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Lecture 25

Lecture 25. Two-Phase Simple Upper Bounded Simplex Algorithm Example: Minimize –2x 1 – x 2 Subject to x 1 + x 2 < 6 0 < x 1 < 5 0 < x 2 < 5. The Graph. x 2 bound. x 2. x 1 bound. optimum. x 1. Convert To An Equality Constraint. Minimize –2x 1 – x 2

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Lecture 25

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  1. Lecture 25 • Two-Phase Simple Upper Bounded Simplex Algorithm • Example: • Minimize –2x1 – x2 • Subject to x1 + x2< 6 • 0 < x1< 5 • 0 < x2< 5

  2. The Graph x2 bound x2 x1 bound optimum x1

  3. Convert To An Equality Constraint • Minimize –2x1 – x2 • Subject to x1 + x2 +x3 = 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3< Note Bound on the slack variable

  4. Add An Artificial Variable • Minimize –2x1 – x2 • Subject to x1 + x2 +x3 + A= 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3< • 0 < A <

  5. Phase I Problem • Minimize A • Subject to x1 + x2 +x3 + A= 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3< • 0 < A <

  6. Step 0. Initialization • BASIC = {A at 6} • NONBASIC = {x1 at 0, x2 at 0, x3 at 0} • B = 1 • B-1 = 1

  7. Step 1. Pricing • v = cBB-1 = 1(1) = 1 • cbar1 = c1 – va1 = 0 – 1(1) = -1 • cbar2 = c2 – va2 = 0 – 1(1) = -1 • cbar3 = c3 – va3 = 0 – 1(1) = -1 • All price favorably!

  8. Step 2. Optimality • Not optimal • Let p = 3 - that is x3 is allowed to increase from 0

  9. Step 3. Direction • y = B-1a3 = 1(1) = 1 •  = 1 Why?

  10. Step 4 Step Size • - = min{, (xj-ℓj)/(yj) : yj > 0} • + = min{, (uj-xj)/(-yj) : yj < 0} •  = min{- , + , up - ℓp} • - = min{, (6-0)/1} = 6 • + = min{} =  •  = min{6, , -0} = 6

  11. Step 5. New Point • XB = xB - y = 6 – 6(1)(1) = 0 • A = 0 • xp = x3 = ℓ3 +  = 0 + 6 = 6 • Current Point: [x1,x2,x3,A] = [0,0,6,0] • BASIC = {x3 at 6} • NONBASIC = {x1 at 0, x2 at 0, A at 0} • B = 1 B-1 = 1

  12. Step 1. Pricing • v = cBB-1 = 0(1) = 0 • cbar1 = c1 – va1 = 0 – 0(1) = 0 • cbar2 = c2 – va2 = 0 – 0(1) = 0 • All price unfavorably!

  13. Step 2. Optimality • Optimal For Phase I

  14. Phase II Problem • Minimize -2x1 – x2 • Subject to x1 + x2 +x3 = 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3<

  15. Step 1. Pricing • Current Point: [x1,x2,x3] = [0,0,6] • BASIC = {x3 at 6} • NONBASIC = {x1 at 0, x2 at 0} • B = 1 B-1 = 1 • v = cBB-1 = 0(1) = 0 • cbar1 = c1 – va1 = -2 – 0(1) = -2 • cbar2 = c2 – va2 = -1 – 0(1) = -1 • All price favorably!

  16. Step 2. Optimality • Not Optimal • Let p = 1

  17. Step 3. Direction • y = B-1a1 = 1(1) = 1 •  = 1 Why?

  18. Step 4 Step Size • - = min{, (xj-ℓj)/(yj) : yj > 0} • + = min{, (uj-xj)/(-yj) : yj < 0} •  = min{- , + , up - ℓp} • - = min{, (6-0)/1} = 6 • + = min{} =  •  = min{6, , 5-0} = 5

  19. Step 5. New Point • XB = xB - y = 6 – 5(1)(1) = 1 • x3 = 1 • xp = x1 = ℓ1 +  = 0 + 5 = 5 • Current Point: [x1,x2,x3] = [5,0,1] • BASIC = {x3 at 1} • NONBASIC = {x1 at 5, x2 at 0} • B = 1 B-1 = 1 Note: The basis stayed the same.

  20. Step 1. Pricing • v = cBB-1 = 0(1) = 0 • cbar1 = c1 – va1 = -2 – 0(1) = -2 Unfavorable Why? • cbar2 = c2 – va2 = -1 – 0(1) = -1 Favorable

  21. The Graph x2 bound x2 x1 bound current point x1

  22. Step 2. Optimality • Not Optimal • Let p = 2

  23. Step 3. Direction • y = B-1a1 = 1(1) = 1 •  = 1 Why?

  24. Step 4 Step Size • - = min{, (xj-ℓj)/(yj) : yj > 0} • + = min{, (uj-xj)/(-yj) : yj < 0} •  = min{- , + , up - ℓp} • - = min{, (1-0)/1} = 1 • + = min{} =  •  = min{1, , 5-0} = 1

  25. Step 5. New Point • XB = xB - y = 1 – 1(1)(1) = 0 • x3 = 0 • xp = x2 = ℓ2 +  = 0 + 1 = 1 • Current Point: [x1,x2,x3] = [5,1,0] • BASIC = {x2 at 1} • NONBASIC = {x1 at 5, x3 at 0} • B = 1 B-1 = 1

  26. Step 1. Pricing • v = cBB-1 = -1(1) = -1 • cbar1 = c1 – va1 = -2 – (-1)(1) = -1 Unfavorable Why? • cbar3 = c3 – va3 = 0 – (-1)(1) = 1 Unfavorable Why? • Optimality Obtained

  27. The Graph x2 bound x2 x1 bound current point x1

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