The correct second tableau and third tableau and some of their calculations follow. The optimal solution, given in the third tableau, is X1=18,X2=4,S1=0,S2=0, and profit =$190.
Stpng 1 and 2 To go from the first to the second tableau, we note that the pivot column (in the first tableau) is X1, which has the highest Cj−Zj value, $9. The pivot row is S1, since 40/2 is less than 30/1, and the pivot number is 2.
Step 3 The new X1 row is found by dividing each number in the old S1 row by the pivot number—that is, 2/2=1,1/2=0.5,1/2=0.5,0/2=0, and 40/2=20.
Step 4 The new values for the S2 row are computed as follows:
This solution is not optimal, and you must perform stpng 1 to 5 again. The new pivot column is X2, the new pivot row is S2, and 2.5 (circled in the second tableau) is the new pivot number.
The final solution is X1=18,X2=4, and profit =$190.
Solved Problem M7-3 Use the final simplex tableau in Solved Problem M7-2 to answer the following questions.
What are the shadow prices for the two constraints?
Perform RHS ranging for constraint 1.
If the right-hand side of constraint 1 were increased by 10, what would the maximum possible profit be? Give the values for all the variables.
Find the range of optimality for the profit on X1.
Solution
Shadow price =−(Cj−Zj)
For constraint 1, shadow price =−(−4)=4.
For constraint 2, shadow price =−(−1)=1.
For constraint 1, we use the S1 column.
Quantity
S1
Ratio
18
0.6
18/(0.6)=30
4
−0.2
4/(−0.2)=−20
The smallest positive ratio is 30, so we may reduce the right-hand side of constraint 1 by 30 units (for a lower bound of 40−30=10). Similarly, the negative ratio of −20 tells us that we may increase the right-hand side of constraint 1 by 20 units (for an upper bound of 40+20=60).
The maximum possible profit= Original profit + 10(shadow price)= 190 + 10(4) = 230
The values for the basic variables are found using the original quantities and the substitution rates:
Original Quantity
S1
New Quantity
18
0.6
18+(0.6)(10)=24
4
−0.2
4+(−0.2)(10)=2
X1 = 24, X2 = 2, S1 = 0, S2 = 0 (both slack variables remain nonbasic variables)profit = 9(24) + 7(2) = 230 (which was also found using the shadow price)
For this solution to remain optimal, the Cj−Zj values must remain negative or zero.
−4 −(0.6)Δ−4−20/3≤≤≤0(0.6)ΔΔ
and
−1 + (0.2)Δ(0.2)ΔΔ≤≤≤015
So the change in profit (Δ) must be between −20/3 and 5. The original profit was 9, so this solution remains optimal as long as the profit on X1 is between 2.33=9−20/3 and 14=9+5.