EXAMPLE 8.2: Assembly-Line Balancing The Model J Wagon is to be assembled on a c

ID: 427235 • Letter: E

Question

EXAMPLE 8.2: Assembly-Line Balancing The Model J Wagon is to be assembled on a conveyor belt. Five hundred wagons are required per day. Production time per day is 420 minutes, and the assembly steps and times for the wagon are given in Exhibit 8.9A. Assignment: Find the balance that minimizes the number of workstations, subject to cycle time and precedence constraints SOLUTION I. Draw a precedence diagram. Exhibit 8.9B illustrates the sequential relationships identi- fied in Exhibit 8.9A. (The length of the arrows has no meaning.) Determine workstation cycle time. Here we have to convert to seconds because our task times are in seconds. 2. Production time per day 60 sec. X 420 min. Output per day 500 wagons 25.200 504 500 Determine the theoretical minimum number of workstations required (the actual num- ber may be greater): T 195 seconds , C-50.4 seconds 3.87 = 4 (rounded up) 4. Select assignment rules. In general, the strategy is to use a rule assigning tasks that either have many followers or are of long duration because they effectively limit the balance achievable. In this case, we use the following as our primary rule: a. Prioritize tasks in order of the largest number of following tasks. NUMBER OF FOLLOWING TASKS TASK Bor D C or E F G, H. or l Our secondary rule, to be invoked where ties exist from our primary rule, is b. Prioritize tasks in order of longest task time (shown in Exhibit 8.10). Note that D should be assigned before B, and E assigned before C due to this tiebreaking rule.

Explanation / Answer

b) Cycle Time= 60*420/500 = 50.4 sec

c) Minimum number of workstations= sum of total task times / cycle time

= 195/50.4 = 3.872 = 4 workstations

Task Table

Task

Task time

Followers

Followers #

Precedence

B,C,F,G,J,K

C,F,G,J,K

F,G,J,K

E,H,I,J,K

H,I,J,K

J,K

F,G,H,I

The total station time is nothing but the cycle time. So, total station time of each station is 50.4 secs.

In order of longest task time and then on the largest no of following tasks, tasks:

D>A>E>H>I>B>C>F>G>J>K

Workstation 1:

First task =D

Time left= 50.4-50 = 0.4 secs

So, workstation 1: D

Workstation 2:

First task =A

Time left= 50.4-45 = 5.4 sec

So, workstation 2: A

Workstation 3:

First task=E

Time left= 50.4-15= 35.4 sec

Second task=H

Time left=35.4-12 = 23.4 sec

Third task=I

Time left=23.4-12 = 11.4 sec

Fourth task=B

Time left=11.4-11 = 0.4 sec

So, workstation 3: E->H->I->B

Workstation 4:

First task= C

Time left=50.4 – 9= 41.4 secs

Second task=F

Time left=41.4-12 = 29.4 sec

Third task=G

Time left=29.4-12 = 17.4 sec

Fourth task=J

Time left=17.4-8 = 9.4 sec

FIFTH task=K

Time left=9.4-9 = 0.4 sec

So, workstation 4: C->F->G->J->K

Work Station

Tasks

Workstation time

Idle Time

0.4

5.4

E->H->I->B

0.4

C->F->G->J->K

0.4

e) Total idle time = 0.4+5.4+0.4+0.4 = 6.6 sec

f) Efficiency with 4 workstations = (sum of all tasks) / (no of workstations * Cycle time)

= (195)/ (4*50.4) = 0.967 or 96.7 %