Question

Consider the following LP problem developed at Zafar Malik’s Carbondale, Illinois, optical scanning firm:

Maximize profit: = $1x1 + $1x2

Subject to: $2x1 + $1x2 ≤ 100

$1x1 + $2x2 ≤ 100

a. What is the optimal solution to this problem?

b. If a technical breakthrough occurred that raised the profit per unit of X₁ to $3, would this affect the optimal solution?

c. Instead of an increase in the profit coefficient X₁ to $3, suppose that profit was overestimated and should only have been $1.25. Does this change the original optimal solution?

d. Based on your answer to c., are there any unused resources?

e. Based on your answer to c. (at $1.25 for variable x1), how much would profit increase in constraint # 2 if 75 was added to the RHS?   

Answer 1

We solve the given problem in Excel using Excel Solver as shown below:

Variables x1 x2 33.3333 33.3333 Objective p $ 66.67 Constraints Condition 100 2 100 <= 100 100

The above table in the form of formulas is shown below for better understanding and reference:

B C D A 1 Variables 2 x1 3 x2 Solver Parameters 33.3333333333333 33.3333333333333 mtoto Set Objective: SBS6 To: Max =1*B2+1*B3 O Min 0 O Value of: 5 Objective 6 p 7 8 Constraints 91 10 2 Condition By Changing Variable Cells: $B$2:$B$3 =2*B2+1*B3 =1*B2+2*B3 100 100 Subject to the Constraints: $B$9:$B$10 <= $D$9:$D$10 Add Change Delete Reset All Load/Save Make Unconstrained variables Non-Negative Select a Solving Method: Simplex LP Options Solving Method Select the GRG Nonlinear engine for Solver Problems that are smooth nonlinear. Select the LP Simplex engine for linear Solver Problems, and select the Evolutionary engine for Solver problems that are non-smooth Help Solve Close

As seen from above, the optimal solution is x1 = 33.33 & x2 = 33.33

b. We will re-solve the problem with revised values as shown below:

Variables x150 x2 0 Objective p U $ 150.00 Constraints 1 2 100 50 Condition <= <= 100 100

Hence, the optimal solution will change as shown above, x1 = 50, x2 = 0, Total profit = $150

c. We will re-solve the problem with revised values as shown below:

Variables x1 x2 33.33 33.33 Objective P $ 75.00 Constraints 1 2 100 100 Condition <= T <= 100 100

Hence, the optimal solution will change as shown above, x1 = 33.33, x2 = 33.33 will remain same, Total profit = $75

d. Based on the above solution, there are no unused constraints as seen above since LHS = RHS from both constraints.

e. We will re-solve the problem with revised values as shown below:

Variables x1 X2 8.33 83.33 Objective p $ 93.75 Constraints 1 Condition <= 100 100 175 175

Increase in profit between part c and part e = 93.75 - 75 = $18.75

_______________________________________________________________________________________

In case of any doubt, please ask through the comment section before Upvote/downvote.

Kindly rate the answer as it will be encouraging for me to keep answering further questions!!!