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Budi Harsanto blogs.unpad.ac.id/budiharsanto
2012
Dual Problem
Course : Quantitative Method / Operations Research
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Dual
Every primal have dual form.
Dual form give us useful economics information.
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Dual Procedure
If PRIMAL objective maximization, DUAL objective become
minimization. Vice versa.
RHS PRIMAL, become DUAL objective function coefficent (OFC).
OFC PRIMAL, become DUAL RHS.
PRIMAL constraint coefficient transpose become DUAL constraint
coefficient.
Adjust the inequality.
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1st Example
PRIMAL Obj: Max profit Z= 14X1 + 5X2 Subject to: 6X1+7X2 <
1000 3X1+ 8X2 < 800
DUAL Obj: Min opportunity cost Z= 1000U1 + 800U2 Subject to:
6U1+3U2 > 14 7U1+8U2 > 5
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2nd Example
PRIMAL Objective: Maximize profit Z= 30X1 + 80X2 Subject to:
2X1+4X2 < 1000 6X1+2X2 < 1200 X2 < 200
DUAL Objective: Minimize opportunity cost Z= 1000U1 + 1200U2 +
200U3 Subject to: 2U1+6U2+0U3 > 30 4U1+2U2+U3 > 80
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Dual Solving Procedure
like primal.
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1st Exercise
PRIMAL Objective: Maximize profit Z= 50X1 + 120 X2 Subject to:
2X1+4X2 < 80 3X1+ X2 < 60
DUAL Objective: Minimize opportunity cost Z= 80U1 + 60U2 Subject
to: 2U1+3U2 > 50 4U1+ U2 > 120
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Dual Equation
Objective: Minimize opportunity cost
Z= 80U1+60U2+0S1+0S2+MA1+MA2
Subject to:
2U1+3U2- S1+A1 = 50
4U1+U2-S2+A2 = 120
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Dual Simplex Equation
Objective: Minimize opportunity cost
Z= 80U1+60U2+0S1+0S2+MA1+MA2
Subject to:
2U1+3U2- S1+A1 = 50
4U1+U2-S2+A2 = 120
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1st Iteration
C 80 60 0 0 M M
Solution Mix U1 U2 S1 S2 A1 A2 Quantity
M A1 2 3 -1 0 1 0 50
M A2 4 1 0 -1 0 1 120
Z 6M 4M -M -M M M 170M
C-Z 80-6M 60-4M M M 0 0 -
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2nd Iteration
C 80 60 0 0 M M
Solution Mix U1 U2 S1 S2 A1 A2 Quantity
80 U1 1 3/2 -1/2 0 1/2 0 25
M A2 0 -5 2 -1 -2 1 20
Z 80 120-5M -40+2M -M 40-2M M 2000+20M
C-Z 0 5M-60 -2M+40 M 3M-40 0 -
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3rd Iteration
C 80 60 0 0 M M
Solution Mix U1 U2 S1 S2 A1 A2 Quantity
80 U1 1 1/4 0 -1/4 0 1/4 30
0 S1 0 -5/2 1 -1/2 -1 1/2 10
Z 80 20 0 -20 0 20 2400
C-Z 0 40 0 20 M M-20 -
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Dual Result
Same as primal, with different
visualization form.
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C 80 60 0 0 M M
Solution Mix U1 U2 S1 S2 A1 A2 Quantity
80 U1 1 1/4 0 -1/4 0 1/2 30
0 S1 0 -5/2 1 -1/2 -1 1/2 10
Z 80 20 0 -20 0 20 2400
C-Z 0 40 0 20 M M-20 -
DUAL optimal solution
C 50 120 0 0
Solution Mix X1 X2 S1 S2 Quantity
X2 1/2 1 1/4 0 20
S2 5/2 0 -1/4 1 40
Z 60 120 30 0 2400
C-Z -10 0 -30 0 -
PRIMAL optimal solution
U1 ~ 1st Resources
