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Lecture 25. Two-Phase Simple Upper Bounded Simplex Algorithm Example: Minimize â€“2x 1 â€“ 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 â€“2x 1 â€“ x 2

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## Lecture 25

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**Lecture 25**• Two-Phase Simple Upper Bounded Simplex Algorithm • Example: • Minimize –2x1 – x2 • Subject to x1 + x2< 6 • 0 < x1< 5 • 0 < x2< 5**The Graph**x2 bound x2 x1 bound optimum x1**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**Add An Artificial Variable**• Minimize –2x1 – x2 • Subject to x1 + x2 +x3 + A= 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3< • 0 < A <**Phase I Problem**• Minimize A • Subject to x1 + x2 +x3 + A= 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3< • 0 < A <**Step 0. Initialization**• BASIC = {A at 6} • NONBASIC = {x1 at 0, x2 at 0, x3 at 0} • B = 1 • B-1 = 1**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!**Step 2. Optimality**• Not optimal • Let p = 3 - that is x3 is allowed to increase from 0**Step 3. Direction**• y = B-1a3 = 1(1) = 1 • = 1 Why?**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**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**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!**Step 2. Optimality**• Optimal For Phase I**Phase II Problem**• Minimize -2x1 – x2 • Subject to x1 + x2 +x3 = 6 • 0 < x1< 5 • 0 < x2< 5 • 0 < x3<**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!**Step 2. Optimality**• Not Optimal • Let p = 1**Step 3. Direction**• y = B-1a1 = 1(1) = 1 • = 1 Why?**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**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.**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**The Graph**x2 bound x2 x1 bound current point x1**Step 2. Optimality**• Not Optimal • Let p = 2**Step 3. Direction**• y = B-1a1 = 1(1) = 1 • = 1 Why?**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**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**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**The Graph**x2 bound x2 x1 bound current point x1

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