Question

2 Jackson Hole Manufacturing Jackson Hole Manufacturing is a small manufacturer of plastic products used in the automotive and computer industries. One of its major contracts is with a large computer company and involves the 2 production of plastic printer cases for the computer companys portable printers. The printer cases are produced on two injection molding machines. The M-100 machine has a production capacity of 25 printer cases per hour, and the M-200 machine has a production capacity of 40 cases per hour. Both machines use the same chemical material to produce the printer cases; the M-100 uses 40 pounds of the raw material per hour and the M-200 uses 50 pounds per hour. The computer company asked Jackson Hole to produce as many of the cases during the upcoming week as possible; it will pay $18 for each case Jackson Hole can deliver. However, next week is a regularly scheduled vacation period for most of Jackson Hole's production employees; during this time, annual maintenance is performed for all equipment in the plant. Because of the downtime for maintenance, the M-100 will be available for no more than 15 hours, and the M-200 will be available for no more than 10 hours. However, because of the high setup cost involved with both machines, management requires that each machine must be operated for at least 5 hours. The supplier of the chemical material used in the production process informed Jackson Hole that a maximum of 1000 pounds of the chemical material will be available for next weeks production; the cost for this raw material is $4 per pound used. In addition to the raw material cost, Jackson Hole estimates that the hourly cost of operating the M-100 and the M-200 are $50 and $70, respectively 2.1 Formulate a linear programming model that can be used to maximize the con- tribution to profit 2.2 Find the optimal solution (full hours) 2.3 What is the profit for this production run?

Answer 1

2.1)

Here, Given Data,

Capacity of the Machine M-100 = 25 Printer cases per Hour,

Capacity of the Machine M -200 = 40 printer cases per Hour,

Material Usage for the machine M-100 = 40 Pounds per Hour

Material Usage for the machine M-200 = 50 Pounds per Hour

The Given cost for the raw material = USD 6 per Pound,

X1 = Number of Hours spent on the Machine -100

X2 = Number of Hours spent on the Machine -200,

Again, Total cost = Cost of Raw Material X Number of Pounds Material usage

Or, 6*40*X1 +6*50*X2 + 50*X1 + 75*X2 = 290*X1 + 375*X2

Again Total Revenue = 25*18*X1 + 40*18*X2 = 450*X1 + 720*X2,

Profit would be Total Revenue minus Total Cost

Or, profit = (450*X1 + 720*X2) – (290*X1 + 375*X2) = 160*X1 + 345*X2,

We will be Maximizing Z or Profit Value,

So, Objective Function is Maximizing Z = 160*X1 + 345*X2,

Decision Variables are X1 and X2,

Constraints or subject to:

For Downtime case,

for the Machine M -100, X1=< 15,

for the Machine M -200, X2=< 10,

for high setup cost,

for the machine M -100, X1>= 5,

For the Machine M-200, X2>=5,

Considering Raw Material, Constraint 40*X1 + 50*X2 =< 1000

X1, X2>=0(Non-negativity)

2.2)

The optimum solution,

When X2 = 10, X1 will be 12.5,

Calculation: 40*X1 + 50X2 = 1000, or, 40*X1 + 50*10 = 1000, or, 40*X1 = 500, or, X1 = 12.5,

Here, optimal point is X1 = 12.5 and X2 = 10,

2.3)

Now Maximizing the Profit or Z = 160*X1 + 345*X2 = 160*12.5 + 345*10 = 5450 or USD 5450,

Now let me calculate slack time for each constraints:

If X1 =< 15, then Slack time would be 15 – 12.5 = 2.5 Hours

If X2 =<10, then Slack time would be 10 – 10 = Zero Hours

If X1 >=5, then Slack time would be 15 – 5 = 10 Hours

If X2 >=5, then Slack time would be 10 – 5 = 5 Hours

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