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Last Name ________________________ First Name ________________________ ID __________________________

Operations Management II 73-431 Winter 2004 Odette School of Business

University of Windsor

Final Exam Solution Monday, April 12, 3:30 – 6:30 pm

Toldo Health Education Centre Room 102

Instructor: Mohammed Fazle Baki Aids Permitted: Calculator, straightedge, and 3 one-sided formula sheets. Time available: 3 hours Instructions: • This solution has 19 pages including this cover page • It’s not necessary to return tables and formula sheets • Please be sure to put your name and student ID number on each page • Show your results up to four decimal places • Show your work Grading:

Question Score Question Score

1 /15 2 /6

3 /3 4 /6

5 /11 6 /6

7 /10 8 /11

9 /8 10 /4

11 /3 12 /6

13 /3 14 /5

15 /3 Total /100

Name:_________________________________________________ ID:_________________________

2

Question 1: (15 points) Circle the most appropriate answer

1.1 Advantages of the cellular layouts include the following, except:

a. Better scheduling b. Easier to control and automate c. Reduced work-in-process inventory d. Low capital investment

1.2 Which one is the gravity problem?

a. Minimize weighted sum of the squared rectilinear distances b. Minimize weighted sum of the squared Euclidean distances c. Minimize weighted sum of the rectilinear distances d. Minimize weighted sum of the Euclidean distances

1.3 Cost of quality is the lowest if the error is detected

a. at the time of production b. after a part is used in some other assembly c. at the final testing d. by a customer

1.4 Which of the following is the best reason to monitor variation using the R chart instead of standard

deviation chart? a. Cheaper to use b. Easier to understand c. Less loss of information d. Range does not necessarily change when the process mean changes

1.5 A make-to-order/assemble-to-order production system is associated with each of the following

except a. high variation b. low volume of production c. high efficiency d. low capital investment

1.6 Recall that λ is the failure rate and t the age. The difference between the conditional probabilities

and the probabilities obtained by the exponential distribution is small for a. small λ and small t b. small λ and large t c. large λ and small t d. large λ and large t

1.7 In the context of economic design of X control chart, if k increases

a. π increases b. Type I error increases c. searching cost increases d. cost of operating in out-of-control-condition increases

Name:_________________________________________________ ID:_________________________

3

Name:_________________________________________________ ID:_________________________

4

1.8 The average outgoing quality increases if a. the proportion of defective, p increases b. the proportion of defective, p decreases c. choice a when AOQL≤≤ p0

d. choice b when AOQL≤≤ p0 1.9 In the context of acceptance sampling, the consumer does not get any defective item if

a. a three-sigma control chart is used b. the consumer’s risk is low c. the sampling and testing show no defective item at all d. the sampling and testing show too many defective items to accept the lot

1.10 The producer’s risk increases if

a. only c increases, not n b. only n increases, not c

c. both nand c increase proportionately

d. none of the above

1.11 For which of the following should we use a cchart to monitor process quality? a. If a sheet of glass has any flaw in it b. Errors in the weight of a cereal box c. The ounce of beer poured in a can d. b and c

1.12 Minimizing maximum rectilinear distance is a suitable objective to find a location for

a. a fire station b. a computer centre c. an airline hub d. phone cable

1.13 Examples of improvement procedure:

a. CRAFT, positional weight method, but not the elimination of intersection b. ALDEP, positional weight method, but not the elimination of intersection c. CRAFT, elimination of intersection, but not the positional weight method d. ALDEP, elimination of intersection, but not the positional weight method

1.14 Which of the following is a Type II error?

a. When assignable causes are assumed, but there exist only common causes b. When assignable causes are assumed and the process is out of control c. When common causes are assumed, but there exists some assignable causes d. When a sample measurement is inside LCL and UCL and the process is in control

1.15 If the components are in series, the reliability of the system increases

a. if the number of components increases b. if the number of components decreases

Name:_________________________________________________ ID:_________________________

5

c. as the system is used more and the components are worn out d. none of the above

Name:_________________________________________________ ID:_________________________

6

Question 2: (6 points)

An amplifier has a constant failure rate of 8% per 1,000 hour.

a. (2 points) What is the probability that the amplifier will survive 4,000 hours?

( ) ( ) ( ) 7164.008.011 4 =−=−=≥ tP λ4000life

b. (2 points) What is the average life in hours? 5.1208.011

===λ

θ in 1000 hours = 12,500 hours

c. (2 points) What failure rate per 1,000 hour is required to have a probability of survival of 85% at 4,000 hour? ( ) hour 1,000 per -1 or, or, %98.30398.085.085.01,85.01 4/14/14 ====−=− λλλ


The next-door neighbours are currently installing a new cistern that will provide water for their household. The different components for the system and their reliability are described in the diagram below. Determine the reliability of the system.

0.96

0.50

0.70

0.95

0.40

0.92

0.85

0.75

0.600.55

0.65

1

23

4

6

7

8

9

( )( )( )( )

( )( )

( )( )( )( ) ( )( )

7608.092.09448.097.094.096.0

92.0

9448.0631875.0185.011111631875.075.08425.0

75.08425.065.0155.011

85.0

97.040.095.0195.094.070.0160.0150.011

96.0

98321

9

748

657

6

5

4

3

2

1

=××××==

=

=−−−=−−−====

==−−−=

=

=−+==−−−−=

=

RRRRRR

R

RRRRRR

RR

R

RRR

system


In one of the acceptance sampling plans, 20 items were to be tested for 200 hours with replacement and with an acceptance number of 3. Plot an Operating Characteristic curve showing probability of acceptance as function of average life. Consider average life, θ = 500, 1000, 2000, 4000, and 8000 hours. For each average life compute and plot the probability of acceptance.

Name:_________________________________________________ ID:_________________________

7

( ) 3,20000,420020 ==== cnT hours,

θ θ

=λ1

Tλ=µ ( )( )

Tmxp

pTcP

TcPPa

λµ

λµ

λµ

===

−===+≥−=

==≤=

,4'

1|411

|3

for 3- ATable from obtained is where

defective of Number

defective of Number

500 0.002 0.002(4000)

= 8

( )( )

8,4'9576.0

0424.09576.018|411

8|3

==

=−===+≥−=

==≤=

mx

cPcPPa


defective of Numberdefective of Number

µµ

1000 0.001 0.001(4000)

= 4

( )( )

4,4'5665.0

4335.05665.014|411

4|3

==

=−===+≥−=

==≤=

mx

cPcPPa



µµ

2000 0.0005 0.0005(4000)

= 2

( )( )

2,4'1429.08571.01429.01

2|411

2|3

===−=

==+≥−=

==≤=

mx

cP

cPPa


defective of Number

defective of Number

µ

µ

4000 0.00025 0.00025(4000)

= 1

( )( )

1,4'0190.0

9810.00190.011|411

1|3

==

=−===+≥−=

==≤=

mx

cPcPPa



µµ

8000 0.000125 0.000125(4000) = .50

( )( )

50.0,4'019.0

9982.00018.0150.0|411

50.0|3

==

=−===+≥−=

==≤=

mx

cPcPPa



µµ

Name:_________________________________________________ ID:_________________________

8

Operating Characteristic Curve, Question #4

0

0.2

0.4

0.6

0.8

1

0 2000 4000 6000 8000 10000

Average life

Pro

bab

ility

of

Ace

epta

nce

, Pa


A company employs the following sampling plan: It draws a sample of 20 percent of the lot being inspected. If 10 percent or less of the sample is defective, the lot is accepted. Otherwise, the lot is rejected.

a. (2 points) If a lot contains 100 items, what are n and c (sampling plan)?

( ) ( ) 22010.010.0,2010020.020.0,100 ======= ncNnN

b. (2 points) Continue from part a. If a lot contains 100 items of which 20 percent are defective, what is the probability that the lot is accepted? Use Binomial approximation table.

( ) ( )( ) ) 2,- A(Table

20,| defective fewer or defective fewer or

20.0,20,22061.020.0,20|3

20.02

=======≤=

====

pncrpnrP

pnPcPPa

c. (2 points) Continue from parts a and b. If the company purchases items in lots of 100, then what is average outgoing quality if the proportion of defectives is 0.10? Assume that the defectives are replaced.

( ) ( )0330.0

1002061.02010020.0

=−

=−

=N

PnNp a AOQ

d. (2 points) Continue from part a. If lot sizes of 100 are used, then what is producer’s risk, α , if AQL=0.02. Use Poisson approximation table.

( ) ( )( ) ( ) 9921.00079.0140.0|3'140.031

40.011

=−==≥−==−=

=×==+−====

mxPmP

npmcPpcPPa

| defective more or

0.0220 | defective more or 0.02 AQL| defective fewer or

(Table A-3 gives cumulative Poisson probabilities for 'x or more defectives. The value 0.0079 is obtained from 40.0,3' == mx )

=α Producer’s risk = 1- aP at p = AQL = 0.06 = 1-0.9921 = 0.0079

e. (3 points) Continue from part a. If lot sizes of 100 are used, then what is consumer’s risk, β , if LTPD=0.25. Use Normal approximation.

Name:_________________________________________________ ID:_________________________

9

=β Consumer’s risk = ( )0.25 LTPD| Acceptance ==pP = ( )25.020 == pncP drawn, items | defective fewer or

= ( )9365.175.025.020,525.0202 =××==×== σµ npP | defective fewer or

=

==

−≤ 9365.1,5

50.2σµ

σµ

| zP (continuity correction)

( ) ( ) ( ) ( )029.15.0029.1029.19365.1

550.2 ≤≤−−=≤≤−−≤≤∞−=−≤=

−≤= zPzPzPzPzP

≈ 0.50 - 0.4015 (From Table A-1) = 0.0985

Name:_________________________________________________ ID:_________________________

10


Control charts for X and R are maintained on the shear strength of spot welds. One hundred and fifty observations divided into subgroups of size six are used as a baseline to construct the charts, and estimates of µ and σ are computed from these observations. Assume that the 150 observations are

15021 XXX ,,, K and the ranges of 25 subgroups are .,,, 2521 RRR K From these baseline data the following quantities are computed:

∑∑==

==25

1

100

1

000100085j

ji

i RX , ,,

a. (2 points) Compute the values of three-sigma limits for both X and R charts.

785.15534.240

,4025000,1

,67.566256

000,85

2

25

1

100

1 =======×

==∑∑

==

dR

k

R

Rnk

XX j

ji

i

σ

0400,80402

33.5476

785.15367.5663,00.586

6

785.15367.5663

34 =×===×==

=−=−==+=+=

RdLCLRdUCL

nXLCL

nXUCL

RR

XX

σσ

b. (2 points) Suppose that it is required to set up an X control chart such that the probability of concluding that process is out of control when it is actually in control is 0.04. Find the upper and lower control limits for the X chart that has the probability of Type I error 0.04.

TablevAzz =====

02.02/

02.02/04.02/,04.0

α

αα

055.2

02.02/04.02/,04.0

02.02/

=

======

0.480.02-0.50 area for value-z 1- ATablezzα

αα

43.5536

785.15055.267.566,91.579

6

785.15055.267.566 2/2/ =−=−==+=+=

nzXLCL

nzXUCL

XX

σσαα

c. (2 points) Suppose that there is a probability of 0.10 that the process shifts from an in-control to an out-of-control state in any period. What is the expected number of periods that the process remains in control? For the X chart obtained in Part b, what is the expected searching cost per cycle if each search costs $200.

( )

( )[ ] [ ] 272$904.012001

910.0

10.011

2 =×+=+=

=−=−=

TEa

TE

απ

π

cycle per cost Search

periods

Name:_________________________________________________ ID:_________________________

11


Frank Brown, an independent TV repairman, is considering purchasing a home in Ames, Iowa, that he will use as a base of operations for his repair business. Frank’s primary sources of business are 5 industrial accounts located throughout the Ames area. He has overlaid a grid on a map of the city and determined the following locations for these clients as well as the expected number of calls per month he receives:

Client 1 2 3 4 5

Grid Location (2,6) (4,8) (5,9) (3,4) (8,7)

Expected Calls/month 3 7 4 6 5

a. (2 points) Find the optimal location of his house, assuming the goal is to minimize the weighted sum of rectilinear distances.

Sort in the order of ix Sort in the order of iy

Building

i ix iw Cumulative iw Building

i iy iw Cumulative iw

1 2 3 3 4 4 6 6

4 3 6 3+6=9 1 6 3 6+3=9

2 4 7 9+7=16 5 7 5 9+5=14

3 5 4 16+4=20 2 8 7 14+7=21

5 8 5 20+5=25 3 9 4 21+4=25

25÷2=12.5 25÷2=12.5

Optimal location = (4,7)

b. (4 points) Find the optimal location of his house, assuming the goal is to minimize the maximum rectilinear distance to any client. What’s the maximum rectilinear distance between an optimal location and a client?

Client ix iy ii yx + ii yx +−

1 2 6 8 4

2 4 8 12 4

3 5 9 14 4

4 3 4 7 1

5 8 7 15 -1

Min =1c 7 =3c -1

Max =2c 15 =4c 4

Name:_________________________________________________ ID:_________________________

12

( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( )( ) ( ) 5.52/84152/

5.52/4152/

72/8172/42/172/

85,8max14,715max,max

5422

422

5311

311

34125

=−+=−+=

=−=−=

=+−=++==−−=−=

==−−−=−−=

cccy

ccx

cccyccx

ccccc

Any point along (4, 7) and (5.5, 5.5) is optimal.

Maximum rectilinear distance from optimal location (4, 7) is computed below:

Max rectilinear distance

= max(|4-2|+|7-6|, |4-4|+|7-8|, |4-5|+|7-9|, |4-3|+|7-4|, |4-8|+|7-7|)

= max(3, 1, 3, 4, 4) = 4

c. (4 points) Find the optimal location of his house, assuming the goal is to minimize weighted sum of squared of the Euclidean distances. What’s the weighted sum of squared Euclidean distances from the optimal location to the clients?

Client, i ix iy iw ii xw ii yw

1 2 6 3 6 18

2 4 8 7 28 56

3 5 9 4 20 36

4 3 4 6 18 24

5 8 7 5 40 35

∑ =iw 25 =∑ ii xw 112 =∑ ii yw 169

i

ii

w

xwX

∑∑=* =112/25=4.48

i

ii

w

ywY

∑∑=* =169/25=6.76

Optimal location = (4.48, 6.76)

Weighted sum of squared Euclidean distance from optimal location (6.8, 4.96) is computed below:

Weighted sum of squared Euclidean distance

= 3[(4.48-2)2+(6.76-6)2] + 7[(4.48-4)2+(6.76-8)2] + 4[(4.48-5)2+(6.76-9)2] + 6[(4.48-3)2+(6.76-4)2]

+ 5[(4.48-8)2+(6.76-7)2]

= 20.184+2.376+21.152+58.848+62.24 = 174.8

Name:_________________________________________________ ID:_________________________

13

A

B

C

AO

E

DX

IU

A A C C C CA A C C D DA A C C D DB B B B D D

C C A A A AC C A A D DC C B B D DC C B B D D


Consider the activity relationship chart shown on the right. a. (1 point) According to ALDEP, what is the order of placement that starts with

department A?

A-B-C-D

Suppose that departments A, B, C and D consume respectively 6, 4, 8 and 6 squares. Based on a sweep width of 2 and a facility layout size of 6 (horizontal) by 4 (vertical), use the technique employed by ALDEP to find a layout. b. (2 points) Show the layout obtained by ALDEP when the departments are placed in the order

obtained in Part a. Compute the layout evaluation score.

Adjacent departments

Rel Chart Score Numeric Score

A-B A 64 A-C O 1 B-C E 16 B-D I 4 C-D X -1024

Total 939 c. (2 points) Show the layout obtained by ALDEP when the departments are placed in the order C, B, A,

D. Compute the layout evaluation score.

Adjacent departments

Rel Chart Score Numeric Score

A-B A 64 A-C O 1 A-D U 0 B-C E 16 B-D I 4

Total 85 d. (1 point) According to ALDEP, which of the two layouts obtained in parts b and c is better?

Layout c is better, C-B-A-D

Name:_________________________________________________ ID:_________________________

14

6 12 18 24 30 36

612

1824

C

B

A1 A2

D

6,1218,6

18,18

30,9

30,21

Consider the better layout obtained in part d to answer parts e, f, g, h. Suppose that the facility is 24 feet horizontal and 16 feet vertical. So, every square represents an area of 6 feet by 6 feet. e. (2 points) Compute the centroid of Department A.

A1: Area = 12(12) = 144, A2: Area = 12(6) = 72

1972144

217218144

2272144

307218144

=+

×+×=

=+

×+×=

y

x

Centroid of A is (22,19)

f. (1 point) Compute the centroid of Department B.

Centroid of B is ( )6,182120

,2

2412=

++

g. (1 point) What is the rectilinear distance between A and B?

Rectilinear distance between A and B = 171346191822 =+=−+−

h. (1 point) According to CRAFT, what are the feasible interchanges?

• C-A

• C-B

• B-A

• B-D

• A-D


All-Weather Oil and Gas Company has two delivery vehicles and is planning delivery routes to four natural gas customers. The customer locations and gas requirements (in gallons) are given below:

Customer Location Requirements (gallons)

1 (0,3) 250

2 (4,7) 400

3 (1,0) 600

4 (4,3) 350

Assume that the depot is located at the origin of the grid and that the delivery trucks have capacity 1,000 gallons. Also assume that the cost of travel between any two locations is the straight-line (Euclidean) distance between them. Find the route schedule obtained from the savings matrix method and the nearest neighbour method.

Name:_________________________________________________ ID:_________________________

15

W

Schedule After Improvement

12

34

56

78

1 2 3 4 5 6

3

1

2

4

W

Schedule Before Improvement

12

34

56

78

1 2 3 4 5 6

3

1

2

4

a. (2 points) Compute the distance matrix

Distance Matrix

Depot Customer 1 Customer 2 Customer 3 Customer 4

Depot ---------- 3 06.874 22 =+ 1 5

Customer 1 ---------- ---------- 5.66 3.16 4

Customer 2 ---------- ---------- ---------- 7.62 4

Customer 3 ---------- ---------- ---------- ---------- 4.24

Customer 4 ---------- ---------- ---------- ---------- ----------

b. (2 points) Compute the savings matrix

Savings Matrix

Customer 1 Customer 2 Customer 3 Customer 4

Customer 1 ---------- 3+8.06-5.66=5.4 0.84 4

Customer 2 ---------- ---------- 1.44 9.06

Customer 3 ---------- ---------- ---------- 1.76

Customer 4 ---------- ---------- ---------- ----------

c. (1 points) Rank the savings

Rank Customer Pair Savings

1 (2,4) 9.06

2 (1,2) 5.40

3 (1,4) 4.00

4 (3,4) 1.76

5 (2,3) 1.44

6 (1,3) 0.84

d. (3 points) Find and show the delivery routes. Eliminate intersection, if any.

Name:_________________________________________________ ID:_________________________

16


The following matrix contains the material handling costs (in thousand dollars) associated with assigning Machines 1, 2 and 3 to Locations A, B and C. Assign machines to locations to minimize material handling costs. State the optimal assignment and the associated cost.

Locations Machines A B C

1 $200 $500 $700 2 100 400 400 3 300 300 500

a. (1 point) Show the matrix obta ined after row reduction Locations Machines

A B C

1 0 300 500

2 0 300 300

3 0 0 200

b. (1 point) Continue from part a and show the matrix obtained after column reduction Locations Machines

A B C 1 0 300 300

2 0 300 100

3 0 0 0

c. (1 points) Continue from part b, and show the optimal solution Locations Machines

A B C

1 0 200 200

2 0 200 0

3 100 0 0

Optimal solution: Machine 1 à A Machine 2 à C Machine 3 à B d. (1 point) What is the cost associated with the optimal assignment obtained in part c?

1àAà200

2àCà400

3àBà300

$900

Name:_________________________________________________ ID:_________________________

17

1 2

3

A(5,5,5)

4

C(4,5,6)

B(7,8,9)

5

6

7D(5,7,9)

E(1,3,5)

F(3,5,7)

G(2,4,6)

H(1,3,5)

1

2

3

B4

Node 1 = starting nodeNode 4 = finish node

Dummy0

4C6

D2

A7

5

Dummy

E 5


Activities of a project and their immediate predecessors are shown below:

Activity Time (Weeks) Immediate Predecessors

A 7 ---

B 4 ---

C 6 A

D 2 A, B

E 5 B, C

Construct a network for the project. Use activity on arc method.


Following network shows PERT time estimates (a,m,b) in weeks:

a. (2 points) Determine a critical path based on most likely times. What is the mean and standard deviation of the length of the critical path?

Find longest path

A-B-D-G: 5+8+7+4=24 weeks (*)

A-B-E-G: 5+8+3+4=20 weeks

A-B-F-G: 5+8+5+3=21 weeks

A-C-E-G: 5+5+3+6=19 weeks

The critical path is the longest path A-B-D-G

Name:_________________________________________________ ID:_________________________

18

Activity 6

4 bma ++=µ ( )

36

22 ab −

=σ

A ( )5

65545

=++

=µ ( )0

3655 2

2 =−

=σ

B 8 0.1111

D 7 0.4444

G 4 0.4444

Total 24 1.0000

24=µ weeks, 112 === σσ week

b. (2 points) Assuming that the critical path is the one identified in a, find the probability that the project will be completed in 26 days.

( ) ( ) 9772.04772.05.021

242626=+=≤=

−

≤=

−

≤===≤ zPzPzPPσ

µσµ 124,|26length

c. (2 points) Assuming that the critical path is the one identified in a, how many weeks are required to complete the project with probability 0.99?.

0.480.50-0.99 area for value 1- ATable === zz

So, 33.2=z

So, the length of the project ( ) 33.26133.224 =+=+= σµ z weeks


Three jobs must be processed on a single machine that starts at 8:30 am. The processing times and due dates are given below:

Job Processing Time (Hours) Due Date

J1 2 11:30 am

J2 4 1:30 pm

J3 1 3:30 pm

Name:_________________________________________________ ID:_________________________

19

a. (1 point) Determine the schedule for the Critical Ratio (CR) rule.

Current Time 0

Job Processing Time (Hours)

Due Date (Hours) Critical Ratio Assign?

J1 2 3 5.12

03=

−

J2 4 5 1.25 Yes (least CR)

J3 1 7 7

Current Time 4

Job Processing Time (Hours)

Due Date (Hours) Critical Ratio Assign?

J1 2 3 5.02

43−=

− Yes (least CR)

J3 1 7 3

Write your final CR sequence here: J2-J1-J3

b. (2 points) Compute the total completion time and average tardiness when the CR rule is used. Jobs

arranged in CR

sequence

Start Time

(Hours)

Processing Time

(Hours)

Completion Time

(Hours)

Due Date

(Hours)

Lateness

(Hours)

Tardiness

(Hours)

J2 0 4 4 5 -1 0

J1 4 2 6 3 3 3

J3 6 1 7 7 0 0

Total 17 3

Total completion time = 17

Average tardiness = 3/3 = 1

Name:_________________________________________________ ID:_________________________

20

10 20 30 40 50

10

20

30

40

50

0

History

English

Science

Dina's time

Min

a's ti

me

15

10

25

10

Total length=15+10+25+10= 60


Dina and Mina are two sisters currently attending university together. Each requires advising in three subjects: history, English and science. The estimate that the time (in minutes) that each will require for advising is given below:

Advising time required by Dina Advising time required by Mina

Subject Time (Min) Subject Time (Min)

English 15 History 25

History 25 English 20

Science 10 Science 5

They think that the three advisers will be available all day. Dina would like to visit the advisers in the order English, history and science. Mina would prefer to see them in the order history, English and science.

a. (3 points) Determine how should Dina and Mina plan their schedule in order to minimize the time required for both to complete all visits. Use the graphical solution procedure for two-job problem. Draw the graph using the space provided on the next page.

Dina’s time Mina’s time

Advisor Start Processing End Advisor Start Processing End

English 0 15 15 History 0 25 25

History 15 25 40 English 25 20 45

Science 40 10 50 Science 45 5 50

Rectangle Lower left corner (start, start) Upper right corner (end, end)

English (0,25) (15,45)

History (15,0) (40,25)

Science (40,45) (50,50)

Name:_________________________________________________ ID:_________________________

21

20 40 600

10 30 50

English

History

Science

Clock time

Dina

Mina

Mina

Dina

M Dina

b. (2 points) Draw the Gantt chart indicating the optimal schedule. Show the Gantt chart using the space provided below:


Irving Bonner, an independent computer programming consultant, has contracted to complete four computer programming jobs.

Job Time required (days) Due date (days)

1 13 20

2 10 25

3 9 30

4 8 35

Assume that some jobs must be completed in a certain sequence because they involve program modules that will be linked. Precedence restrictions:

3→2 4→1

a. (2 points) Using Lawler’s algorithm, find the sequence in which he should be performing the jobs in order to minimize the maximum lateness subject to the precedence constraints.

Iteration 1:

Job Processing time

Due date Candidate? Completion time if scheduled

Lateness if scheduled

1 13 20 Yes 40 40-20=20

2 10 25 Yes 40 40-25=15*

3 9 30 No

4 8 35 No

Decision in iteration 1:

2

Name:_________________________________________________ ID:_________________________

22

Iteration 2:

Job Processing time



1 13 20 Yes 30 30-20=10

3 9 30 Yes 30 30-30=0*

4 8 35 No


3 2

Iteration 3:

Job Processing time



1 13 20 Yes 13+8=21 21-20 = 1*

4 8 35 No


1 3 2

Final decision after iteration 4

4 1 3 2

b. (1 point) Find the optimal maximum lateness

Max (15,0,1,8-35)=Max(15,0,1,-27)=15

Job Start Processing time Finishing time Due date Lateness

4 0 8 8 35 -27

1 8 13 21 20 1

3 21 9 30 30 0

2 30 10 40 25 15*

Max lateness = Max(-27,1,0,15) = 15

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