3.5 Types of Layout
3.5.3 Repetitive and Product-Oriented Layout
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52 A Modern Approach to Operations Management Two types of product layout are fabrication and assembly lines. The fabrication line builds components (viz. car tires, parts of a refrigerator, etc.) on a series of machines. An assembly line puts the fabricated parts together at a series of workstations. Both are repetitive processes, and in both cases, the line must be ‘balanced’- that is, the time spent to perform work on one machine must equal or
‘balance’ the time spent to perform work on the next machine in the fabrication line.
Assembly lines can be balanced by moving tasks from one individual to another. The central problem then in product layout planning, is to balance the output at each workstation on the production line so that it is nearly the same, while obtaining the desires amount of output. A well-balanced assem- bly line has the advantage of high personnel and facility utilization and equity between employees’
work loads.
3.5.3.1 Assembly-line Balancing
Line-balancing is done to minimize imbalance between machines or personnel while meeting a re- quired output from the line. For this, the management must know the tools, equipment, and work methods used. Then the time needed for each assembly task (e.g. drilling a hole, tightening a nut, or painting a part) must be determined. Management also needs to know the precedence relationship among the activities (i.e. the sequence in which different tasks must be performed). Example-3.2 shows how to turn these task data into a precedence diagram.
Example 3.2. We want to develop a precedence diagram for an electrostatic copier that requires a total assembly time of 66 minutes. Table 3.7 and Figure 3.7 give the tasks, assembly times, and sequence requirements for the copier.
Table 3.7. Precedence data
Task Performance time (minutes) Preceding activity
A 10 —
B 11 A means B cannot be done before A.
C 5 B
D 4 B
E 12 A
F 3 C, D
G 7 F
H 11 E
I 3 G, H
Total time = 66
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A B
C
G I
F
H E
D
10 11
5
7 3
12 11
3 4
Figure 3.7
Once we have constructed a precedence chart summarizing the sequences and performance times, we will concentrate on grouping tasks into job stations so that we can meet the specified production rate. This will consist of the following steps:
• Calculate the cycle time - the maximum time that the product is available at each workstation if the production rate is to be achieved.
Cycle time = (Production time available per day/Units required per day)
• Compute the minimum number of workstations.
minimum number of workstations = Σ i
n
=1
Time for task i / Cycle time (3.2)
where n = number of assembly tasks.
• Balance the line by assigning specific assembly tasks to each workstation. An efficient balance is one that will complete the required assembly, follow the specified sequence, and keep the idle time at each workstation to a minimum. A formal method to do this is:
n Identify a master list of tasks.
n Eliminate the tasks that have been assigned.
n Eliminate the tasks whose precedence relationship has not been satisfied.
n Eliminate the tasks for which inadequate time is available at the workstation.
n Use one of the line-balancing heuristics described in Table 3.8. The five choices are: (i) longest task time, (ii) most following tasks, (iii) ranked positional weight, (iv) shortest time task, and (v) least number of following tasks. It is to be noted that heuristics provide solutions, but they do not guarantee an optimal solution.
Table 3.8. Layout Heuristics for Assembly-line Balancing
(i) Longest task time, From the available tasks, choose the one with the largest time.
(ii) Most following tasks, From the available tasks, choose the one with the largest number of following tasks.
(iii) Ranked positional weight, From the available tasks, choose the one for which the sum of the times for each following task is longest. From example 3.3, we see that the ranked positional weight of task C = 5(C) + 3(F) + 7(G) + 3(I) = 18; task D = 4(D) + 3(F) + 7(G) + 3(I) = 17.
Therefore, C would be selected first.
(iv) Shortest time task, and From the available tasks, choose the one with the shortest task time.
(v) Least number of following tasks. From the available tasks, choose the one with the least number of subsequent tasks.
54 A Modern Approach to Operations Management Example 3.3. On the basis of the precedence diagram and activity times given in Example-3.2, the firm deter- mines that there are 480 productive minutes of work available per day. Furthermore, the production schedule requires that 40 units be completed as output from the assembly line each day.
Thus: Cycle time (in minutes) = 480 minutes/40 units =12 minutes /unit Minimum number of workstations = total task time/cycle time = 66/12
= 5.5 or 6 stations.
A B
C
G I
F
H E
D
10 11
5
7 3
12 11
3 4
Figure 3.8
Use the most following tasks heuristic to assign to workstations.
Figure 3.8 shows one solution that does not violate the sequence requirements and that group tasks into six stations shown by six different colors. To obtain this solution, activities with the most following tasks were moved into workstations to use as much of the available cycle time of 12 minutes as possible.
Workstation Tasks included in the workstation Time (minutes)
1 A 10
2 B 11
3 E 12
4 C, D, F 12
5 H 11
6 G, I 10
• The first workstation consumes 10 minutes and has an idle time of 2 minutes.
• The second workstation uses 11 minutes, and the third consumes the full 12 minutes.
• The fourth workstation groups three small tasks and balances perfectly at 12 minutes.
• The fifth has 1 minute of idle time, and the sixth (consisting of tasks G and I) has 2 minutes of idle time per cycle.
Total idle time for this solution is 6 minutes per cycle.
We can calculate the balance efficiency for Example 3.3 as follows:
Efficiency = Σ task times/(actual no. of workstations) × (assigned cycle time) (3.3) Efficiency = 66 minutes/(6 stations × 12 minutes) = 66/72 = 91.7 %
If we open a seventh workstation (for whatever reason), will decrease the efficiency of the balance to 78.6%.
Efficiency = 66 minutes/(7 stations × 12 minutes) = 66/84 = 78.6 %
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