chap 3 simplex
TRANSCRIPT
CHAPTER 3: SIMPLEX METHOD
Introduction:
In the simplex method, the model is put into the form of table,
& then a number of mathematical steps are performed on the
table (simplex tableau).
The simplex method moves from one better solution to another
until the best one is found, & then it stops.
Example of simplex method:
Max Z = 40 X1 + 50X2
s.t X1 + 2X2 ≤ 40 (labor)
4X1 + 3X2 ≤ 120 (clay)
X1, X2 ≥ 0
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Initial tableau:
Basic Quantity 40 50 0 0Cj Variables (Zj) X1 X2 S1 S2
0 S1 40 1 2 1 00 S2 120 4 3 0 1
Zj 0 0 0 0 0Cj - Zj 40 50 0 0
Basic Quantity 40 50 0 0Cj Variables (Zj) X1 X2 S1 S2
50 X2 20 0.5 1 0.5 00 S2 60 2.5 0 -1.5 1
Zj 1000 25 50 25 0Cj - Zj 15 0 -25 0
Optimal tableau:
Basic Quantity 40 50 0 0Cj Variables (Zj) X1 X2 S1 S2
50 X2 8 0 1 0.8 -0.240 X1 24 1 0 -0.6 0.4
Zj 1360 40 50 16 6Cj - Zj 0 0 -16 -6
Solution:
X1 = 24, X2 = 8
S1 = 0, S2 = 0
Z = 1360
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Steps:
1. Transform the LP model into standard form.
2. Set up the initial tableau for the basic feasible solution at
the origin & compute the zj & cj – zj (max) @ zj – cj (min)
value.
3. Determine the pivot column by selecting the column with
the highest positive value in the cj – zj @ zj – cj row.
4. Determine the pivot row by dividing the quantity column
values by the pivot column values & select the row with
the minimum nonnegative quotient.
5. Compute the new pivot row values using the formula:
new tableau pivot row values =
old tableau pivot row valuespivot number
6. Compute all other row values using the formula:
7. Compute the new zj & cj – zj @ zj – cj row.
New tableau row values
Old tableau row values
Corresponding coefficient in pivot column
Corresponding new tableau in pivot row value
= X
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8. Determine whether the new solution is optimal by
checking the cj – zj @ zj – cj row. If all the values are 0 or
negative, the solution is optimal. If not, return to step 3 &
repeat the simplex steps.
Converting the Model into Standard Form
Standard form requires that all constraints be in the form of
equations rather than inequalities.
A set of rules for transforming all 3 types of model constraints:
Constraint Adjustment Objective function coefficientMaximization Minimization
≤ (+) a slack variable
0 0
= (+) an artificial variable
- M M
≥ (-) a surplus variable
0 0
(+) an artificial variable
- M M
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Artificial variable (Ai) does not have a meaning as slack variable
or surplus variable does.
It is inserted into the equation simply to give a positive solution
at the origin.
M is a very large positive number.
Example 1: Maximize Z = 400x1 + 200x2
s.t x1 + x2 = 30
2x1 + 8x2 ≥ 80
x1 ≤ 20
x1, x2 ≥ 0
Standard form:
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Example 2: Minimize Z = 400x1 + 200x2
s.t x1 + x2 = 30
2x1 + 8x2 ≥ 80
x1 ≤ - 20
x1, x2 ≥ 0
Standard form:
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Example 3:
Z = 3X1 + 5X2 + 10X3 + 0S1 + 0S2 – MA1 – MA2
s.t X1 – X3 + A1 = 50
6X2 + S1 = 380
- 40X2 + 60X3 – S2 + A2 = 300
X1, X2, X3, S1, S2, A1, A2 ≥ 0
LP model:
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Exercises 1:
Transform the following LP model into standard form model.
1. Max Z = 3X1 + 6X2 + 5X3
s.t 3X1 + 2X2 ≤ 18
X1 + X2 ≥ 5
X1 + X3 ≤ 10
X1, X2, X3 ≥ 0
2. Max Z = 8X1 + 7X2
s.t 10X1 + 8X2 = 40
6X1 + 14X2 ≤ 48
X2/X1 ≥ 7/8
X1, X2 ≥ 0
3. Min Z = 4X1 + 8X2
s.t -6X1 + 5X2 ≥ -16
7X1 + 15X2 ≥ 18
X1, X2 ≥ 0
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Exercises 2:
Transform the following standard form model into LP model.
1. Z = 70X1 + 50X2 + 0S1 + MA1 + MA2 + MA3
s.t 5X1 + 3X2 + A1 = 300
4X1 – S1 + A2 = 40
7X2 + A3 = 500
X1, X2, S1, A1, A2, A3 ≥ 0
2. Z = 150X1 + 200X2 + 180X3 + 0S1 + 0S2 + MA1
s.t X1 + S1 = 40
-7X1 – 3X2 + 5X3 + A1 = 350
X2 + X3 + S2 = 100
X1, X2, X3, S1, S2, A1 ≥ 0
3. Z = 4X1 + 8X2 + 0S1 + 0S2 – MA1 – MA2
s.t -6X1 + 5X2 – S1 + A1 = 16
7X1 + 15X2 – S2 + A2 = 18
X1, X2, S1, S2, A1, A1 ≥ 0
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Simplex Method For ≤ Constraints
Example:
Max Z = 5X1 + 10X2 + 15X3
s.t X1 + X2 ≤ 7
X2 + X3 ≤ 10
2X1 + 2X2 + 2X3 ≤ 30
X1, X2, X3 ≥ 0
Standard form:
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Initial tableau:
Cj BasicVar.
Quantity
Zj
Cj BasicVar.
Quantity
Zj
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Cj BasicVar.
Quantity
Zj
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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Simplex Method For ≥ Constraints
Example:
Min Z = 6X1 + 3X2
s.t 2X1 + 4X2 ≥ 16
4X1 + 3X2 ≥ 24
X1, X2 ≥ 0
Standard form:
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Initial tableau:
Cj BasicVar.
Quantity
Zj
Cj BasicVar.
Quantity
Zj
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Cj BasicVar.
Quantity
Zj
Cj BasicVar.
Quantity
Zj
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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Simplex Method For Mixed Constraints ( ≥, =, ≤ )
Example:
Max Z = 400X1 + 200X2
s.t X1 + X2 = 30
2X1 + 8X2 ≥ 80
X1 ≤ 20
X1, X2 ≥ 0
Standard form:
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Initial tableau:
Cj BasicVar.
Quantity
Zj
Cj BasicVar.
Quantity
Zj
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Cj BasicVar.
Quantity
Zj
Cj BasicVar.
Quantity
Zj
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
Corresponding New tableau
Old row - Coefficient in X Pivot = New tableau
Column Value Pivot column Row value Row value
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The basic feasible solution in the initial simplex tableau is the
origin where all the decision variables equal zero.
The quantity column values are the solution values for the
variables in the basic feasible solution.
The cj values are the contribution to profit (or cost) for each
variable.
The zj row values are computed by multiplying the cj column
values by the variable column values and summing.
The variable with the largest positive cj – zj @ zj – cj value is the
entering variable (pivot column).
The leaving variable is determined by dividing the quantity
values by the pivot column values & selecting the minimum
possible value or zero (pivot row).
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The pivot number is the number at the intersection of the pivot
column & row.
The solution is optimal when all cj – zj @ zj – cj ≤ 0.
Artificial variables are assigned a large cost in the objective
function to eliminate them from the final solution.
Once an artificial variable is selected as the leaving variable, it
will never reenter the tableau, so it can be eliminated.
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Irregular types of LP Problems
1. Multiple Optimal Solutions
Alternate optimal solutions have the same Z value but
different variables values.
Example:
Max Z=40 X1+30 X2
s .t :
X1+2 X2≤404 X1+3 X2≤120X1 , X2≥0
The optimal tableau for this problem is as follows:
Basic Quantity 40 30 0 0Cj Variables (Zj) X1 X2 S1 S2
0 S1 10 0 5/4 1 -1/440 X1 30 1 ¾ 0 ¼
Zj 1200 40 30 0 10Cj - Zj 0 0 0 -10
To get the alternate solution, proceed the tableau i.e. X2 enter
the tableau and X1 leave.
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2. An Infeasible Problem
An infeasible problem has an artificial variable in the final
simplex tableau.
Example:
Basic Quantity 5 3 0 0 0 -M -MCj Var. X1 X2 S1 S2 S3 A1 A2
3 X2 4 2 1 ½ 0 0 0 0-M A1 4 1 0 0 -1 0 1 0-M A2 2 -2 0 -1/2 0 -1 0 1
Zj 12-6M 6 + M 3 3/2 +
M/2
M M -M -M
Cj – Zj -1-M 0 -3/2 –
M/2
-M -M 0 0
3. An Unbounded Problem
A pivot row cannot be selected for an unbounded problem.
Basic 4 2 0 0 MCj Variables Quantit
yX1 X2 S1 S2 A1
M A1 4 1 0 -1 0 10 S2 2 0 1 0 1 0
Zj 4M M 0 -M 0 MCj – Zj 4-M 2 M 0 0
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4. Tie For the Pivot Column
Sometimes when selecting the pivot column, the greatest
positive cj – zj (or zj – cj) row values are the same.
5. Tie For the Pivot Row
Example:
Basic Quantity 4 6 0 0 0Cj Var. X1 X2 S1 S2 S3
0 S1 12 6 0 1 -4 06 X2 3 0 1 0 1 00 S3 10 5 0 0 -10 1
Zj 18 0 6 0 6 0Cj – Zj 4 0 0 -6 0
6. Negative Quantity Values
Standard form for simplex solution requires positive (right-
hand-side) RHS values.
Example:
-6x1 + 2x2 ≥ -30
(-1) (-6x1 + 2x2 ≥ -30)
6x1 - 2x2 ≤ 30
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