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Mixing problem with 5 tanks

Mixing problem with 5 tanks . Team2. Contents. - Problem. - Codes / Results. - Conclusion. - Further. Problem. Find the amount of salt in each tank at time t = 5, 10, 15, 20, 30 (min) , and plot the salt content function at each time step t. r (gal/min). t ank 1. t ank 2. t ank 3.

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Mixing problem with 5 tanks

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  1. Mixing problem with 5 tanks Team2

  2. Contents - Problem - Codes / Results - Conclusion - Further

  3. Problem Find the amount of salt in each tank at time t = 5, 10, 15, 20, 30(min), and plot the salt content function at each time step t. r (gal/min) tank 1 tank 2 tank 3 tank 4 tank 5 r r r r r V1 (gal) V2 (gal) V3 (gal) V4 (gal) V5 (gal) With V1 = 20, V2 = 25, V3 = 30, V4 = 35, V5 = 40, r = 10(gal/min), and x1(0) = 10, x2(0) = x3(0) = x4(0) = x5(0) = 0

  4. Problem

  5. Problem where

  6. QR - Code A=[-1/2,0,0,0,0;1/2,-2/5,0,0,0;0,2/5,-1/3,0,0; 0,0,1/3,-2/7,0;0,0,0,2/7,-1/4]; [Q,R]=qr(A); A(:,6:10)=R*Q; k=2; pause while norm(diag(A(:,5*k-4:5*k))-diag(A(:,5*k-9:5*k-5))) /norm(diag(A(:,5*k-4:5*k)))>0.0001 [Q,R]=qr(A(:,5*k-4:5*k)); A(:,5*k+1:5*k+5)=R*Q; k=k+1; end

  7. QR - Result tol

  8. QR - Result tol

  9. QR - Result tol

  10. QR - Result tol

  11. QR - Result tol

  12. Power - Code A=[-1/2,0,0,0,0;1/2,-2/5,0,0,0;0,2/5,-1/3,0,0; 0,0,1/3,-2/7,0;0,0,0,2/7,-1/4]; q=[1,1,1,1,1]'; eg=q'*A*q; z=A*q; q(:,2)=z/norm(z); eg(2)=q(:,2)'*A*q(:,2); E1=norm(q(:,2)-q(:,1))/norm(q(:,2)); E2=(eg(2)-eg(1))/eg(2); k=2; t=0

  13. Power - Code whilet==0 z=A*q(:,k); q(:,k+1)=z/norm(z); eg(k+1)=q(:,k+1)'*A*q(:,k+1); k=k+1; if norm(q(:,k)-q(:,k-1))/norm(q(:,k))<0.0001 break elseif norm(q(:,k)+q(:,k-1))/norm(q(:,k))<0.0001 break end end end

  14. Power - Result tol Conv. rate

  15. Inverse - Code A=[-1/2,0,0,0,0;1/2,-2/5,0,0,0;0,2/5,-1/3,0,0; 0,0,1/3,-2/7,0;0,0,0,2/7,-1/4]; q=[1,1,1,1,1]'; eg=q'*A*q; z=A\q; q(:,2)=z/norm(z); eg(2)=q(:,2)'*A*q(:,2); k=2; t=0

  16. Inverse - Code whilet==0 z=A\q(:,k); q(:,k+1)=z/norm(z); eg(k+1)=q(:,k+1)'*A*q(:,k+1); k=k+1; if norm(q(:,k)-q(:,k-1))/norm(q(:,k))<0.0001 break elseif norm(q(:,k)+q(:,k-1))/norm(q(:,k))<0.0001 break end end end

  17. Inverse - Result tol Conv. rate

  18. Shifted Inverse - Code A=[-1/2,0,0,0,0;1/2,-2/5,0,0,0;0,2/5,-1/3,0,0; 0,0,1/3,-2/7,0;0,0,0,2/7,-1/4]; q=[1,1,1,1,1]'; B=A+0.375*eye(5); eg=q'*A*q; z=B\q; q(:,2)=z/norm(z); eg(2)=q(:,2)'*A*q(:,2); k=2; t=0

  19. Shifted Inverse - Code whilet==0 z=B\q(:,k); q(:,k+1)=z/norm(z); eg(k+1)=q(:,k+1)'*A*q(:,k+1); k=k+1; if norm(q(:,k)-q(:,k-1))/norm(q(:,k))<0.0001 break elseif norm(q(:,k)+q(:,k-1))/norm(q(:,k))<0.0001 break end end end

  20. Shifted Inverse – Result1 sigma=-0.375 tol Conv. rate

  21. Shifted Inverse – Result2 sigma=-0.325 tol Conv. rate

  22. Shifted Inverse – Result3 sigma=-0.2915 tol Conv. rate

  23. Conclusion

  24. Conclusion Tank1

  25. Conclusion Tank2

  26. Conclusion Tank3

  27. Conclusion Tank4

  28. Conclusion Tank5

  29. Further whilet==0 z=B\q(:,k); q(:,k+1)=z/norm(z); eg(k+1)=q(:,k+1)'*A*q(:,k+1); k=k+1; if norm(q(:,k)-q(:,k-1))/norm(q(:,k))<0.0001 break elseif norm(q(:,k)+q(:,k-1))/norm(q(:,k))<0.0001 break end end end

  30. Further whilet==0 z=B\q(:,k); q(:,k+1)=z/norm(z); eg(k+1)=q(:,k+1)'*A*q(:,k+1); B=A-eg(k+1)*eye(5); k=k+1; if norm(q(:,k)-q(:,k-1))/norm(q(:,k))<0.0001 break elseif norm(q(:,k)+q(:,k-1))/norm(q(:,k))<0.0001 break end end end

  31. Further tol = - Shifted Inverse with fixed sigma # of iterations = 21 - Shifted Inverse with updated sigma # of iterations = 9

  32. Q&A

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