for statsolver to link

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 two-week period

Let N be the number of

New

Customer contacts per two-week period

Regular

1

Customer

Revenue/Hour

Hours/Contact

$25

50/60

New

$8

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

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 non-negativity 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 non-negativity 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 non-negativity 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 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

Still stressed with your coursework?
Get quality coursework help from an expert!