state l university 3 rd edition applied management science for decision making, 2e © 2014 pearson...

STATE l UNIVERSITY

3rd EDITION

Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro , PhD

Multiple ChoiceMultiple Choice

1. Intermittent flow operations: a. involve close management of workers. b. generate a high product mix. c. have few schedule changes.have few schedule changes. d. have high turnovers of raw materials and work-in-process inventories.

2. Continuous flow operations:

a. utilize general-purpose equipment.utilize general-purpose equipment. b. are capital-intensive operations. c. have unpredictable material flows. d. have work centers grouped together by function ( department ).


3. Repetitive manufacturing:

a. assembles pre-made parts into finished goods. b. produces one basic product with minor variations. c. Is characterized by very low equipment utilization.Is characterized by very low equipment utilization. d. utilizes highly trained, flexible labor.

4. Design capacity is:

a. sustainable capacity.sustainable capacity. b. capacity dictated by the firm’s built-in organizational constraints. c. theoretical capacity. d. actual output.


5. The stepping-stone method is used to:

a. obtain an initial ( 1st ) solution. b. balance source supply and destination demand. c. evaluate occupied cells ( routes ) for possible costevaluate occupied cells ( routes ) for possible cost reductions.reductions. d. evaluate empty cells ( routes ) for possible cost reductions.

6. Which of the following statements is true?

a. evaluation paths selected must be as short as possible. b. evaluation paths may contain diagonal movements. c. occupied cells may not be bypassed if turning move-occupied cells may not be bypassed if turning movements are desired.ments are desired. d. all occupied cells of each solution must be evaluated.


7. The intent of the load-distance model is:

a. to minimize the number of workers. b. to minimize idle time per cycle. c. to minimize materials and/or information movementto minimize materials and/or information movement costs within a process layout.costs within a process layout. d. none of the above.d. none of the above.

8. Which of the statements about line-balancing is true?

a. if a precedence relationship exists between tasks A and B, they cannot be assigned to the same work station. b. if a line’s balance delay factor is maximized, its efficiencyif a line’s balance delay factor is maximized, its efficiency is maximized.is maximized. c. the theoretical minimum number of work stations can never be achieved, hence the name “theoretical”. d. if the cycle time is reduced, productivity will increase.


9. Hybrid or combination layouts:

a. were developed by Henry Gantt in 1901.were developed by Henry Gantt in 1901. b. require the use of work-in-process inventories between the pure layouts. c. are similar in their characteristics to fixed-position layouts. d. are rarely encountered in industry.

10. Which of the following statements is true?

a. line-balancing attempts to eliminate bottleneck tasks via product redesign and better worker training. b. the efficiency of an assembly line cannot be improved by selecting a different task assignment heuristic for the line-balancing process. c. an assembly line is perfectly balanced when there are an equal number of tasks in each work station. d. “efficiency” is defined as an assembly line’s ability to meet “efficiency” is defined as an assembly line’s ability to meet desired daily output.desired daily output.


11. The Behavioral School approach to job design which attempts to make a worker a “co-manager” is:

a. job enlargement.job enlargement. b. b. job enhancement. c. c. job rotation. d. job enrichment.

12.12. The 5 categories of activities ( operation, transportation, inspection, delay, and storage ) are used in which of the following methods of process analysis?

a. a. gang chart. b. b. simo chart. c. multiple activity chart.c. multiple activity chart. d. d. motion economy chart.


13. The procedure that involves performance ratings and allowance factors is :

a. historical experience.historical experience. b. work sampling. c. direct time study. d. pre-determined time study.

True or FalseTrue or False

14. In cross-docking, labeled and presorted loads are received directly at the warehouse dock for immediate re-routing.

TRUETRUE FALSE

15. A job shop helps a firm follow a differentiation marketing strategy.

TRUETRUE FALSE

16. Group technology layouts are used to convert assembly lines into job shops.

TRUE TRUE FALSE


17.17. The Northwest-Corner technique produces a deliberate cost- efficient solution for the transportation algorithm.

TRUE FALSEFALSE

18.18. Work sampling is widely used to analyze repetitive jobs.

TRUE TRUE FALSE

19.19. The maximum allowable cycle time guarantees that the firm will meet its daily production quota.

TRUETRUE FALSEFALSE


20. The direct time study method makes allowances for unscheduled interruptions, unusual delays, and unusual mistakes on the part of the worker.

TRUE FALSEFALSE

21. In the transportation problem, the number of cells in an evaluation path should always be an odd number such as “5”, ”7”, or “9”.

TRUETRUE FALSE

22. One of the limitations of the transportation algorithm is that it cannot minimize shipping costs between two levels of the supply chain.

TRUETRUE FALSE

True or FalseTrue or False Nine ( 9 ) welders who perform the identical short-cycle job were observed by a time and motion engineer over six ( 6 ) cycles each. The total time recorded was three-hundred-ten ( 310 ) minutes. The performance rating for a particular welder was established at ninety-two percent ( 92% ). Additionally, each welder is granted a twelve percent ( 12% ) allowance for personal needs, fatigue, and routine delays.

23. The observed time ( OT ) for this job is 5.833 minutes. TRUETRUE FALSE

24. The normal time ( NT ) for the selected welder ≈ 5.280 minutes. TRUE FALSEFALSE

25. The standard time ( ST ) for the selected welder ≈ 6.000 minutes.

TRUE FALSEFALSE

True or FalseTrue or False Nine ( 9 ) welders who perform the identical short-cycle job were observed by a time and motion engineer over six ( 6 ) cycles each. The total time recorded was three-hundred-ten ( 310 ) minutes. The performance rating for a particular welder was established at ninety-two percent ( 92% ). Additionally, each welder is granted a twelve percent ( 12% ) allowance for personal needs, fatigue, and routine delays.

observed time ( OT ) = 310 / ( 9 x 6 ) = 5.7407

normal time ( NT ) = 5.7407 x .92 = 5.2814

standard time ( ST ) = 5.284 / ( 1 - .12 ) = 6.0045

Assembly Line Balancing ProblemAssembly Line Balancing Problem

A firm desires to achieve a production rate of seventy-five ( 75 ) units per six ( 6 ) hour day, utilizing an assembly line. The table below lists the nine ( 9 ) tasks that must be performed to produce each unit as well as the sequence in which they must be performed.

Problem 1


Task Predecessor ( s ) Time ( in seconds )

A none 60

B A 60

C A 120

D C 60

E C 180

F C 60

G D,E,F 60

H B 120

I G,H 60

Assembly Line Balancing ProblemAssembly Line Balancing ProblemFor your convenience, a precedence relationship ( s ) diagram has beenincluded below. The tasks are shown as nodes. The sequence require-ments are shown by arrows. The required standard task times are shownwithin each task’s respective node:

A60

B60

C120

D60

E180

F60

H120

I60

G60


REQUIREMENT:

A. The maximum allowable cycle time.

B. The theoretical minimum number of work stations.

C. The balanced assembly line under the longest operating time ( LOT ) assignment heuristic.

D. The balance delay factor of the assembly line.

E. The efficiency of the assembly line.

Problem 1


MaximumAllowable

Cycle Time=

Total available production time

Daily production quota

= 6 hours x 60 x 60

75 units

= 21,600 seconds

75 units

= 288 seconds


26. In problem one, the maximum allowable cycle time is:

a. 288 seconds.288 seconds. b. 315 seconds. c. 348 seconds. d. 384 seconds.


TheoreticalMinimum

Number ofWork Stations

=Total time to produce one unit

Maximum allowable cycle time

=780 seconds

288 seconds= 2.7083 ≈ 3


27. In problem one, the theoretical number of work stations is:

a. 11 b. 2 c. 3 d. 4


A60

B60

C120

D60

E180

F60

H120

I60

G60

CANDIDATE(S) FOR TASK 1 IN STATION 1


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60)

Production Time

Idle Time

Balancing under the LOT assignment heuristicBalancing under the LOT assignment heuristic


A60

B60

C120

D60

E180

F60

H120

I60

G60


XX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60)

C (120)

Production Time

Idle Time

Balancing under the LOT assignment heuristic


A60

B60

C120

D60

E180

F60

H120

I60

G60


XXXX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60)

C (120)

B (60)

Production Time 240

Idle Time 48



A60

B60

C120

D60

E180

F60

H120

I60

G60


XX

XX

XX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60) E (180)

C (120)

B (60)

Production Time 240Idle Time 48



A60

B60

C120

D60

E180

F60

H120

I60

G60


XX

XX

XX XX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60) E (180)

C (120) D (60) D (60)

B (60)

Production Time 240 240240Idle Time 48 4848



A60

B60

C120

D60

E180

F60

H120

I60

G60


XX

XX

XXXX

XX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60) E (180) H (120)

C (120) D (60) D (60)

B (60)




A60

B60

C120

D60

E180

F60

H120

I60

G60


XXXX

XXXX

XX

XX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60) E (180) H (120)

C (120) D (60) D (60) F (60)

B (60)




A60

B60

C120

D60

E180

F60

H120

I60

G60


XX

XX

XXXX

XX

XX

XX


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60) E (180) H (120)

C (120) D (60) D (60) F (60)

B (60) G (60)

Production Time 240240 240240 240240Idle Time 48 4848 4848



A60

B60

C120

D60

E180

F60

H120

I60

G60


XX

XX

XX

XX

XX

XX

XX

XX


Cycle time

288

seconds

288

seconds

288

seconds

288

seconds

1,152

seconds

Station

1

Station

2

Station

3

Station

4

Total

A (60) E (180) H (120) I (60)

C (120) D (60) D (60) F (60)

B (60)

G (60)

Production Time

240 240 240 60 780

Idle Time 48 4848 4848 228228 372



Cycle time

288

seconds

288

seconds

288

seconds

288

seconds

1,152

seconds

Station

1

Station

2

Station

3

Station

4

Total

A (60) E (180) H (120) I (60)

C (120) F (60) F (60) D (60)

B (60)

G (60)

Production Time

240 240 240 60 780

Idle Time 48 4848 4848 228228 372


2nd SOLUTION

Assembly LineAssembly Line Balancing Problem Balancing Problem

LineEfficiency =

240 sec + 240 sec + 240 sec + 60 sec

288 sec x 4 stations

= 780 seconds

1,152 seconds

= 67.71%


28. In problem one, the efficiency of the assembly line is:

a. 64%64% b. 68% c. 70% d. 72%

Assembly LineAssembly Line Balancing Problem Balancing Problem

BalanceDelayFactor

=48 sec + 48 sec + 48 sec + 228 sec

288 sec x 4 stations

= 372 seconds

1,152 seconds

= 32.29%


29. In problem one, the balance delay factor is:

a. 28%28% b. 30% c. 32% d. 36%


30. In problem one, the tasks assigned to work station ‘one’ are:

a. A,CA,C b. A,C,E c. A,C,D d. none of the above.


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60)

C (120)

B (60)

Production Time 240

Idle Time 48



31. In problem one, the tasks assigned to work station ‘two’ are:

a. E,DE,D b. E,F c. E,D or E,F d. F,B,H,D


Cycle time

288

seconds

288

seconds

288

seconds

864

seconds

Station

1

Station

2

Station

3

Total

A (60) E (180)

C (120) D (60) D (60)

B (60)




Cycle time

288

seconds

288

seconds

288

seconds

288

seconds

1,152

seconds

Station

1

Station

2

Station

3

Station

4

Total

A (60) E (180) H (120) I (60)

C (120) F (60) F (60) D (60)

B (60)

G (60)

Production Time

240 240 240 60 780

Idle Time 48 4848 4848 228228 372


2nd SOLUTION

Transportation ProblemTransportation Problem

WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

1010 2020 4040

3030

FROM

TOProblem 2


REQUIREMENT:

Evaluate the empty cells via the stepping-stone technique.

a. A-3b. B-1c. B-3

1. Do not attempt to evaluate any other empty cells2. Do not attempt a transfer operation3. Show all supporting calculations via “T” accounts

Problem 2


WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

1010 2020 4040

3030

FROM

TO

EVALUATE EMPTY CELL A-3

+

+

-

-

Transportation ProblemTransportation ProblemEVALUATE EMPTY ROUTE A-3

A3 + $25 C3 - $21

C2 + $16 A2 - $22

+ $41 - $43

- $ 2

GROSSINCREASE

GROSSDECREASE

NET DECREASE


32. In problem two, the evaluation number for empty cell ‘A-3’ is:

a. - 4- 4 b. + 4 c. -10 d. +10 e. none of the above.


WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

1010 2020 4040

3030

FROM

TO

EVALUATE EMPTY CELL B-1

+

+

-

-

Transportation ProblemTransportation ProblemEVALUATE EMPTY ROUTE B-1

B1 + $24 A1 - $19

A2 + $22 B2 - $21

+ $46 - $40

+ $ 6

GROSSINCREASE

GROSSDECREASE

NET INCREASE


33. In problem two, the evaluation number for empty cell ‘B-1’ is:

a. - 9- 9 b. + 9 c. - 7 d. + 7 e. none of the above.


WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

1010 2020 4040

3030

FROM

TO

EVALUATE EMPTY CELL B-3

+

+ -

-

Transportation ProblemTransportation ProblemEVALUATE EMPTY ROUTE B-3

B3 + $19 C3 - $21

C2 + $16 B2 - $21

+ $35 - $42

- $ 7

GROSSINCREASE

GROSSDECREASE

NET DECREASE


34. In problem two, the evaluation number for empty cell ‘B-3’ is:

a. - 4- 4 b. + 4 c. - 3 d. + 3 e. none of the above.

Transfer Operation ProblemTransfer Operation Problem

WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

1010 2020 4040

3030

FROM

TO

Given the followingGiven the following transportation algorithm intermediate solution::

Problem 3


REQUIREMENT:

1. Perform a transfer operation on empty cell A-4.2. What are the new allocations for A-4’s evaluation path?

DO NOT COMPUTE THE NEXT INTERMEDIATE SOLUTION OR ITS COSTS

Problem 3


WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

1010 2020 4040

3030

FROM

TO

CELL A - 4 TRANSFER OPERATION

+

+

-

-


35. In problem three, the evaluation number for empty cell ‘A-4’ is:

a. + 2+ 2 b. - 32 c. + 15 d. - 15 e. none of the above.


WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

3030 2020 2020

3030

FROM

TO

CELL A - 4 TRANSFER OPERATION

+

+

-

-


WAREHOUSE

1

WAREHOUSE

2

WAREHOUSE

3

WAREHOUSE

4

PLANT

AVAILABILITY

PLANT APLANT A 50

PLANT BPLANT B 30

PLANT CPLANT C 70

WAREHOUSE

DEMAND30 60 20 40

$19$19 $22$22 $25$25 $17$17

$24$24 $21$21 $19$19 $21$21

$29$29 $16$16 $21$21 $26$26

150

150

2020

3030

3030 2020 2020

3030

FROM

TO

NEW ALLOCATIONS FOR CELL A-4 EVALUATION PATH


36. In problem three, the new allocation for cell ‘A-4’ is:

a. 15 b. 20 c. 25 d. 30 e. none of the above.


37. In problem three, the new allocation for cell ‘C-2’ is:



38. In problem three, the new allocation for cell ‘A-2’ is:


Gravity Location ProblemGravity Location ProblemA garden center has three ( 3 ) retail locations in Massachusetts.A garden center has three ( 3 ) retail locations in Massachusetts.The firm wants to construct a new central distribution center forThe firm wants to construct a new central distribution center forservicing its three retail sites:servicing its three retail sites:

RetailRetail

LocationLocation

Map CoordinatesMap Coordinates

( X , Y )( X , Y )

Daily Truck Daily Truck

Round TripsRound Trips

Acton ( 19,10 ) 6

Haverhill ( 24,14 ) 8

Northampton ( 7,6 ) 10

Problem 4

Gravity Location ProblemGravity Location Problem

REQUIREMENT:

A. Compute the location of the proposed facility in terms of the ‘x’ and ‘y’ coordinates.

NOTE: 1. Do not attempt to draw a graph. 2. The supporting calculations are sufficient.

Problem 4


( 19 x 6 ) + ( 24 x 8 ) + ( 7 x 10 )

6 + 8 + 10

114 + 192 + 70

24

376

24

X =

= 15.67

0 5 10 15 20 250 5 10 15 20 25

XX++


39. In problem four, the ‘X’ coordinate for the new central distribution center is:

a. 9.67 b. 14.00 c. 15.67 d. 30 e. none of the above.

Gravity LocaGravity Location Problemtion Problem

( 10 x 6 ) + ( 14 x 8 ) + ( 6 x 10 )

6 + 8 + 10

60 + 112 + 60

24

232

24

Y =

= 9.67

0 5 10 15 20 250 5 10 15 20 25XX

++1515

1010

55

00

YY


40. In problem four, the ‘Y’ coordinate for the new central distribution center is:

a. 9.67 b. 14.00 c. 15.67 d. 30 e. none of the above.


Y

X0 5 10 15 20 25 30 35

15

10

5

0

NORTHAMPTONNORTHAMPTON

ACTONACTON

HAVERHILLHAVERHILL

++THE NEW

DISTRIBUTIONCENTER

( 15.67, 9.67 )

++


41. The sources of a transportation matrix are either factories or outside vendors.

TRUE FALSE

42. Evaluation numbers are either ‘positive’ or ‘negative’.

TRUE FALSE

43. When total source units exceed total destination units, an additional row must be included on the transportation matrix.

TRUE FALSE


44. Qualitative factors are considered in the use of the transportation algorithm by the operations staff when locating retail facilities.

TRUE FALSE

45. The gravity location model uses shipping costs to locate a central distribution center. TRUE FALSE

46.The gravity location model locates the central distribution center at the geo- graphic center of the distribution system.

TRUE FALSE


47. The transshipment model can accommodate multiple levels of the supply chain simultaneously. TRUE FALSE

48. The transshipment model can accommodate multiple modes of transport. TRUE FALSE

49. The transshipment model utilizes a sophisticated version of the transportation algorithm.

TRUE FALSE

50. The gravity location model draws part of its input from the current monthly or quarterly forecasted demands at each facility in the distribution system.

TRUE FALSE

state l university 3 rd edition applied management science for decision making, 2e © 2014 pearson...

Documents

multiple choice

occupied cells routes

phd slide

theoretical capacity

sustainable capacity

possible cost reductions

process layout

process inventories