Question 6 options: Hartman Company is trying to determine how much of each of t

ID: 447121 • Letter: Q

Question

Question 6 options:

Hartman Company is trying to determine how much of each of two products should be produced over the coming planning period. The only serious constraints involve labor availability in three departments. Shown below is information concerning labor availability, labor utilization, overtime, and product profitability.

Product 1

Product 2

Regular Hours Available

Overtime Hours Available

Cost of Overtime per Hour

Profit per Unit

Dept A hours/Unit

0.35

107

$17

Dept B hours/Unit

0.3

0.2

$26

Dept C hours/Unit

0.2

0.5

If all production is done in a standard workweek, then Profit per Unit includes the cost to pay for the workforce. But, if overtime is needed in each department, then the Profit Function needs to be reduced by the Cost per Hour of Overtime in Each Department multiplied by the Number of Overtime Hours Used in Each Department. For example, if we used 5 hours of Overtime in Department A, we would need to Subtract $17*5 from our Profit equation.

Setup and Solve the Linear Programming Problem and determine the number of units of Product 1 and Product 2 to produce to Maximize Profit. Add an Additional Constraint to your LP to make sure that ALL of the Variables are INTEGERS

Hint: You will need 5 Decision Variables, 2 of them to determine the production quantities, and 3 of them to determine how much overtime to use in each of the departments.

Product 1

Product 2

Regular Hours Available

Overtime Hours Available

Cost of Overtime per Hour

Profit per Unit

Dept A hours/Unit

0.35

107

$17

Dept B hours/Unit

0.3

0.2

$26

Dept C hours/Unit

0.2

0.5

Explanation / Answer

Let x1 = units of product 1 produced

x2 = units of product 2 produced

Max

30x1

15x2

s.t.

0.35x2

100

Dept. A

0.30x1

0.20x2

Dept. B

0.20x1

0.50x2

Dept. C

x1, x2 ³ 0

Entering = X2, Departing = S1, Key Element = 3
R1(new) = R1(old) / 3 = R1(old) × 1/3
R2(new) = R2(old) - 2 R1(new)
R3(new) = R3(old) - 2 R1(new)

Entering = X3, Departing = S2, Key Element = 5
R2(new) = R2(old) / 5 = R2(old) × 1/5
R1(new) = R1(old)
R3(new) = R3(old) - 4 R2(new)

Entering = X1, Departing = S3, Key Element = 41/15
R3(new) = R3(old) / 41/15 = R3(old) × 15/41
R1(new) = R1(old) - 2/3 R3(new)
R2(new) = R2(old) + 4/15 R3(new)

Since all Cj - Zj 0,
Optimum Solution is arrived with value of variables as :
X1 = 89/41
X2 = 50/41
X3 = 62/41
Maximise Z = 765/41

Max