The ENM 541 Final Exam Review

So begins a very fine practice

exam…

A precursor, warm-up, harbinger, and forerunner to the final exam.

More of what on our very last exam

Question #1 Assembly line balancing problem or project

scheduling problem Question #2

Facility layout or location problem Question #3

Equipment replacement problem or question on bonus topic

Problem 1 – Assembly Line Balancing

The assembly of a manufactured thing requires the following 12 tasks:

Task A B C D E F G H I J K L

Time(minutes)

20 6 5 21 8 35 15 10 15 5 46 16

ImmediatePredecessor(s)

A B C D

G EH

C F,IJ

K

a. Using the ranked positional weight (RPW) method, determine the resulting balance using a cycle time of 70 minutes. b. Find the minimum cycle time that results in a four-station balance.

Problem 1 – Assembly Line Balancing

Assigned Tasks

Workstation (a) Positional Weight Method (b) 4-station balance

I

II

III

IV

V

VI

VII

VIII

Total Idle Time Line Efficiency (LE) Smoothness Index (SI)

Part (a)

Part (b)

Precedence Diagram

A

K

J

I

HG

F

E

D

CB

L

4615

10

15

21

8

35

6 5 520

16

Positional Weights

Task A B C D E F G H I J K L

Time 20 6 5 21 8 35 15 10 15 5 46 16

PW 138 118 112 123 85 97 102 87 77 67 62 16

rank 1 3 4 2 8 6 5 7 9 10 11 12

A B C J K L G H IPWA = 20 + 6 + 5 + 5 + 46 + 16 + 15 + 10 + 15 = 138

(a) PW METHOD

Assigned Tasks

Workstation (a) Positional Weight Method Station time

I A,D,B,C,G 67

II F,H,E,I 68

III J,K,L 67

CYCLE TIME = 68

Task A B C D E F G H I J K L

Time 20 6 5 21 8 35 15 10 15 5 46 16

PW 138 118 112 123 85 97 102 87 77 67 62 16

rank 1 3 4 2 8 6 5 7 9 10 11 12

Performance Measures

Total Idle Time

Line Efficiency (LE)

Smoothness Index (SI)

Part (a) 2 99.0% 1.414

1

3(68) 202 2n

ii

IT k C t

1 202100% 99%

3(68)

k

jj

S

LE xk C

2 2max

1

2(68 67) 1.414k

jj

SI S S

(B) 4-STATION BALANCE

Assigned Tasks

Workstation (a) Positional Weight Method Station time

I A, D, B 47

II

III

IV

CYCLE TIME =

Min cycle time= 202/4 = 50.5=51

Task A B C D E F G H I J K L

Time(minutes)

20 6 5 21 8 35 15 10 15 5 46 16

ImmediatePredecessor(s)

A B C D

G EH

C F,IJ

K

(B) 4-STATION BALANCE

Assigned Tasks

Workstation (a) Positional Weight Method Station time

I A, D, B 47

II C, E, F 48

III

IV

CYCLE TIME =

Min cycle time= 202/4 = 50.5=51

Task A B C D E F G H I J K L

Time(minutes)

20 6 5 21 8 35 15 10 15 5 46 16

ImmediatePredecessor(s)

A B C D

G EH

C F,IJ

K

(B) 4-STATION BALANCE

Assigned Tasks

Workstation (a) Positional Weight Method Station time

I A, D, B 47

II C, E, F 48

III G, J, H, I 45

IV

CYCLE TIME =

Min cycle time= 202/4 = 50.5=51

Task A B C D E F G H I J K L

Time(minutes)

20 6 5 21 8 35 15 10 15 5 46 16

ImmediatePredecessor(s)

A B C D

G EH

C F,IJ

K

(B) 4-STATION BALANCE

Assigned Tasks

Workstation (a) Positional Weight Method Station time

I A, D, B 47

II C, E, F 48

III G, J, H, I 45

IV K, L 62

CYCLE TIME = 62

Min cycle time= 202/4 = 50.5=51

Task A B C D E F G H I J K L

Time(minutes)

20 6 5 21 8 35 15 10 15 5 46 16

ImmediatePredecessor(s)

A B C D

G EH

C F,IJ

K

Performance Measures

Total Idle Time

Line Efficiency (LE)

Smoothness Index (SI)

Part (a) 2 99.0% 1.414Part (b) 46 81.5% 26.65

1

4(62) 202 46n

ii

IT kC t

1 202

100% 81.5%4(62)

k

jj

S

LE xkC

2 2 2 2max

1

(62 47) (62 48) (62 45) 26.65k

jj

SI S S

Problem 2 – Project SchedulingA maintenance project consists of 10 activities as defined in Table 1.

Activity Duration (in days)

Precedes Number workers

Activity Duration (in days)

Precedes Number workers

A 7 B,C 5 F 6 I 3

B 2 D,F 5 G 1 J 1

C 4 E,G 4 H 5 - 4

D 3 H 6 I 9 - 7

E 5 I 2 J 8 - 4

(a) Develop the project network(b) Find the critical path:(c) If activity B takes 6 days instead of two, what change(s) in project completion occurs?d) If there are 10 workers available, attach a resource-feasible schedule and its profile.

(a) The Project Network

1 2

3

4

7

6

5

8A7

B2

D3

H5

C4

E5

F6

G1 J8

I9

Critical Path

1 2

3

4

7

6

5

8A7

B2

D3

H5

C4

E5

F6

G1 J8

I9ES=0

ES=7

ES=9

ES=11

ES=12

ES=12

ES=17

EF=26

Critical Path

1 2

3

4

7

6

5

8A7

B2

D3

H5

C4

E5

F6

G1 J8

I9ES=0

ES=7

ES=9

ES=11

ES=12

ES=12

ES=17

LF=26

LS=21

LS=18

EF=26

LS=7LS=17

LS=18

LS=17LS=11

LS=10

LS=0LS=12

ES=11

ES=7

(b) Find Critical PathActivity Early start Early finish Late start Late finish slack

A 0 7 0 7 0

B 7 9 10 12 3

C 7 11 7 11 0

D 9 12 18 21 9

E 9 14 12 17 3

F 11 17 11 17 0

G 11 12 17 18 6

H 12 17 21 26 9

I 17 26 17 26 0

J 12 20 18 26 6

A - C - F - I

(c) Activity B takes 6 days Change Description

1 Completion time becomes 27 days

2 New critical path is A-B-E-I

3 Activities C and F now have 1 day slack

1 2

3

4

7

6

5

8A7

B6

D3

H5

C4

E5

F6

G1 J8

I9

(d) 10 workers available, attach a

resource-feasible schedule and its profile.

5 10 15 20 25 34

10

5

A

B

C

D

Act day work

A 7 B,C 5

B 2 D,F 5

C 4 E,G 4

D 3 H 6

E 5 I 2

F 6 I 3

G 1 J 1

H 5 - 4

I 9 - 7

J 8 - 4

E

G

F

7 11 17

H

9 12 17 26

I

J

A Project Network Problem A guidance and detection system is being built as

part of a large defense project. The detection portion consists of radar and sonar subsystems. Separate equipment is required for each of the subsystems. In each case, the equipment must be calibrated prior to production. After production, each subsystem is tested independently. The radar and sonar are combined to form the detection system, which also must be tested prior to integration with the guidance system. The final test of the entire system requires complex equipment. Relevant data is given below:

Relevant Data

Activity Description Time (days)A Calibrate machine 1 (for radar) 2.0B Calibrate machine 2 (for sonar) 3.5C Calibrate machine 1 (for guidance) 1.5D Assemble and prepare final test gear 7.0E Make radar subsystem 4.5F Make sonar subsystem 5.0G Make guidance subsystem 4.5H Test radar subsystem 2.0I Test sonar subsystem 3.0J Test guidance subsystem 2.0K Assemble detection subsystem (radar and sonar) 1.5L Test detection subsystem 2.5M Final assembly of three subsystems 2.5N Testing of final assembly 3.5

Develop the project network

9 10 1211I

6 831

G4 7

E2 5

A

B

C

F K L M N

D

J

H

Earliest Earliest Latest Latest Activity Start Finish Start Finish Slack Critical  

A 0 2.0 3.0 5.0 3.0 B 0 3.5 0 3.5 0 * C 0 1.5 7.5 9.0 7.5 D 0 7.0 11.0 18.0 11.0 E 2.0 6.5 5.0 9.5 3.0 F 3.5 8.5 3.5 8.5 0 * G 1.5 6.0 9.0 13.5 7.5 H 6.5 8.5 9.5 11.5 3.0 I 8.5 11.5 8.5 11.5 0 * J 6.0 8.0 13.5 15.5 7.5 K 11.5 13.0 11.5 13.0 0 * L 13.0 15.5 13.0 15.5 0 * M 15.5 18.0 15.5 18.0 0 * N 18.0 21.5 18.0 21.5 0 *

 

Develop the project network

9 10 1211I

6 831

G4 7

E2 5

A

B

C

F K L M N

D

J

H

Solutions

(b) Find the critical path B-F-I-K-L-M-N (21.5 DAYS)

(c) How much time (in days) is available for assembling and calibrating the final test gear without delaying the project?

11 days (slack) or 18 total days

(d) What are the activities that must be completed by the end of 10 days to guarantee that the project is not delayed?

A, B, C, E, F

(e) What are the activities that must be started by the end of 10 days to guarantee that the project is not delayed?

A, B, C, E, F, G, H, I

Earliest Earliest Latest Latest Activity Start Finish Start Finish Slack Critical  

A 0 2.0 3.0 5.0 3.0 B 0 3.5 0 3.5 0 * C 0 1.5 7.5 9.0 7.5 D 0 7.0 11.0 18.0 11.0 E 2.0 6.5 5.0 9.5 3.0 F 3.5 8.5 3.5 8.5 0 * G 1.5 6.0 9.0 13.5 7.5 H 6.5 8.5 9.5 11.5 3.0 I 8.5 11.5 8.5 11.5 0 * J 6.0 8.0 13.5 15.5 7.5 K 11.5 13.0 11.5 13.0 0 * L 13.0 15.5 13.0 15.5 0 * M 15.5 18.0 15.5 18.0 0 * N 18.0 21.5 18.0 21.5 0 *

 

(d) Completed by end of 10 days

Earliest Earliest Latest Latest Activity Start Finish Start Finish Slack Critical  

A 0 2.0 3.0 5.0 3.0 B 0 3.5 0 3.5 0 * C 0 1.5 7.5 9.0 7.5 D 0 7.0 11.0 18.0 11.0 E 2.0 6.5 5.0 9.5 3.0 F 3.5 8.5 3.5 8.5 0 * G 1.5 6.0 9.0 13.5 7.5 H 6.5 8.5 9.5 11.5 3.0 I 8.5 11.5 8.5 11.5 0 * J 6.0 8.0 13.5 15.5 7.5 K 11.5 13.0 11.5 13.0 0 * L 13.0 15.5 13.0 15.5 0 * M 15.5 18.0 15.5 18.0 0 * N 18.0 21.5 18.0 21.5 0 *

 

(e) Started within 10 days

Discrete LocationThe Hi N. Mitey Company has recently purchased three new

copying machines and has identified five possible locations for the machines. A survey was conducted to determine the usage each department would make of each machine. The following data applies:

DepartmentMachine

1 2 3 4 5

A 5 3 1 8 0B 1 4 0 3 6C 0 3 2 8 7

DepartmentSite 1 2 3 4 5U 8 4 5 3 1V 1 6 8 5 3X 9 8 4 1 4Y 5 3 8 6 1Z 4 4 1 7 8

A = Trips per day

B = Distances in tenth of a mile

solution

Site U V X Y ZA 81 71 81 90 89

Machine B 39 58 68 41 89C 53 95 68 80 126

Site U V X Y Z cost = 165A 0 1 0 0 0 1B 0 0 0 1 0 1C 1 0 0 0 0 1

1 1 0 1 0

Tenths of a mile per day

C = ABt

A Comparison

Site U V X Y Z cost = 265A 0 0 1 0 0 1

MachineB 0 0 0 0 1 1C 0 1 0 0 0 1

0 1 1 0 1

Site U V X Y Z cost = 165A 0 1 0 0 0 1B 0 0 0 1 0 1C 1 0 0 0 0 1

1 1 0 1 0

optimal

Management’s proposal:

Cost = 26.5 – 16.5 = 10 miles per day x 250 days/year = 2500 miles

Location

Given the data below, solve a two facility (m = 2) rectilinear distance problem where v12 = 3.

i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16Wi,1 9 7 4 3 2 12 4 5 11 17 14 6 5 8 15 4wi,2 5 16 0 3 4 1 1 22 24 18 17 8 6 0 4 3ai 0 10 8 12 4 18 4 5 14 19 20 14 3 6 9 10bi 0 3 8 20 9 16 1 3 6 0 4 25 14 6 21 10

Formulation:

MIN 3 C12 + 3 D12 + 9 E11 + 9 F11 + 7 E21 + 7 F21 + 4 E31 + 4 F31

+ 3 E41 + 3 F41 + 2 E51 + 2 F51 + 12 E61 + 12 F61 + 4 E71 + 4F71 + 5 E81 + 5 F81 + 11 E91 + 11 F91 + 17 E101 + 17 F101 + 14 E111 + 14 F111 + 6 E121 + 6 F121 + 5 E131 + 5 F131 + 8 E141 + 8 F141 + 15 E151 + 15 F151 + 4 E161 + 4 F161 + 5 E12 + 5 F12 + 16 E22 + 16 F22 + 3 E42 + 3 F42 + 4 E52 + 4 F52 + E62 + F62 + E72 + F72 + 22 E82 + 22 F82 + 24 E92 + 24 F92 + 18 E102 + 18 F102 + 17 E112 + 17 F112 + 8 E122 + 8 F122 + 6 E132 + 6 F132 + 4 E152 + 4 F152 + 3 E162 + 3 F162

2) - C12 + D12 + X1 - X2 = 0

3) - E11 + F11 + X1 = 0 4) - E21 + F21 + X1 = 10 5) - E31 + F31 + X1 = 8 6) - E41 + F41 + X1 = 12 7) - E51 + F51 + X1 = 4 8) - E61 + F61 + X1 = 18 9) - E71 + F71 + X1 = 4 10) - E81 + F81 + X1 = 5 11) - E91 + F91 + X1 = 14 12) - E101 + F101 + X1 = 19 13) - E111 + F111 + X1 = 20 14) - E121 + F121 + X1 = 14 15) - E131 + F131 + X1 = 3

16) - E141 + F141 + X1 = 6

17) - E151 + F151 + X1 = 9 18) - E161 + F161 + X1 = 10 19) - E12 + F12 + X2 = 0 20) - E22 + F22 + X2 = 10 21) - E32 + F32 + X2 = 8 22) - E42 + F42 + X2 = 12 23) - E52 + F52 + X2 = 4 24) - E62 + F62 + X2 = 18 25) - E72 + F72 + X2 = 4 26) - E82 + F82 + X2 = 5 27) - E92 + F92 + X2 = 14 28) - E102 + F102 + X2 = 19 29) - E112 + F112 + X2 = 20 30) - E122 + F122 + X2 = 14 31) - E132 + F132 + X2 = 3 32) - E142 + F142 + X2 = 6 33) - E152 + F152 + X2 = 9 34) - E162 + F162 + X2 = 10

constraints

X-Solution:OBJECTIVE FUNCTION VALUE 1)

1392.000 Nonzero values only: VARIABLE VALUE REDUCED

COST D12 2.000000 0.000000 E11 12.000000 0.000000 E21 2.000000 0.000000 E31 4.000000 0.000000 E51 8.000000 0.000000 F61 6.000000 0.000000 E71 8.000000 0.000000 E81 7.000000 0.000000 F91 2.000000 0.000000 F101 7.000000 0.000000

F111 8.000000 0.000000

F121 2.000000 0.000000

E131 9.000000 0.000000

E141 6.000000 0.000000

E151 3.000000 0.000000

E161 2.000000 0.000000

E12 14.000000 0.000000

E22 4.000000 0.000000

E32 6.000000 0.000000

E42 2.000000 0.000000

E52 10.000000 0.000000

F62 4.000000 0.000000

E72 10.000000 0.000000

E82 9.000000 0.000000

F102 5.000000 0.000000

F112 6.000000 0.000000

E132 11.000000 0.000000

E142 8.000000 0.000000

E152 5.000000 0.000000

E162 4.000000 0.000000

X1 12.000000 0.000000

X2 14.000000 0.000000

Y-Solution: VARIABLE VALUE REDUCED COST C12 2.000000 0.000000 E11 6.000000 0.000000 E21 3.000000 0.000000 F31 2.000000 0.000000 F41 14.000000 0.000000 F51 3.000000 0.000000 F61 10.000000 0.000000 E71 5.000000 0.000000 E81 3.000000 0.000000 E101 6.000000 0.000000 E111 2.000000 0.000000 F121 19.000000 0.000000 F131 8.000000 0.000000 F151 15.000000 0.000000

E12 4.000000 0.000000 E22 1.000000 0.000000 F32 4.000000 0.000000 F42 16.000000 0.000000 F52 5.000000 0.000000 F62 12.000000 0.000000 E72 3.000000 0.000000 E82 1.000000 0.000000 F92 2.000000 0.000000 E102 4.000000 0.000000 F122 21.000000 0.000000 F132 10.000000 0.000000 F142 2.000000 0.000000 F152 17.000000 0.000000 F162 6.000000 0.000000 Y1 6.000000

0.000000 Y2 4.000000

0.000000(x1,y1) = (12,6)(x1,y1) = (14,4)

The Euclidean Solution Euclidean Distance weighted Euclideani Wi,1 wi,2 ai bi facility 1 facility 2 facility 1 facility 2

1 9 5 0 0 14.17729 14.3714 127.5956 71.8572 7 16 10 3 4.582724 4.162883 32.07907 66.606133 4 0 8 8 4.505032 5.7268 18.02013 04 3 3 12 20 13.08526 14.41097 39.25578 43.23295 2 4 4 9 8.628129 9.811246 17.25626 39.244996 12 1 18 16 10.68178 11.40976 128.1814 11.409767 4 1 4 1 10.25504 10.32039 41.02016 10.320398 5 22 5 3 8.350938 8.632006 41.75469 189.90419 11 24 14 6 1.868544 0.860455 20.55398 20.65092

10 17 18 19 0 9.581 8.077488 162.877 145.394811 14 17 20 4 8.166247 6.97743 114.3275 118.616312 6 8 14 25 18.15292 19.37517 108.9175 155.001313 5 6 3 14 11.74697 13.20192 58.73483 79.2115514 8 0 6 6 6.439747 7.227134 51.51798 015 15 4 9 21 14.47847 15.92809 217.177 63.7123416 4 3 10 10 3.888482 5.418581 15.55393 16.25574

V12 = 3 sums 148.5886 155.9117 1194.823 1031.418facility 1 facility 2 Fac1Fac2 1.533045 4.599136

x y x y 12.37 6.92 13.218 5.64 totals 306.0333 2230.84

Equipment Replacement

6. A complex machine used in the manufacture of brake pads has an instantaneous repair cost rate given by C(u) = 10u2 where u is measured in years. The cost of replacing the machine is $25,000. It has a salvage value of $5,000 regardless of its age at replacement. Determine the replacement age that will minimize the average annual repair cost.

2 2

0

25,000 1 5000 25,000 10 5000( ) 10

3

t

G t u du tt t t t t

0

1 ( )( ) ( )

tK S tG t C u du

t t t

The Solutiont G(t)

10 2,333 11 2,222 12 2,147 13 2,102 14 2,082 15 2,083 16 2,103 17 2,140 18 2,191 19 2,256 20 2,333

t G(t)14.4225 2,080

Trial and error

Solver

Repair or Replace?

7. A fuel pump used in a jet engine if repaired will incur a fixed cost of $500 for a special set of tools and a variable cost of $330 per failure. If discarded and replaced there is no fixed cost but a variable cost of $55. A condemnation rate of 5 percent is anticipated. Determine the indifference curve between repairing and replacing the pump.

Repair or Replace?repair cost

Discard cost

repair cost = discard cost

5000 330 (.05)r ra b f ckf f c f

0 55d da cf b f c f f

5000 330 (.05) 55

(.05) 55 5000 330

275 5000

.95

f c f cf f

c f f f f

fc

f

Repair or Replace?

repair

discard

275 5000

.95

fc

f

f

c

7 (b)

If repair results in a failure rate given by (t) = .05 t1.9 with t measured in years and the life of the engine is estimated to be 20 years, what is the largest unit cost for which repair is still optimal?

2020 2.91.9

0 0

.005[ (20)] .005 10.22

2.9

tE N t dt

275 5000

.95

fc

f

275 10.22 5000$804.46

.95 10.22c

8. When to replace?

Assume the jet engine pump which costs $950 is (minimally) repaired upon failure with the failure rate as given in problem 7. If the cost per repair is $330, when should the pump be replaced?

1/ 1/2.9950

* 4.68 yr.( 1) 330(.05 / 2.9)(2.9 1)

b

u

f

Ct

C a b

(t) = abtb-1 = .05 t1.9

b = 2.9 and a = .05/2.9 = .017241379

Block replacement

Kettering Labs has 420 florescent light fixtures each containing 2 light bulbs.

When a bulb fails, a maintenance person must be dispatched and the University is charged for one hour of labor at the rate of $78 (fully burdened) per hour.

The University pays 3 dollars for each light bulb (large quantity discount).

An engineering case study has determined that all 840 lights could be replaced by a single maintenance person in 8 hours.

Given the following fraction of failures each month, determine the optimum block replacement strategy and the savings that will be incurred.

Fraction failing each month

nbr mo. fraction failing1 0.012 0.013 0.034 0.035 0.036 0.047 0.058 0.129 0.13

10 0.1811 0.1812 0.19

sums 1

a1 = $78 + $3 = $81 per failure

a2 = $3 x 840 + 8 x $78 = $3,144

1

3144 81

Min ( )

k

jj

n

G kk

Solution…nbr mo.

fraction failing 1 2 3 4 5 6 7 8 9 10 11 12 nj cumulative Gk

1 0.01 8.4 8.4 8.43824.

4

2 0.01 8.4 0.08 8.5 16.92255.

8

3 0.03 25.2 0.08 0.1 25.4 42.32188.

8

4 0.03 25.2 0.25 0.1 0.3 25.8 68.02163.

9

5 0.03 25.2 0.25 0.3 0.3 0.3 26.2 94.32155.

8

6 0.04 33.6 0.25 0.3 0.8 0.3 0.3 35.4 129.62274.

3

7 0.05 42 0.34 0.3 0.8 0.8 0.3 0.4 44.7 174.42467.

1

8 0.12 100.8 0.42 0.3 0.8 0.8 0.8 0.4 0.4 104.7 279.13218.

6

9 0.13 109.2 1.01 0.4 1.0 0.8 0.8 1.1 0.4 1.0 115.8 394.83902.

9

10 0.18 151.2 1.09 1.0 1.3 1.0 0.8 1.1 1.3 1.0 1.2 161.0 555.84816.

7

11 0.18 151.2 1.51 1.1 3.0 1.3 1.0 1.1 1.3 3.1 1.2 1.6 167.5 723.45612.

3

12 0.19 159.6 1.51 1.5 3.3 3.1 1.3 1.4 1.3 3.1 3.5 1.6 1.7 183.0 906.36379.

9sums 1 840

Savings = $7,419.85 - $2,155.80 = $5,264.05 per month

1

3144 81

Min ( )

k

jj

n

G kk

0 1840 81

Cost per time period $7,419.85( ) 9.17

n a

E T

E(T) = 9.17 or 1/9.17 = .10905 failures per month