Project Question 7-14

The Electrocomp Corporation manufactures two electric products: air conditioners and large fans. The assembly process for each is similar in that both require a certain among of wiring and drilling. Each air conditioners takes 3hours of wiring and drilling. Each air conditioner takes 3 hours of wiring and 2 hours of drilling. Each fan must go through 2 hours of wiring and 1 hour of drilling. During the next production period, 240 hours of wiring time are available and up to 140 hours of drilling time may be used. Each air conditioner sold yields a profit $25. Each fan assembled may be sold for a $15 profit. Formulate and solve this LP production mix situation to find the best combination of air conditioners and fans that yields the highest profit.


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Let X1 and X2 be the decision variables denoting air conditioners and large fans produced by the company respectively.

Step: Formulation of the L.P model.

ConstraintsAir conditionersLarge fanAvailability
Hiring3 hours2 hours240 hours
Drilling2 hours1 hours140 hours

Step 2

Using the corner point graphical method:

            Simplify the constraints by dividing them with a common factor. This now becomes.

The pointsZ=25x1+15x2Total
A(0, 0)0
B(70, 0)1750
C(60, 40)2100

            X1 and X2 is 60 and 40 respectively. Therefore, the company should 60 air conditioners and 40 fans to realize maximum profit of 2100.

            Problem 7-15

Electrocamp’s management realizes that it forgot to include two critical constraints. In particular, management decides that there should be a minimum number of air conditioners produced in order to fulfil a contract. Also, due to an oversupply of fans in the preceding period, a limit should be placed on the total number of fans produced.

  1. If Electrocamps decides that at least 20 air conditioners be produced but not more than 80 fans should be produced, what would be the optimal solution? How much slack or Surplus is there for each of the four constraints?
  2. If Electrocamps decides that at least 30 air conditioners should be produced but no more than 50 fans shold be produced, what should be the optimal solution? How much slack or surplus is there for each of the four constraints at the optimal solution?


Step: Standard form

Step 2: initial simplex tableau


 Since it is a minimization problem we pick the most negative Cj-Zj. Entering X1, departing S1 and pivot 2.

Step 3: New simplex tableau


            The solution is optimal but not feasible. Pick the most negative Cj-Zj. Entering variable X2, departing R1 and pivot ½.

            Step 4: New simplex tableau


            Optimality has been reached.

X1=80 and X2=60

Max Z=2900 is the optimal solution of the question.

  1. Introducing new constraints

X1≥20 and X2≤80

Take the value of X1 and X2, then substitutes to the new constraints.

80≥20 and 60≤80 which is true.

Conclusion: The current solution will still be optimal since the optimal solution is not affected.

  • Introducing

X1≥30 and X2≤50

Fixing the values of X1 and X2 to the new constraints to be added we yield:

80≥30 true and 60≤50 which is false.

Problem 7-18

            The dean of the western college of Business must plan the school course offering for the fall semester. Students make it necessary to offer at least 30 undergraduates and 20 graduates in the term. Faculty contracts also dictate that at least 60 courses be offered in total. Each undergraduate course taught costs the college and average $2500 in faculty wages, and each graduate course costs $3000. How many graduates and undergraduates should be taught in the fall so that total faculty salaries are kept to a minimum?


Let X1 and X2 be the decision variable denoting undergraduates and graduates courses offered in the college.

Step 1: Formulate the problem




X1, X2≥0

minZ =2500X1+300X2

Step 2: Standard form




Min Z= 2500X1 +3000X2+0S1+0S2+MR1+MR2+MR3

            Using the Big M-Method the initial tableau will be


            Since it is a minimizing problem we pick the most negative Cj-Zj

Entering variable=X2

Departing variable=R2


Step 4: New optimal solution


            The solution is not optimal and feasible. We pick most negative Cj-Zj. Entering=X1, departing R1 and pivot=1

Step 5: New optimal solution


            The solution is optimal but not feasible. Pick the most negative Cj-Zj




Step 6:

 Zj250030000-2500160, 000

            Optimality has been achieved

Feasible solution




Min Z=$160, 000

Conclusion: There should be 40 undergraduates’ course and 20 graduate courses to be offered in the college.

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