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Simplex method : Tableau Form
Maximise Z= 10X1+9X2+0S1+0S2+0S3+0S4
subject to:
7/10X1+1X2+1S1 =630
1/2X1+5/6X2+ +1S2 =600
1X1+2/3X2 +1S3 =708
1/10X1+1/4X2 +1S4 =135
X1,X2,X3,X4,S1,S2,S3,S4 > 0
Constraints equations make more variables (six) than equations (four), the simplex method find solutions for these equations by assigning zero values to variables
A basic solution can be either feasible or infeasible. A basic feasible solution is a solution which is both basic and satisfy non negativity condition.
Simplex Form
cj = objective function coefficient for variables j
bj = right hand side coefficient for constraints i
aij = Coefficient associated with variable j in constraint i c1c2 …………………………………………….cn
--------------------------------------------------------------------------------------
a11 a12 ……………………………………………… a1n b1
a21 a22 ………………………………………………..a2n b2
………………………………………………………………………….
………………………………………………………………………….
am1 am2 amn bm
C(row) = row of objective function coefficient
b(column)= column of right hand side value of the constraint equation
A(matrix) = m-row and n-column of coefficient of the variables in the constraint equations
Simplex Tableau
X1 X2 S1 S2 S3 S4 bj/aij
Basis Cj 10 9 0 0 0 0
S1 0 7/10 1 1 0 0 0 630
S2 0 ½ 5/6 0 1 0 0 600
S3 0 1 2/3 0 0 1 0 708
S4 0 1/10 ¼ 0 0 0 1 135
Profit Zj
0 0 0 0 0 0 0
Cj-Zj 10 9 0 0 0 0
Improving the Solution
Zj represents the decrease in profit that will result if one unit of X1 is brought into the solution
Cj-Zj - net evaluation row
Values in Zj row is calculated by multiplying the element in the cj column by
the corresponding elements in the column of the A matrix and summing
them.
The process of moving from one basic feasible solution to another is called iteration.
We select the entering variable that has the highest coefficient in the net evaluation row.
Criteria of moving a variable from the current basic:
For each row i compute the ratio bj/aij> 0. This ratio tells us the maximum amount of the variable Xj that can be brought into the solution and still satisfy the constraint equation represented by that row.
Select the basic variable corresponding to the minimum of these ratios as the variable to leave.
Optimality
Optimality Condition: The optimal solution to a LP problem has been reached when there are no positive values in the net evaluation row of the simplex tableau.
Pivot Row : Replace the leaving variable In the basic column with the entering
variable. New pivot row = current row divided by pivot element
All other rows including Z :
New row : Current row – (pivot column element) X ( New pivot row)
X1 X2 S1 S2 S3 S4 bj/aij
Basis Cj 10 9 0 0 0 0
S1 0 0 16/30 1 0 -7/10 0 134.4
S2 0 0 ½ 0 1 -1/2 0 246
X1 10 1 2/3 0 0 1 0 708
S4 0 0 22/120
0 0 -1/10 1 64.2
Profit Zj
10 20/3 0 0 10 0 7080
Cj-Zj 0 7/3 0 0 -10 0
X1 X2 S1 S2 S3 S4 bj/aij
Basis Cj 10 9 0 0 0 0
X2 9 0 1 30/16 0 -21/16 0 252
S2 0 0 0 -15/16 1 5/32 0 120
X1 10 1 0 -20/16 0 50/16 0 540
S4 0 0 0 -11/32 0 9/64 1 18
Profit Zj
10 9 70/16 0 111/16 0 7668
Cj-Zj 0 0 -70/16 0 -111/16
0
Condition for Optimal Solution
Cj-Zj< 0 for both of the non-basic variables S1 & S2, an attempt to bring non- basic variables into basic will lower the current value of the objective function. The optimal solution of a LP problem has been reached when there are no positive values at the net evaluation row.
X1 540X2 252S1 0S2 120S3 0S4 18
Sample Problem
Max. 4X1+6X2+3X3+1X4
s.t 3/2X1+2X2+4X3+3X4<550
4X1+1X2+2X3+1X4<700
2X1+3X2+1X3+2X4<200
X1, X2, X3, X4 >0
Problem in standard form:
Max> 4X1+6X2+3X3+1X4+0S1+0S2+0S3
s t 3/2X1+2X2+4X3+3X4+1S1 = 550
4X1+1X2+2X3+1X4 +1S2 = 700
2X1+3X2+1X3+2X4 +1S3 = 200
X1, X2, X3, X4, S1, S2, S3 > 0
Initial Simplex Tableau
X1 X2 X3 X4 S1 S2 S3
Basis
Cj 4 6 3 1 0 0 0
S1 0 3/2 2 4 3 1 0 0 550
S2 0 4 1 2 1 0 1 0 700
S3 0 2 3 1 2 0 0 1 200
Zj 0 0 0 0 0 0 0 0Cj-Zj 4 6 3 1 0 0 0
Iteration :1…
X1 X2 X3 X4 S1 S2 S3
Basis Cj 4 6 3 1 0 0 0
S1 0 1/6 0 10/3 5/3 1 0 -2/3 416(2/3)
S2 0 10/3 0 5/3 1/3 0 1 -1/3 633(1/3)
x2 6 2/3 1 1/3 2/3 0 0 1/3 66(2/3)
Zj 12/3 6 6/3 12/3 0 0 6/3 400
Cj-Zj 0 0 3/3 -9/3 0 0 -6/3
Iteration 2….
x1 x2 x3 x4 S1 S2 S3
Basis
Cj 4 6 3 1 0 0 0
x3 3 3/60 0 1 5/10 3/10 0 -2/10 125
S2 0 39/12 0 0 -15/30
-5/10 1 0 425
x2 6 39/60 1 0 15/30 -1/10 0 12/30 25
Zj 81/20 6 3 9/2 3/10 0 54/30 525Cj-Zj -1/20 0 0 -7/2 -3/10 0 -54/30
Treating Constraints Eliminating negative right hand side:
The number of std bag had to be less or
X1 < X2-25 equal to the number of deluxe bag after 25
1X1 - 1X2 < -25 bags had been set aside for displaying
-X1 + X2 > 25 purpose
Greater than equal to constraint :
6X1 + 3 X2 -4X3 > -20
-6X1 – 3X2 + 4 X3 < 20
Simplex ProblemMax. 10X1 + 9 X2
s t 7/10X1+1X2 < 6301/2X1+5/6X2 < 6001X1+2/3X2 < 7081/10X1+1/4X2 <1351X1 >100 a5, a6 (artificial variables)
1X2 >100 very large cost -M Max. 10X1+9X2+ 0S1+S2+0S3+0S4+0S5+0S6-Ma5-Ma6
s t. 7/10X1+1X2+1S1 = 6301/2X1+5/6X2 +1S2 =6001X1+2/3X2 +1S3 =7081/10X1+1/4X2 +1S4 =1351X1 -1S5+1a5 =100 1X2 -S6+1a6 =100
X1 X2 S1 S2 S3 S4 S5 S6 a5 a6Basis
Cj 10 9 0 0 0 0 0 0 -M -M
S1 0 7/10 1 1 0 0 0 0 0 0 0 630
S2 0 1/2 5/6 0 1 0 0 0 0 0 0 600
S3 0 1 2/3 0 0 1 0 0 0 0 0 708
S4 0 1/10 ¼ 0 0 0 1 0 0 0 0 135
a5 -M 1 0 0 0 0 0 -1 0 1 0 100
a6 -M 0 1 0 0 0 0 0 -1 0 1 100
Zj
Cj-Zj
-M
10+M
-M
9+M
0
0
0
0
0
0
0
0
M
-M
M
-M
-M
0
-M
0
-200M
X1 X2 S1 S2 S3 S4 S5 S6 a5 a6Basis Cj 10 9 0 0 0 0 0 0 -M -M
S1 0 0 1 1 0 0 0 7/10 0 -7/10 0 560
S2 0 0 5/6 0 1 0 0 1/2 0 -1/2 0 550
S3 0 1 2/3 0 0 1 0 1 0 -1 0 608
S4 0 0 ¼ 0 0 0 1 1/10 0 -1/10 0 125
X1 10 1 0 0 0 0 0 -1 0 1 0 100
a6 -M 0 1 0 0 0 0 0 -1 0 1 100
ZjCj-Zj
10
0
-M
9+M
0
0
0
0
0
0
0
0
-10
-10
M
-M
10-M-10
-M
0
1000-
100M
X1 X2 S1 S2 S3 S4
S5 S6 a5 a6
Basis Cj 10 9 0 0 0 0 0 0 -M -M
S6 0 0 0 30/16 0-210/160 0 0 1 0 -1 152
S2 0 0 0 -15/16
125/160
0 0 0 0 0 120
S3 0 0 0 -20/16 0 300/1
60 0 1 0 -1 0 440
S4 0 0 0 -11/32 0
45/320
1 0 0 0 0 18
X1 10 1 0 -20/16 0 300/1
60 0 0 0 0 0 540
X2 9 0 1 30/16 0
-210/160
0 0 0 0 0 252
ZjCj-Zj
10
0
9
9+M
70/16
-70/16
0
0
111/16
-111/16
0
0
0
0
0
0
0
-M
0
-M
7668
Equality Constraints
Max. 6X1+3X2+4X3+1X4s t -2X1-1/2X2+1X3+-6X4 = -60 1X1 +1X3+2/3X4<20
-1X2-5X3 <-50X1, X2, X3, X4> 0
Max. 6X1+3X2+4X3+1X4+0S2+0S3-Ma1-Ma2st 2X1+1/2X2-1X3+6X4 +1a1 = 60
1X1 +1X3-2/3X4 +1S1 = 20 1X2+5X3 -1S3 +1a5= 50X1, X2, X3, X4, S2, S3, a1, a3 > 0
Sensitivity Analysis
Study of how optimal solution and value of optimal solution to a LP given a changes in the various coefficients of the problem:
1. What effect on the optimal solution will a change in a coefficient of the objective function (Cj) have?
2. What effect on the optimal solution will a change in constraint in right hand side value (bj) have?
3. What effect on the optimal solution will a change in a coefficient of a constraint equation (aij) have?