Operations Research - The Dual Simplex Method

1. CHAPTER 09 – THE DUAL SIMPLEX METHOD
OPERATIONS RESEARCH

2. PRIMAL SIMPLEX
 Maximize or Minimize Z= 𝑗=1
𝑛
𝑐𝑗 𝑥𝑗
 Subject to 𝑗=1
𝑛
𝑎𝑖𝑗 𝑥𝑗 = 𝑏𝑖, i= 1,2,…,m. 𝑥𝑗 ≥ 0. j= 1,2,…,n

3. RULES OF CONSTRUCTING DUALITY
 For every primal constraint there is a dual variable.
 For every primal variable there is a dual constraint.
 The constraint coefficients of a primal variable from the left-hand side
coefficients of the corresponding dual constraint and its objective function
coefficient of the same variable becomes the right-hand side of the dual
constraints.
 If the primal objective function is maximize (minimize), then the dual problem
is minimize (maximize).
 If the primal problem is maximize (minimize), then the dual constraints must
be >= (<=).
 All dual variables are originally unrestricted.

4. EXAMPLE 1
 Primal
 Min Z= 15x1 + 12x2
 Subject to: x1 + 2x2 >= 3
2x1 - 4x2 <= 5
x1, x2 >= 0

5. EXAMPLE 1
 Standard from
 Min Z= 15x1 + 12x2
 Subject to: x1 + 2x2 – x3 = 3
2x1 - 4x2 + x4 = 5
x1, x2, x3, x4 >= 0

6. EXAMPLE 1
 The dual problem
 Max W = 3y1 + 5y2
 Subject to: y1 + 2y2 <= 15
2y1 – 4y2 <= 12
-y1 + 0y2 <= 0 y1 >= 0
0y1 + y2 <= 0 y2 <= 0, y2 is unrestricted

7. EXAMPLE 2
 Primal
 Max Z= 5x1 + 12x2 +4x3
 Subject to: x1 + 2x2 + x3 <= 10
2x1 - x2 + 3x3 = 8
x1, x2, x3 >= 0

8. EXAMPLE 2
 Standard form
 Max Z= 5x1 + 12x2 +4x3
 Subject to: x1 + 2x2 + x3 + x4 = 10
2x1 - x2 + 3x3 + 0x4 = 8
x1, x2, x3, x4 >= 0

9. EXAMPLE 2
 The dual problem
 Min W= 10y1 + 8y2
 Subject to: y1 + 2y2 >= 5
2y1 - y2 >= 12
y1 + 3y2 >= 4
y1 + 0y2 >= 0 y1 >= 0, y2 is unrestricted

10. EXAMPLE 2 BY TWO PHASE METHOD
 By two phase method
 Min W= 10y1 + 8y2
 Subject to: y1 + 2y2 >= 5
2y1 - y2 >= 12
y1 + 3y2 >= 4
y1 + 0y2 >= 0 y1 >= 0, y2 is unrestricted
 When y is unrestricted its equal to y’-y’’

11. EXAMPLE 2 BY TWO PHASE METHOD
 Chanocial form
 Min W= 10y1 + (8y2’ – 8y2’’)
 Subject to: y1 + (2y2’ – 2y2’’) – y3 + y6 = 5, (y3 is surplus, y6 is artificial)
2y1 – (y2’ – y2’’) – y4 + y7 = 12, (y4 is surplus, y7 is artificial)
y1 + (3y2’ – 3y2’’) – y5 + y8 = 4, (y5 is surplus, y8 is artificial)

12. EXAMPLE 2 BY TWO PHASE METHOD
 Phase 1:
 Min W’= y6 + y7 + y8
 Subject to: y1 + 2y2’ – 2y2’’ – y3 + y6 = 5
2y1 – y2’ + y2’’ – y4 + y7 = 12
y1 + 3y2’ – 3y2’’ – y5 + y8 = 4

13. EXAMPLE 2 BY TWO PHASE METHOD
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14. EXAMPLE 2 BY TWO PHASE METHOD
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15. EXAMPLE 2 BY TWO PHASE METHOD
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16. EXAMPLE 2 BY TWO PHASE METHOD
 Phase 2:
 Min W’= 10y1 + 8y2’ – 8y2’’
 Subject to: y1 + 2y2’ – 2y2’’ – y3 = 5
2y1 – y2’ + y2’’ – y4 = 12
y1 + 3y2’ – 3y2’’ – y5 = 4
 Basic: y5=3/5, y2’’=2/5, y1=29/5, W=54.8

17. EXAMPLE 2 BY TWO PHASE METHOD
-2/52/5-0'y2'-y2'y2
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18. OPTIMAL DUAL SOLUTION
 How to find the optimal solution of the dual problem, when the optimal
solution of the primal problem is known? By 2 methods.
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19. EXAMPLE 3
 Consider this problem
 Max Z= 5x1 + 12x2 + 4x3
 Subject to: x1 + 2x2 + x3 <= 10
2x1 - x2 + 3x3 = 8
x1, x2, x3 >= 0

20. EXAMPLE 3
 Standard form
 Max Z= 5x1 + 12x2 +4x3 – Mx5
 Subject to: x1 + 2x2 + x3 + x4 = 10, x4 is a slack
2x1 - x2 + 3x3 + x5 = 8, x5 is artificial
x1, x2, x3, x4, x5 >= 0

21. EXAMPLE 3
 The dual problem
 Min W= 10y1 + 8y2
 Subject to: y1 + 2y2 >= 5
2y1 - y2 >= 12
y1 + 3y2 >= 4
y1 + 0y2 >= 0 y1 >= 0, y2 is unrestricted
We now show how the optimal dual values are determined using the two methods
described before.

22. EXAMPLE 3
   
   
     29/5,-2/5
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