MANAGERIAL DECISION ANALYSIS(MGT 780)

ASSIGNMENT 5INTEGER AND GOAL PROGRAMMING

PREPARED FOR:

DR. TAN PECK LEONG

PREPARED BY:

SYURGA FATHONAH BINTI ARIS@AZIS 2013343955

DATE OF SUBMISSION:31 OCTOBER 2014

10-19 Triangle Utilities provides electricity for three cities.The company has four electric generators that are used to provide electricity. The main generator operates 24 hours per day, with an occasional shutdown for routine maintenance. Three other generators (1, 2, and 3) are available to provide additional power when needed. A startup cost is incurred each time one of these generators is started. The startup costs are $6,000 for 1, $5,000 for 2, and $4,000 for 3. These generators are used in the following ways: A generator may be started at 6:00 A.M. and run for either 8 hours or 16 hours, or it may be started at 2:00 P.M. and run for 8 hours (until 10:00 P.M.). All generators except the main generator are shut down at 10:00 P.M. Forecasts indicate the need for 3,200 megawatts more than provided by the main genera- tor before 2:00 P.M., and this need goes up to 5,700 megawatts between 2:00 and 10:00 P.M. Generator 1 may provide up to 2,400 megawatts, generator 2 may provide up to 2,100 megawatts, and generator 3 may provide up to 3,300 megawatts. The cost per megawatt used per eight hour period is $8 for 1, $9 for 2, and $7 for 3.

(a) Formulate this problem as an integer programming problem to determine the least-cost way to meet the needs of the area.

(b) Solve using computer software.

SOLUTION

Objective: To minimize cost 6000(G1X1+G1X2+G1X3) + 5000(G2X1+G2X2+G2X3) + 4000(G3X1+G3X2+G3X3) + 8(G1Y1+G1Y2) + 9(G2Y1+G2Y2) + 7(G3Y1+G3Y2)

Variables:

G1X1=Generator 1 functioning at 6am-2pmG1X2=Generator 1 functioning at 2pm-10pmG1X3=Generator 1 functioning at 6am-10pmG2X1=Generator 2 functioning at 6am-2pmG2X2=Generator 2 functioning at 2pm-10pmG2X3=Generator 2 functioning at 6am-10pmG3X1=Generator 3 functioning at 6am-2pmG3X2=Generator 3 functioning at 2am-10pmG3X3=Generator 3 functioning at 6am-10pmG1Y1=Megawatt produced by Generator 1 functioning at 6am-2pmG1Y2=Megawatt produced by Generator 1 functioning at 2pm-10pmG2Y1=Megawatt produced by Generator 2 functioning at 6am-2pmG2Y2=Megawatt produced by Generator 2 functioning at 2pm-10pm

Constraint:

G1Y1 + G2Y1 + G3Y1 ≥ 3200 Requirement for megawatts at 6am-2pmG1Y2 + G2Y2 + G3Y2 ≥ 5700 Requirement for megawatts at 2pm-10pm

G1Y1 ≤ 2400 (G1X1 + G1X3)G1Y1 – 2400G1X1 – 2400G1X3 ≤ 0 Max megawatts generator 1 (6am-2pm)

G1Y2 ≤ 2400(G1X2 + G1X3)G1Y2 – 2400G1X2 -2400G1X3 ≤ 0 Max megawatts generator 1 (2pm-10pm)

G2Y1 ≤ 2100(G2X1 + G2X3)G2Y1 – 2100G2X1 – 2100G2X3 ≤ 0 Max megawatts generator 2 (6am-2pm)

G2Y2 ≤ 2100(G2X2 + G2X3)G2Y2 – 2100G2X2 – 2100G2X3 ≤ 0 Max megawatts generator 2 (2pm -10pm)

G3Y1 ≤ 3300(G3X1 + G3X3)G3Y1 – 3300G3X1 – 3300G3X3 ≤ 0 Max megawatts generator 3 (6am-2pm)

G3Y2 ≤ 3300(G3X2 + G3X3)G3Y2 – 3300G3X2 – 3300G3X3 ≤ 0 Max megawatts generator 3 (2pm-10pm)

G1X1 + G1X2 + G1X3 ≤ 1 Generator 1 start upG2X1 + G2X2 + G2X3 ≤ 1 Generator 2 start up G3X1 + G3X2 + G3X3 ≤ 1 Generator 3 start up

Data

Orignal problem with answer

Solution

G1X2=Generator 1 functioning at 2-10G3X3=Generator 3 functioning at 6am-10pm

From the solution, generator 1 will be utilized in the period 2pm-10pm (G1X2) which generate 2,400 megawatts of electricity. Meanwhile, for generator 3, it will be utilized for 16 hours in the period of 6am to 10pm(G3X3). It generates 3,200 megawatt at 6am-2pm (G3Y1)and 3,300 megawats at 2pm-10pm (G3Y2).

10-21 Geraldine Shawhan is president of Shawhan File Works, a firm that manufactures two types of metal file cabinets. The demand for her two-drawer model is up to 600 cabinets per week; demand for a three- drawer cabinet is limited to 400 per week. Shawhan File Works has a weekly operating capacity of 1,300 hours, with the two-drawer cabinet taking 1 hour to produce and the three-drawer cabinet requiring 2 hours. Each two-drawer model sold yields a $10 profit, and the profit for the large model is $15. Shawhan has listed the following goals in order of importance:(a) Attain a profit as close to $11,000 as possible each week.(b) Avoid underutilization of the firm’s production capacity.(c) Sell as many two- and three-drawer cabinets as the demand indicates.

Set this up as a goal programming problem

10-22 Solve Problem 10-21. Are any goals unachieved in this solution? Explain.

SOLUTION

10-21:

X1 : two-drawer cabinet produced per weekX2 : three-drawer cabinet produced per week

Minimize deviation: P1D1- + P1D1

+ +P2D2-+ P3D3

-+P3D4-

1. 10X1 + 15X2 +D1- - D1

+ = 110002. 1X1 + 2X2 + D2

- - D2+ = 1300

3. X1 + D3- - D3

+ = 600X2 + D4

- - D4+ = 400

All Xi, Di variables ≥ 0

10-22

Data

The solution is to produce 500 two-drawer (X1) , and 400 three-drawer (X2) . Based on priority analysis, priority 3 =100. This means that the sales goal is underachieved by 100. 100, meaning the two-drawer sales goal is underachieved by 100.

10-23 Hilliard Electronics produces specially coded com- puter chips for laser surgery in 64MB, 256MB, and 512MB sizes. (1MB means that the chip holds 1 mil- lion bytes of information.) To produce a 64MB chip requires 8 hours of labor, a 256MB chip takes 13 hours, and a 512MB chip requires 16 hours. Hilliard’s monthly production capacity is 1,200 hours. Mr. Blank, the firm’s sales manager, esti- mates that the maximum monthly sales of the 64MB, 256MB, and 512MB chips are 40, 50, and 60, respectively. The company has the following goals (ranked in order from most important to least important):

1. Fill an order from the best customer for thirty 64MB chips and thirty-five 256MB chips.2. Provide sufficient chips to at least equal the sales estimates set by Mr. Blank.3. Avoid underutilization of the production capacity.

Formulate this problem using goal programming.

SOLUTION

X1 : 64MB chipX2 : 256MB chipX3 : 512MB chip

Minimize deviation: P1D1- + P1D2

+ +P2D3-+ P2D4

-+P2D5- +P3D6

-

1. X1 +D1- - D1

+ = 30 X2 + D2

- - D2+ = 35

2. X1 + D3- - D3

+ = 40X2 + D4

- - D4+ = 50

X3 + D5- - D5

+ = 60

3. 8X1+13X2+16X3 + D6- - D6

+ = 1200

All Xi, Di variables ≥ 0

Using goal programming:

Data

Solution

Solution is to produce 40 64MB chip (X1) , 50 256MB chip (X2) and 60 512MB chip (X3) .

10-26 Mick Garcia, a certified financial planner (CFP) has been asked by a client to invest $250,000. This money may be placed in stocks, bonds, or a mutual fund in real estate. The expected return on investment is 13% for stocks, 8% for bonds, and 10% for real estate. While the client would like a very high expected re- turn, she would be satisfied with a 10% expected re- turn on her money. Due to risk considerations, several goals have been established to keep the risk at an ac- ceptable level. One goal is to put at least 30% of the money in bonds. Another goal is that the amount of money in real estate should not exceed 50% of the money invested in stocks and bonds combined. In ad- dition to these goals, there is one absolute restriction. Under no circumstances should more than $150,000 be invested in any one area.(a) Formulate this as a goal programming problem.Assume that all of the goals are equally

important. (b) Use any available software to solve this problem.

How much money should be put in each of the investment options? What is the total return? Which of the goals are not met?

SOLUTION

X1 : StockX2 : BondX3 : Real EstateExpected return :

Minimize deviation: P1D1- + P1D2

- + P1D3+

1. 0.13X1 + 0.08X2 + 0.10X3 +D1- - D1

+ = 0.1 ( Return on investment )2. X2 + D2

- - D2+ = 75000 (30% of the money in the bond)

3. X3 + D3- - D3

+ = 0.5 (X1+X2) ( not exceed 50% money invested in stock and bonds combine)X3 - 0.5X1 - 0.5X2 + D3

- - D3+ = 0

Condition

X1 + X2 + X3 ≤ 250000

X1 ≤ 150000

X2 ≤ 150000

X3 ≤ 150000

All Xi, Di variables ≥ 0

Data

Solution

The solution is to invest $150000 in stock (X1), $75000 in bonds (X2) and $25000 in real estate (X3). The total return of 10% is $25000 (10/100* $250000). Based on the solution, all goal are met.