Question

Consider the linear program max z = 5x, + 3x2 + xz st. x + x₂ + x₂ <6 5xı + 3x2 +6xz S15 X, X2, X, 20 and an associated tableau Z X1 X2 X3 S1 S2 RHS 1 0 0 5 0 1 15 0 0 0.4 -0.2 1 -0.2 3 0 1 0.6 1.2 0 0.2 3 (a) What basic solution does this tableau represent? Is this solution optimal? Why or why not? (b) Does this tableau represent a unique optimum? If not, find an alternative optimal solution.

Answer 1

(a)

Given tableau represents the following basic solution:

x1 = 3 (x1 column has only 1 and 0, its RHS is 3)

s1 = 3 (s1 column has only 1 and 0, its RHS is 3)

x2, x3, s2 = 0

Objective value = 15

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Cj-Zj for x1 column = 5-(0*0+5*1) = 0

Cj-Zj for x2 column = 3-(0*0.4+5*0.6) = 0

Cj-Zj for x3 column = 1-(0*(-0.2)+5*1.2) = -5

Cj-Zj for s1 column = 0-(0*1+5*0) = 0

Cj-Zj for s2 column = 0-(0*(-0.2)+5*0.2) = -1

0 Ratio 1 Cj 5 3 1 0 Xg СВі BV x1 x2 x3 s1 s2 Solution 0 s1 0 0.4 -0.2 -0.2 3 5 x1 1 0.6 0 0.2 3 Zj = 2CBi*aijCj - Zj 0 0 -5 0 -1 15 We see that values of Cj - Zj satisfy the optimality condition Cj - Zj <= 0 1.2

For this solution, Value of Cj-Zj for each column is <= 0

Therefore, this solution is optimal.

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(b)

In the optimal tableau, we see that value of Cj-Zj for x1 and x2 column are equal, i.e. 0, which means the solution is NOT unique.

There is an alternative solution.

Because, Cj-Zj for x2=0, therefore, x2 is the next entering variable.

Ratio for s1 row = 3/0.4 = 7.5

Ratio for x1 row = 3/0.6 = 5

Minimum ratio is 5, corresponding to x1 , so, x1 is the leaving variable.

s1 1 3 Optimal tableau 1 Сj 5 3 1 0 0 Xp СВі BV x1 x2 x3 s2 Solution 0 s1 0 0.4 -0.2 -0.2 5 x1 1 0.6 1.2 0 0.2 3 Zj = ECBi*aij cj - Zj 0 0 -5 0 -1 15 We see that values of Cj - Zj satisfy the optimality condition Cj - Zj <= 0 Therefore, we have reached the optimal solution Ratio 7.5 5 To find the alternative solution, Entering Variable: Corresponding to maximum value of Cj-Zj, entering variable is x2 Leaving Variable: Corresponding to non-negative min-ratio, leaving variable is s3 Ratio 1 1 Optimal tableau 2 Cj 5 1 0 0 XB СВі BV x1 x2 x3 s1 s2 Solution 0 s1 -0.6667 0 -1 -0.3333 3 x2 1.66667 1 2 0 0.33333 5 Cj - Zj 0 0 -5 0 -1 15 We see that values of Cj - Zj satisfy the optimality condition Cj - Zj<= 0 Therefore, we have reached the optimal solution Optimal Solution: x1 = 0 x2 = 5 x3 = 0 Objective Value: Za 15

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Step-wise process - Construct the Initial Simplex tableau Identify Key Column - for maximisation problem, Key Column is the one with the highest value of Cj-Zj Calculate Ratio Ratio is obtained by dividing Solution values by Key column values Identify Key Row - for maximisation problem, Key Row is the one with the least non-negative Ratio Identify key Element - Element lying at the cross section of Key Row and Key Column Construct a new simplex table by the following method. The new table will be same as the initial tableau, except the rows highlighted in yellow. Basic Variable and corresponding CBI value will change as below Incoming Variable - Variable pertaining to Key Column Leaving Variable Variable pertaining to Key Row New Values of aij For Key Row, New Value = Old Value / Key Element For Non-Key Rows, New Value = Old Value - (Corres. Key Row Value * Corres. Key Column Value / Key Element) Check for optimality condition, if condition is satisfied, optimal solution is reached, otherwise proceed for next iteration Optimality condition - for maximisation problem, Cj-Zj so