06_ReviewProblemSolution[1]

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Review Problems and Solutions for

Chapter 6: Process Selection and Facility Layout

For the following three problems (1, 2, 7), we assume that parallel workstations are

not allowed.

1. An assembly line with 17 tasks is to be balanced. The longest task is 2.4 minutes, andthe total time for all tasks is 18 minutes. The line will operate for 450 minutes per day.

a. What are the minimum and maximum cycle times?b. What range of output is theoretically possible for the line?c. What is the minimum number of workstations needed if the maximum output

rate is to be sought?d. What cycle time will provide an output rate of 125 units per day?e. What output potential will result if the cycle time is (1) 9 minutes? (2) 15

minutes?

Solu t ion :

OT = 450 minutes

a. Minimum cycle time = length of longest task, which is 2.4 minutes.

Maximum cycle time = task times = 18 minutes.

b. Range of output:

units2518

450:.min18@

units5.1874.2

450:.min4.2@

=

=

c. 8toroundswhich,5.7450

)18(5.187OT

tDxN ===

d. cycleperminutes6.3125

450CTCT,forSolving

CT

OTOutput ===

e. Potential output:

(1) units509

450

CT

OT:.min9CT ===

(2) units3015

450:.min15CT ==

2. A manager wants to assign tasks to workstations as efficiently as possible, andachieve an hourly output of 33 units. Assume the shop works a 60-minute hour.Assign the tasks shown in the accompanying precedence diagram (times are inminutes) to workstations using the following rules:

a. In order of most following tasks. Tiebreaker: greatest positional weight.b. In order of greatest positional weight.c. What is the efficiency?


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Solu t ion :

Desired output = 33.33 units per hour

Operating time = 60 minutes per hour

unitperminutes80.1hourperunits33.33

hourperminutes60

outputDesired

timeOperatingCT ===

a.

Task Number of following tasks Positional Weight

A 7 6

B 6 4.6

C 2 1.6

D 2 2.2

E 2 2.3

F 1 1.0

G 1 1.5

H 0 0.5

Assembly Line Balancing Table (CT = 1.8)

Work Station Task Task Time Time RemainingFeasible tasks

Remaining

I A 1.4 0.4 B 0.5 1.3 C, D, EII

E 0.8 0.5

D 0.7 1.1 C

C 0.6 0.5 F

III

F 0.5 0


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G 1.0 0.8 HIV

H 0.5 0.3

b.

Assembly Line Balancing Table (CT = 1.8)


Remaining

I A 1.4 0.4

B 0.5 1.3 C, D, EII

E 0.8 0.5

D 0.7 1.1 C

C 0.6 0.5 F

III

F 0.5 0

G 1.0 0.8 HIV

H 0.5 0.3

c. %3.832.7

0.6

stationsofno.xCT

timeTotalEfficiency ===

7. For the set of tasks given below, do the following:

a. Develop the precedence diagram.b. Determine the minimum and maximum cycle times in seconds for a desired

output of 500 units in a 7-hour day. Why might a manager use a cycle time of50 seconds?

c. Determine the minimum number of workstations for output of 500 units perday.

d. Balance the line using the largest positional weightheuristic. Break ties withthe most following tasksheuristic. Use a cycle time of 50 seconds.

e. Calculate the percentage idle time for the line.


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Solu t ion :

a.

b. ===500

7(60)

D

OTCT .84 minutes = 50.4 seconds (maximum cycle time)

Minimum cycle time = maximum task time = 45 seconds (results in 560 units of

production)

c. stations4or83.34.50

193

CT

tN ==

=

d.

Task Number of followers *PW

A 6 106

B 5 61

C 4 50D 4 106

E 3 56

F 2 30

G 2 31

H 2 29

I 1 19

J 0 10

*Positional weight

CT = 50 seconds


Remaining

I A 45 5

III D 50

III B 11 39 C, E

a b c

ed

h

g

f

i j


5/6

E 26 13 C, F

C 9 4

G 12 38 H, F

F 11 27 H

H 10 17 I

IV

I 9 8

V J 10 40

e. %8.22)5)(50(

1931 ==I

Optional Problem (Parallel workstations are allowed).

Consider the following precedence diagram:

Suppose that OT (operating time per day) = 480 minutes, and our target output rate D is

960 for each day. Balance the line.

Solution:

CT = OT/D = 0.5 minutes. Suppose that we use the number of followers to break ties,

then


Remaining

I a 0.1 0.4

II 0.7 min (task time for c) >0.5 min, new workstations are needed

II & III c 0.7 0.3 -


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IV 1.0 min (task time for b) >0.5 min, new workstations are needed

IV & V b 1.0 0 -

VI d 0.5 0 -

VII e 0.2 0.3 -

Notice that we can use smaller number of workstations. In fact, we can use twoworkstations (instead of three) on tasks a and c, i.e., each of these two work station works

on tasks a and c. In other words, we can treat tasks a and c as one task, with a task cycle

time = 0.8 min < (0.5 min x 2 = 1.0 min). Then it is clear that two workstations are

enough for the task (a and c together).

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06_ReviewProblemSolution[1]