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	$0.30	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

Optimal Solution Gallons of Regular Gasoline: 40,000 Gallons of Premium Gasoline: 10,000 Total Profit Contribution: $17,000Parts 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

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

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