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QAS 19 Assignment 2 1
QAS Decision Analysis
Note:
• You should do additional exercises from the textbook to improve your understanding of the material.
• Show your work to get credit. See the course outline for a description of how assignments should be done.
• The assignment should be typed.
Excel Solver can be used to solve the linear programming models in problems 1, 2, 3 and 4.
19. Part a
Max. 3A + 4B
s.t. –1A + 2B + S1 = 8
1A + 2B + S2 = 12
2A + 1B + S3 = 16
A, B, S1, S2, S3 ≥ 0
NOTE: 19. Part b follows 19. Part c
19. Part c
−1� + 2� + �� = 8 ; 20 3 ,
8 3� ⟹ −
20 3 + 2 ∙
8 3 + �� = 8 ⟹ 3 �−
20 3 +
16 3 + �� = 8� ⟹
−20 + 16 + 3�� = 3 ∙ 8 ⟹ −4 + 3�� = 24 ⟹ 3�� = 28 ⟹ �� = �� �
1� + 2� + �� = 12 ; 20 3 ,
8 3� ⟹
20 3 + 2 ∙
8 3 + �� = 12 ⟹ 3 �
20 3 +
16 3 + �� = 12� ⟹
20 + 16 + 3�� = 36 ⟹ 36 + 3�� = 36 ⟹ 3�� = 0 ⟹ �� = �
QAS 19 Assignment 2 2
19. Part c (Continued)
2� + 1� + �� = 16 ; 20 3 ,
8 3� ⟹ 2 ∙
20 3 +
8 3 + �� = 16 ⟹ 3 �
40 3 +
8 3 + �� = 16� ⟹
40 + 8 + 3�� = 48 ⟹ 48 + 3�� = 48 ⟹ 3�� = 0 ⟹ �� = �
19. Part b
−2 1� + 2� = 12! 2� + 1� = 16 −2� − 4� = −24 −3� = −8 ⟹ −3�−3 =
−8 −3 ⟹ � =
8 3 ⟹ 2� + 1� = 16 ⟹ 2� + 1 ∙
8 3 = 16 ⟹ 3 �2� +
8 3 = 16� ⟹
6� + 8 = 48 ⟹ 6� = 40 ⟹ 6�6 = 40 6 ⟹ � =
20 3 ⟹
20 3 ,
8 3� ⟹ 3� + 4� = 3 ∙
20 3 + 4 ∙
8 3 =
60 3 +
32 3 =
"� �
QAS 19 Assignment 2 3
Let S be the number of units of the stock fund.
Let M be the number of units of the money market fund.
Fund Cost/Unit ARR Risk/Unit
S.F. $50 10% 8
M.M.F. $100 4% 3
Objective: Minimize risk subject to…
Let D be the maximum dollars Innis has been authorized to invest = $1,200,000; D ≤ 1,200,000. However, D = 50S + 100M; therefore, 50S + 100M ≤ 1,200,000
Let I be the minimum annual income the investor requires = $60,000; I ≥ 60,000. However, I = 0.10x50S + 0.04x100M; therefore, 0.10x50S + 0.04x100M ≥ 60,000 or 5S + 4M ≥ 60,000
Let T be the minimum amount the investor wants invested in the M.M.F. = $300,000; T ≥ 300,000. However, T = 100M; therefore: 100M ≥ 300,000
Part a & b
Min. 8S + 3M
s.t. 50S + 100M ≤ 1,200,000
5S + 4M ≥ 60,000
100M ≥ 300,000
S, M ≥ 0
Units of Stock Fund: 4,000
Units of Money Market Fund: 10,000
Annual Income: $60,000
QAS 19 Assignment 2 4
Part c
Max. 5S + 4M
s.t. 50S + 100M ≤ 1,200,000
5S + 4M ≥ 60,000
100M ≥ 300,000
S, M ≥ 0
Let R be the number of gallons of regular gasoline.
Let P be the number of gallons of premium gasoline.
Gasoline Type
Profit/ Gallon
Gallons of Gr.
A Crude Oil
Regular $0.30 0.3
Premium $0.50 0.6
Objective: Maximize total profit; i.e., 0.30R + 0.50P
Grade A crude oil ≤ 18,000 gallons; i.e., 0.30R + 0.60P ≤ 18,000
Gallons of regular gasoline plus gallons of premium gasoline ≤ 50,000; i.e., R + P ≤ 50,000
Demand for premium gasoline ≤ 20,000 gallons; i.e., P ≤ 20,000
Units of Stock Fund: 18,000
Units of Money Market Fund: 3,000
Annual Income: $102,000
QAS 19 Assignment 2 5
Parts a & b
Max. 0.30R + 0.50P
s.t. 0.30R + 0.60P ≤ 18,000
R + P ≤ 50,000
P ≤ 20,000
R, P ≥ 0
Check
Part c
Max. 0.30R + 0.50P
s.t. 0.30R + 0.60P + S1 = 18,000
R + P + S2 = 50,000
P + S3 = 20,000
R, P, S1, S2, S3 ≥ 0
Part d
Optimal Solution
Gallons of Regular Gasoline: 40,000
Gallons of Premium Gasoline: 10,000
Total Profit Contribution: $17,000
S1, the slack variable for Grade A Crude Oil, is 0, which means all of it will be consumed.
S2, the slack variable for Production Capacity, is 0, which means all Production Capacity will be utilized.
S3, the slack variable for Premium Gasoline Demand, is 10,000, which means we will not be able to supply 100% of our distributors’ demand if that demand exceed P = 10,000 gallons.
The binding constraints are those constraints whose slack variables are zero; namely, the availability of Grade A Crude Oil and Manufacturing Capacity.
QAS 19 Assignment 2 6
Let R be the number of Regular Customer contacts per two-week period
Let N be the number of New Customer contacts per two-week period
Customer Revenue/Hour Hours/Contact
Regular $25 50/60
New $8 1
Let T be the amount of available technician time during a two-week period, i.e., 80 hours; therefore: T ≤ 80. However, T = R x 50/60 + N x 1; therefore, (5/6)R + 1N ≤ 80
Let D be the minimum amount of revenue generated by a technician during a two-week period, i.e., $800; therefore: D ≥ 800. However, D = R x 50/60 x 25 + N x 1 x 8 = (125/6)R + 8N; therefore, (125/6)R + 8N ≥ 800
Time spent on New Customers must be at least 60% of the time spent on Regular Customers; therefore, N x 1 ≥ 0.6 x R x (50/60) or 1N – R x (6/10) x (50/60) ≥ 0 or 1N – R x (1/1) x (5/10) ≥ 0 or –(1/2)R + 1N ≥ 0
Objective: Maximize the total number of customers contacted during a two-week period; in other words…
Objective: Maximize R + N
Parts a & b
Max. R + N
s.t. # $ R + 1N ≤ 80
��# $ R + 8N ≥ 800
– � � R + 1N ≥ 0
R, N ≥ 0
Optimal Solution
Regular Customer Contacts / 2 wks: 60
New Customer Contacts / 2 wks: 30
FYI… Total Revenue: $1,490
QAS 19 Assignment 2 7
Check
QAS 19 Assignment 2 8
5. Refer to the model and graph shown below. Answer questions (a) to (f).
Max –3X + 6Y s.t. 6X – 2Y ≤ 3 –2X + 3Y ≤ 6 X + Y ≥ 3 X , Y ≥ 0
Point A (0.5, 0) Point B (1.5, 3) a. Identify the feasible region.
_The dark purple region below._
b. Which point is optimal?
_Point “B”_
c. What is the optimal solution? What is the optimal objective function value?
3 −2% + 3& = 6! 6% − 2& = 3 −6% + 9& = 18 7& = 21 ⟹ & = 3 ⟹ 6% − 2 ∙ 3 = 3 ⟹ 6% − 6 = 3 ⟹ 6% = 9 ⟹ % = 96 ⟹ % =
3 2 ⟹
)*+,-./ �0/1+,02: �� , ��
−3% + 6& ; 32 , 3� ⟹ −3 3 2� + 6435 ⟹
− 92 + 18 ⟹ − 9 2 +
36 2 ⟹
�6 �
d. Which constraints are binding?
The binding constraints are 78 − �9 ≤ � and −�8 + �9 ≤ 7
e. Write the linear program in standard form.
Max. –3X + 6Y
s.t. 6X – 2Y + S1 = 3
–2X + 3Y + S2 = 6
X + Y – S3 = 3
X, Y, S1, S2, S3 ≥ 0
f. What are the values of the slack/surplus variables at the optimal solution?
% + & − �� = 3 ; 3 2 , 3� ⟹
3 2 + 3 − �� = 3 ⟹
2 �32 + 3 − �� = 3� ⟹ 3 + 6 − 2�� = 6 ⟹
3 = 2�� ⟹ �� = � �
… and sand sand sand since the binding constraints ?@A 78 − �9 ≤ � ?BC – �8 + �9 ≤ 7, we we we we know GH?G I� ?BC I� AJK?L MA@N
O
A
1 2 3
3
2
1
B
C
D
F
G
H
X
Y
