LINEAR PROGRAMMING PROBLEMS Harvard Case Solution & Analysis

Linear Programming Problems Case Solution  

Products Profit per item Cutting hour Finishing hour Packaging hour Demand Minimum Production
Running 30 2 0.5 1.5 300 100
Hiking 40 3 1.5 0.125   100
Casual 20 1 0.5 1.5 300 100

 

Limitations Per month
Cutting hour 1000
Finishing hour 600
Packaging hour 500

P5

Limitations

Limitations
Plastic 72
Labor 80
Machine 60

Goal 1: The company should use the available plastic completely

Product Cost Units Produced Profit Per toy Total Profits Plastic Labor Machine
Truck Toy 2,560 2 500 1,000 12 20 20
Car Toy 4,125 8 500 3,750 60 60 30
4,750 72 80 50

Goal 2: The company should not underutilized labor and machine hours available

Product Cost Units Produced Profit Per toy Total Profits Plastic Labor Machine
Truck Toy 4,440 4 500 2,000 24 40 40
Car Toy 3,000 5 500 2,500 40 40 20
4,500 64 80 60

Goal 3: The company should minimize production cost

Product Cost Units Produced Profit Per toy Total Profits Plastic Labor Machine
Truck Toy 4,440 4 500 2,000 24 40 40
Car Toy 3,000 5 500 2,500 40 40 20
7,440 4,500 64 80 60

Decision

The company should focus on its first goal because it is maximizing the company’s overall profit to 4750 dollars.

P5

  • The economic order quantity has been found by using the excel solver and by using the EOQ formula for each appliance.
    1. The optimal inventory level for each appliance by using the EOQ formula and excel solver is presented below.
Products EOQ EOQ by Formula
Microwave               57               81
Range               30               42
Washer               35               49
Dryer               35               49
Dishwasher               48               68
  1. The Objective function is

Products Total Cost Total Cost by formula
Microwave         1,715         1,819
Range         5,916         6,275
Washer         6,928         7,348
Dryer         6,928         7,348
Dishwasher         3,847         4,080
Total       25,334       26,871
  • After the incorporation of the discount factor in the model, the profit has been maximized by using the excel solver which is presented below.

Products Demand SP HC OC Storage space EOQ Total capacity Total Cost Cost of purchases Total Profit
Microwave 700 300 15 70 3 57 171 1,715 150 103,285
Range 500 2,000 100 175 18 30 532 5,916 1,100 444,084
Washer 600 2,000 100 200 20 35 693 6,928 1,100 533,072
Dryer 600 2,000 100 200 22 35 762 6,928 1,100 533,072
Dishwasher 500 800 40 185 18 48 866 3,847 400 196,153
3,024 25,334 1,809,666
5,000

By analyzing the answer report which can be seen in the attached excel workbook, it has been identified that there is enough capacity to store the raw material along with the enough capacity to purchase at a time too much raw material for the production.

P7

  • The problem has been addressed as the simple linear problem where the goal is to minimize the total cost and all of the constraints has been addressed according to the five goal of the company. The simple linear problem has located in the excel workbook.
  • The deviational variables are the three-room types which are constrained with the at least expansion of 5, 10 and 15 rooms for small, medium and large rooms construction respectively.
  • The problem has been solved with solver and found that company should expand 28 small rooms, 10 medium rooms and 15 large rooms.
Room Type Area per room Cost per room Needed to expand Total Area Total Cost
Small          1,400          20,000               28          39,200              560,000
Medium          1,800          30,000               10          18,000              300,000
Large          2,200          40,000               15          33,000              600,000
Total                90,200          1,460,000
Constrained                90,000          2,000,000

P7

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