 S3 S1-1

  23. IP to S1 Initial point S1

  24. S1-1 Initial point S1

  25. S1-2 Initial point S2 S1 S1-1

  26. S1-1 S1-3 Initial point S2 S1 S3 S1-1

  27. S1-4 Initial point S2 S1 S3 S1-1

  28. S2-1 S4 Initial point S2 S1 S3

  29. S2-2 S4 Initial point S2 S1 S3

  30. S2-3 S4 Initial point S2 S1 S3

  31. S2-4 S4 Initial point S2 S1 S3

  32. S3-1 S4 Initial point S2 S1 S3

  33. S3-2 S4 Initial point S2 S1 S3

  34. S3-3 S4 Initial point S2 S1 S3

  35. S3-4 S4 Initial point S2 S1 S3

  36. S4-1 S4 Initial point S2 S1 S3

  37. S4-2 S4 Initial point S2 S1 S3

  38. S4-3 S4 Initial point S2 S1 S3

  39. S4-4 S4 Initial point S2 S1 S3

  40. Problem Definition • Case 3-1 (i=1,2,3) X1 X2 X3 (0.0) (0.5) (1.0) Collocation method

  41. Maximum point Saddle point Minimum point Result of Case1 B B S4 B S2 S3 S1 B B Result table Connection Diagram

  42. Maximum point Saddle point Minimum point Result of Case1 ############################################################ Start new :k=1, starting point is (1.5959,0.9521) eigenvalue is (2.5184,46.8559) 1. Direction is (0.3394,-0.9406)near saddle, point found (-0.5734,-2.2449) 2. Direction is (-0.3394,0.9406)near saddle, point found (0.5610,2.1741) 3. Direction is (-0.9406,-0.3394)near saddle, point found (0.5607,2.1742) duplicated with S3! 4. Direction is (0.9406,0.3394)near saddle, point found (0.5605,2.1743) duplicated with S3! ############################################################ Start new :k=2, starting point is (-0.5734,-2.2449) eigenvalue is (-1.0893,29.6883) 1. Direction is (-0.8948,0.4465)found point (negative eig) is (-0.8181,2.1590) 2. Direction is (0.8948,-0.4465)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 3. Direction is (0.4465,0.8948)near saddle, point found (-0.5734,-2.2448) duplicated with S2! 4. Direction is (-0.4465,-0.8948)meet boundary ############################################################ Start new :k=3, starting point is (0.5610,2.1741) eigenvalue is (-2.3162,27.3930) 1. Direction is (-0.8976,0.4409)found point (negative eig) is (-0.8181,2.1590) duplicated with S4! 2. Direction is (0.8976,-0.4409)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 3. Direction is (0.4409,0.8976)near saddle, point found (0.5746,2.2789) duplicated with S3! 4. Direction is (-0.4409,-0.8976)near saddle, point found (-0.2369,-0.8328) ############################################################ Start new :k=4, starting point is (-0.8181,2.1590) eigenvalue is (3.8710,30.4831) 1. Direction is (-0.8399,-0.5427)meet boundary 2. Direction is (0.8399,0.5427)near saddle, point found (0.5295,2.1891) duplicated with S3! 3. Direction is (-0.5427,0.8399)near saddle, point found (0.5320,2.1880) duplicated with S3! 4. Direction is (0.5427,-0.8399)near saddle, point found (0.5319,2.1880) duplicated with S3! ############################################################ Start new :k=5, starting point is (-0.2369,-0.8328) eigenvalue is (-19.6756,-2.8165) 1. Direction is (-0.9761,0.2175)found point (negative eig) is (-0.8181,2.1590) duplicated with S4! 2. Direction is (0.9761,-0.2175)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 3. Direction is (-0.2175,-0.9761)found point (negative eig) is (1.5959,0.9521) duplicated with S1! 4. Direction is (0.2175,0.9761)found point (negative eig) is (-0.8181,2.1590) duplicated with S4! B B S4 B S2 S3 S1 B B Result table Connection Diagram

  43. Maximum point Saddle point Minimum point Result of Case1 B B S4 B S2 S3 S1 B B Result table Connection Diagram

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