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Linear Programming (Simplex Min Method Exercise) - SOLUTIONS

1) The simplex minimization method involves converting inequality constraints into equations using slack and artificial variables, with the artificial variables having coefficients much larger than other variables. 2) The initial table is formed using the coefficients of slack and artificial variables, and the optimum column is found by selecting the most negative entry in the Cj-Zj row. 3) The example problem is minimized using this method, converting the constraints to equations and performing simplex iterations until an optimal solution is found with objective value C = 146.

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96 views8 pages

Linear Programming (Simplex Min Method Exercise) - SOLUTIONS

1) The simplex minimization method involves converting inequality constraints into equations using slack and artificial variables, with the artificial variables having coefficients much larger than other variables. 2) The initial table is formed using the coefficients of slack and artificial variables, and the optimum column is found by selecting the most negative entry in the Cj-Zj row. 3) The example problem is minimized using this method, converting the constraints to equations and performing simplex iterations until an optimal solution is found with objective value C = 146.

Uploaded by

Melody Dacuya
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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LP: Simplex Minimization Method

Steps for Solving:


1. The Cj column of the initial table begins with the coefficients of artificial variables and
of slack variables in the objective with positive coefficients in the constraints.
2. Instead of looking for the most positive quantity in Cj-Zj row for the optimum column,
look for the most negative entry.
3. The optimum table or final table has entries in the Cj-Zj row which are either zero or
positive.

Summary of Converting Constraints to Equations in a Minimization Problem


4. Add an artificial variable if the symbol is =
5. Add a slack variable if the symbol is ≤
6. Subtract a slack variable but add an artificial variable if the symbol is ≥

Note: The slack variables have zero coefficients in the objective function while the
artificial variable contributes a value, which is power of ten greater than any
coefficient in the program.
The initial table has entries in the product column, which are the slack and artificial
variables with positive coefficients in the constraints.
Example:
Minimize: C = 20X1 + 10X2 + 80X3
Subject to:
X 1 + X2 + X 3 ≥6
2X1 + 4X2 + X3 = 20
2X1 + X2 ≤5
X1, X2, X3 ≥0

Converting Constraints to Equations


Minimize: C = 20X1 + 10X2 + 80X3 + 0S1 + 0S2 + 100A1 + 100A2
Subject to:
X 1 + X2 + X 3 - S 1 + A1 = 6
2X1 + 4X2 + X3 + A2 = 20
2X1 + X2 + S2 =5
X1 , X2 , X3 , S1 , S2 , A1 , A2 ≥0
Cj 20 10 80 0 0 100 100
Prod Qty X1 X2 X3 S1 S2 A1 A2
0 S2 6 1 1 1 -1 0 1 0 6/ 1 = 6

100 A1 20 2 4 1 0 0 0 1 20 / 4 = 5

100 A2 5 2 1 0 0 1 0 0 5/1=5

Zj 2500 400 500 100 0 100 0 100

Cj - Zj -380 -490 -20 0 -100 100 0


Cj 20 10 80 0 0 100 100
Prod Qty X1 X2 X3 S1 S2 A1 A2
0 S2 1 ½ 0 ¾ -1 0 1 -¼ 1/ ½ = 2

10 X2 5 ½ 1 ¼ 0 0 0 ¼ 5 / ½ = 10

100 A2 0 3/2 0 -¼ 0 1 0 -¼

Zj 50 155 10 -45/2 0 100 0 -45/2

Cj - Zj -135 0 205/2 0 -100 100 245/2

NR2 = OR2/4 = 20 2 4 1 0 0 0 1
4
=5 ½ 1 ¼ 0 0 0 ¼

NR1 = OR1-NR2 = 6 1 1 1 -1 0 1 0
5 ½ 1 ¼ 0 0 0 ¼
1 ½ 0 ¾ -1 0 1 -¼

NR3 = OR3-NR2 = 5 2 1 0 0 1 0 0
5 ½ 1 ¼ 0 0 0 ¼
0 3/2 0 -¼ 0 1 0 -¼
Cj 20 10 80 0 0 100 100
Prod Qty X1 X2 X3 S1 S2 A1 A2
20 X1 2 1 0 3/2 -2 0 2 -½ 2/2 = 1

10 X2 4 0 1 -½ 1 0 -1 ½

100 A2 3 0 0 5/2 -3 -1 3 -½ 3/3 = 1

Zj 380 20 10 275 -330 -100 330 -55

Cj - Zj 0 0 -195 330 100 -230 45

NR1 = 2OR1 = 2 ( 1 ½ 0 ¾ -1 0 1 -¼)


= 2 1 0 3/2 -2 0 2 -½
NR2 = OR2-OR1 or ½ NR1 = 5 ½ 1 ¼ 0 0 0 ¼
1 ½ 0 ¾ -1 0 1 -¼
4 0 1 -½ 1 0 -1 ½

NR3 = OR3-3/2NR1 = 0 3/2 0 -¼ 0 1 0 -¼


3 3/2 0 9/4 -3 0 3 -¾
-(-3 0 0 -5/2 3 1 -3 ½)
= 3 0 0 5/2 -3 -1 3 -½
Cj 20 10 80 0 0 100 100
Prod Qty X1 X2 X3 S1 S2 A1 A2
20 X1 0 1 0 -1/6 0 2/3 0 -1/6

10 X2 5 0 1 1/3 0 -1/3 0 1/3 5 / 1/3 = 15

100 A1 1 0 0 5/6 -1 -1/3 1 -1/6 1 / 5/6 = 6/5

Zj 150 20 10 250/3 -100 -90 100 -70/3

Cj - Zj 0 0 -10/3 100 90 0 230/3

NR3 = OR3 / 3 = 3 0 0 5/2 -3 -1 3 -½


3
=1 0 0 5/6 -1 -1/3 1 -1/6

NR1 = OR1-2NR3 = 2 1 0 3/2 -2 0 2 -½


2 0 0 5/3 -2 -2/3 2 -1/3
0 1 0 -1/6 0 2/3 0 -1/6

NR2 = OR2+NR3 = 4 0 1 -½ 1 0 -1 ½
1 0 0 5/6 -1 -1/3 1 -1/6
5 0 1 1/3 0 -1/3 0 1/3
Cj 20 10 80 0 0 100 100
Prod Qty X1 X2 X3 S1 S2 A1 A2
20 X1 1/5 1 0 0 -1/5 11/15 1/5 -1/5

10 X2 23/5 0 1 0 2/5 -1/5 -2/5 2/5

80 X3 6/5 0 0 1 -6/5 -2/5 6/5 -1/5

Zj 146 20 10 80 -96 -58/3 96 -16

Cj - Zj 0 0 0 96 58/3 4 84

NR3 = 6/5 x OR3 = 6 (1 0 0 5/6 -1 -1/3 1 -1/6)


5
= 6/5 0 0 1 -6/5 -2/5 6/5 -1/5

NR1 = OR1+1/6NR3 = 0 1 0 -1/6 0 2/3 0 -1/6


1/5 0 0 1/6 -1/5 -1/15 1/5 -1/30
1/5 1 0 0 -1/5 11/15 1/5 -1/5

NR2 = OR2-1/3xNR3 = 5 0 1 1/3 0 -1/3 0 1/3


6/15 0 0 1/3 -6/15 -2/15 6/15 -1/15
23/5 0 1 0 2/5 -1/5 -2/5 2/5
Conclusion
Minimum value of C = 146 which is obtained
by using X1 = 1/5, X2 = 23/5, X3 = 6/5

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