Vollmer Manufacturing makes three components for sale to refrigeration companies. The components are processed on two machines: a shaper and a grinder. The times (in minutes) required on each machine are as follows: Component Shaper 6 1 2 3 4 Machine 4 Grinder 4 5 2

Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter12: Queueing Models
Section: Chapter Questions
Problem 59P
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Vollmer Manufacturing makes three components for sale to refrigeration companies. The components are processed on two machines: a shaper and a grinder. The times (in minutes) required on
each machine are as follows:
Component
1
2
3
1
2
Shaper
6
4
3
Machine
4
Grinder
4
The shaper is available for 120 hours, and the grinder for 110 hours. No more than 250 units of component 3 can be sold, but up to 1,000 units of component 1 and up to 1,000 units of
component 2 can be sold. In fact, the company already has orders for 550 units of component 1 that must be satisfied. The per unit selling price and per unit variable costs for each of the three
components are as follows:
5
Component Selling price Material Cost Labor Cost
$26
T⠀⠀
$18
$26
2
$12
$8
$13
$5
$4
$5
(a) For each component, calculate the profit margin (profit margin
Component 1: $
600
Component 2: $
700
Component 3: $
200
=
selling price - material cost - labor cost).
Formulate and solve the model to determine the maximum profit. If the constant is "1", it must be entered in the box. If an amount is zero, enter "0".
Let C1
= units of component 1 manufactured
C2 = units of component 2 manufactured
C3 = units of component 3 manufactured
Transcribed Image Text:Vollmer Manufacturing makes three components for sale to refrigeration companies. The components are processed on two machines: a shaper and a grinder. The times (in minutes) required on each machine are as follows: Component 1 2 3 1 2 Shaper 6 4 3 Machine 4 Grinder 4 The shaper is available for 120 hours, and the grinder for 110 hours. No more than 250 units of component 3 can be sold, but up to 1,000 units of component 1 and up to 1,000 units of component 2 can be sold. In fact, the company already has orders for 550 units of component 1 that must be satisfied. The per unit selling price and per unit variable costs for each of the three components are as follows: 5 Component Selling price Material Cost Labor Cost $26 T⠀⠀ $18 $26 2 $12 $8 $13 $5 $4 $5 (a) For each component, calculate the profit margin (profit margin Component 1: $ 600 Component 2: $ 700 Component 3: $ 200 = selling price - material cost - labor cost). Formulate and solve the model to determine the maximum profit. If the constant is "1", it must be entered in the box. If an amount is zero, enter "0". Let C1 = units of component 1 manufactured C2 = units of component 2 manufactured C3 = units of component 3 manufactured
Max $
s.t.
Maximum profit: $
Variable
C1
C2
C3
8 C1
5.00
6 C1
6.00
4 C1
0 C1
0 C₁
No Lower Limit
+
+
C1, C2, C3
6 C2
4 C2
to 9.00
5 C₂
0 C2
✓to 10.00
✓to No Upper Limit
+
Interpret these ranges for company management.
Individual changes in the profit coefficients within
+
+
9 C3
4 C3
2 C3
(b) What are the objective coefficients ranges for the three components? If required, round your answers to two decimal places.
Objective Coefficient Range
0 C3
VI
VI
VI
VI
VI
IV
IV
7200 Shaper
6600
200
1000
1000
600
0
Grinder
Maximum units of component 3
Maximum units of component 1
Maximum units of component 2
Minimum units of component 1
these ranges will not cause a change in the optimal number of components to produce.
Transcribed Image Text:Max $ s.t. Maximum profit: $ Variable C1 C2 C3 8 C1 5.00 6 C1 6.00 4 C1 0 C1 0 C₁ No Lower Limit + + C1, C2, C3 6 C2 4 C2 to 9.00 5 C₂ 0 C2 ✓to 10.00 ✓to No Upper Limit + Interpret these ranges for company management. Individual changes in the profit coefficients within + + 9 C3 4 C3 2 C3 (b) What are the objective coefficients ranges for the three components? If required, round your answers to two decimal places. Objective Coefficient Range 0 C3 VI VI VI VI VI IV IV 7200 Shaper 6600 200 1000 1000 600 0 Grinder Maximum units of component 3 Maximum units of component 1 Maximum units of component 2 Minimum units of component 1 these ranges will not cause a change in the optimal number of components to produce.
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