Question

Please solve the following problem related to Chapter 6

1. As part of a major plant renovation project, the industrial

engineering department has been asked to balance a revised

assembly operation to achieve an output of 240 units per eight-

hour day. Task times and precedence relationships are as you

know you can follows:

Task                 Duration (minutes)             Immediate Predecessor

_____________________________________________________________

A                             0.3                               -

B                             0.6                                A

C                              0.3                                B

D                             0.5                                  -

E                              1.2                                  D

F                              0.8                                  C

G                              1.2                                 E,F

                      

Do each of the following:

a. Draw the precedence diagram.

b. Determine the minimum cycle time, the maximum cycle time, and

the calculated cycle time.

c. Determine the minimum number of stations needed.

d. Assign tasks to workstations on the basis of most following

tasks. Use shortest processing time as a tiebreaker. If ties

still exist, assume indifference in choice.

e. Compute the percentage of idle time for the assignment in part d.

[Hint: Please read page 262-264 in the textbook. Please read Example

1 on page 264-266, and Example 2 on page 267-269.]

Please type your answer and the steps of computation in a Word or

Excel file.

Answer 1

(a)

(b)

CT_min = the longest task time = 1.2 minutes
CT_max = sum of all the task times = 4.9 minutes

Calculated cycle time = 8 x 60 / 240 = 2 minutes

(c)

Minimum number of stations = CT_max / Calculated cycle time = 4.9/2 = 2.45 i.e. 3 stations

(d)

Sattions	Task	Duration (minutes)	Time left (minutes)	Ready tasks (# followers)
				A(4),D(2)
1	A	0.3	1.7	D(2),B(3)
	B	0.6	1.1	D(2),C(2)
	C	0.3	0.8	D(2),E(1)
	D	0.5	0.3	E(1),F(1)
2	F	0.8	1.2	E(1)
	E	1.2	0	G(0)
3	G	1.2	0.8

(e)

% idle time = 1 - (Flow time / (# stations x cycle time)) = 1 - (4.9/(3*2)) = 18.33%