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.
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.
|Constraints||Air conditioners||Large fan||Availability|
|Hiring||3 hours||2 hours||240 hours|
|Drilling||2 hours||1 hours||140 hours|
Using the corner point graphical method:
Simplify the constraints by dividing them with a common factor. This now becomes.
X1 and X2 is 60 and 40 respectively. Therefore, the company should 60 air conditioners and 40 fans to realize maximum profit of 2100.
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.
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.
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.
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.
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
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
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
Optimality has been achieved
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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