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
19. Part c (Continued)
19. Part b
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
Units of Stock Fund: 4,000
Units of Money Market Fund: 10,000
Annual Income: $60,000Min. 8S + 3M
s.t. 50S + 100M ≤ 1,200,000
5S + 4M ≥ 60,000
100M ≥ 300,000
S, M ≥ 0
Part c
Units of Stock Fund: 18,000
Units of Money Market Fund: 3,000
Annual Income: $102,000Max. 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 
$ 0.3 0 
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
Optimal Solution
Gallons of Regular Gasoline:
40,000
Gallons of Premium Gasoline:
10,000
Total Profit Contribution:
$17,000
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
S
1, the slack variable for Grade
A
Crude Oil, is 0, which means all of
it
will be consumed.
S
2
, the slack variable for Production Capacity, is 0, which means all Production
Capacity
will be utilized.
S
3
, 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 de
mand exceed P = 10,000 gallons.
Part d
The binding
constraints are those constraints whose slack variables are zero; namely, the availability of Grade A Crude Oil and Manufacturing Capacity.
Let R be the number of Regular
Customer
contacts per twoweek period
Let N be the number of
New
Customer contacts per twoweek period
Regular1
Customer 
Revenue/Hour 
Hours/Contact 
$25 
50/60 

New 
$8 
Let T be the amount of available technician time during a twoweek 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 twoweek 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 twoweek period; in other words…
Objective: Maximize R + N
Parts a & b
Optimal Solution
Reg
ular
Customer Contacts / 2 wks:
60
New Customer Contacts / 2 wks:
30
FYI…
Total
Revenue:
$1,490Max. R + N
s.t. R + 1N ≤ 80
R + 8N ≥ 800
– R + 1N ≥ 0
R, N ≥ 0
Check
5. Refer to the model and graph shown below. Answer questions (a) to (f).
1
QAS 19 Assignment 2
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?
d. Which constraints are binding?
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?
OA123321BCDFGHXY
Salexia Timmerman
QAS 19
Assignment 1
1) d = 800 – 10p
a) When p = $20, d = 800 – 10(20) = 800 – 200 = 600
When p = $70, d = 800 – 10(70) = 800 – 700 = 100
Thus, the firm can sell 600 units at the $20 price per unit and 100 units at the $70 price per unit.
b) The mathematical model for the total revenue is
TR = dp = (800 – 10p)p
That is, TR = 800p – 10p2
c) When p = $30, TR = 800(30) – 10(30^2) = 24000 – 9000 = 15,000
When p = $40, TR = 800(40) – 10(40^2) = 32000 – 16000 = 16,000
When p = $50, TR = 800(50) – 10(50^2) = 40000 – 25000 = 15,000
Thus, total revenue is maximized at the $40 price.
d) The expected annual demand corresponding to the recommended price is
d = 800 – 10(40) = 800 – 400 = 400
and TR = 800(40) – 10(40^2) = 32000 – 16000 = 16,000
Thus, the expected annual demand is 400 units and the total revenue is $16,000 corresponding to the recommended price.
3) a) The decision variables are
X = Number of units purchased in the stock fund
Y = Number of units purchased in the money market fund
b) Since each unit invested in the stock fund has a risk index of 8, and each unit invested in the money market fund has a risk index of 3; the total risk index is
8 X + 3Y.
Thus, an objective function that will minimize the total risk index for the portfolio is
Minimize Z = 8 X + 3 Y
c) Since each unit of stock fund costs $50 and each unit of money market fund costs $100, the constraint corresponding to the available funds is
50 X + 100 Y ≤ 1,200,000
Since the annual income from stock fund is ($50)(10%) = $5 and the annual income from the money market fund is ($100)(4%) = $4, the constraint corresponding to the annual income is
5 X + 4 Y ≥ 60,000
Since at least $300,000 to be invested in the money market, at least ($300,000/$100) = 3,000 units must be invested in the money market. Thus, the constraint corresponding to the minimum units in money market is
Y ≥ 3,000
Finally, the nonnegativity constraints, X, Y ≥ 0.
Thus, the constraints of the problem are
50 X + 100 Y ≤ 1,200,000
5 X + 4 Y ≥ 60,000
Y ≥ 3,000
X, Y ≥ 0
4) The decision variables are
X = Number of gallons of regular gasoline produced
Y = Number of gallons of premium gasoline produced
Since the proﬁt contributions are $0.30 per gallon for regular gasoline and $0.50 per gallon for premium gasoline; the total proﬁt contribution is 0.30 X + 0.50 Y.
Thus, the objective function is
Maximize Z = 0.30 X + 0.50 Y
Since each gallon of regular gasoline contains 0.3 gallons of grade A crude oil and each gallon of premium gasoline contains 0.6 gallons of grade A crude oil, the constraint corresponding to the available Grade A crude oil is
0.30 X + 0.60 Y ≤ 18,000
The constraint corresponding to the production capacity is
X + Y ≤ 50
,000
The constraint corresponding to the demand for the premium gasoline is
Y ≤ 20,000
Finally, the nonnegativity constraints, X, Y ≥ 0.
Thus, the linear programming model is
Maximize Z = 0.30 X + 0.50 Y
Subject to
0.30 X + 0.60 Y ≤ 18,000
X + Y ≤ 50,000
Y ≤ 20,000
X, Y ≥ 0
5) The decision variables are
X = Number of necklaces sold
Y = Number of earrings sold
Since the necklaces make the designer a profit of $25 each piece and the earrings make a profit of $10 each piece; the designer’s weekly profit is 25 X + 10 Y.
Thus, the objective function is
Maximize Z = 25 X + 10 Y
Since the boutique will buy between 10 and 40 necklaces each week, the constraint corresponding to requirement of necklaces is
10 ≤ X ≤ 40
Since the boutique bought at least twice the number of earrings as necklaces, the corresponding constraint is
2Y ≥ X
or
2Y – X ≥ 0
Since the designer cannot make more than 50 pieces of jewelry per week, the constraint corresponding to this limitation is
X + Y ≤ 50
Finally, the nonnegativity constraints, X, Y ≥ 0.
Thus, the linear programming model is
Maximize Z = 25 X + 10 Y
Subject to
10 ≤ X ≤ 40
2Y – X ≥ 0
X + Y ≤ 50
X, Y ≥ 0
QAS 19 Assignment 2
1
QAS Decision Analysis
Note:
•
Y
ou 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 twoweek period
Let N be the number of New Customer contacts per twoweek period
Customer Revenue/Hour Hours/Contact
Regular $25 50/60
New $8 1
Let T be the amount of available technician time during a twoweek 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 twoweek 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 twoweek 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