Question 5 of 7: Assembly-Line Balancing

Question 5: Assembly-Line Balancing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Eight assembly tasks with their times and their technological precedence requirements, to be arranged on a paced line meeting a forecast demand of 400 units in an eight-hour day.

Find. (a) the precedence diagram; (b) the theoretical minimum number of workstations for the required output; and (c) a feasible balance achieving that number, together with its efficiency.

Approach. Convert the daily demand into a cycle time, divide the total work content by that cycle time to obtain the theoretical minimum number of stations, then assign tasks to stations in precedence order using the longest-task-time rule until the cycle time would be exceeded.

1. Part (a) — read the immediate predecessors out of the table. The table lists all predecessors, but the diagram needs only the immediate ones. Removing the transitive entries leaves A: none; B: none; C: A; D: C; E: C; F: D; G: E; H: B, E and F. For example H is listed as requiring A, B, C, D, E and F, but A, C and D are already implied through F, so the arrows into H come only from B, E and F. This yields the network drawn in Figure 5.1.
2. Part (b) — convert demand into a cycle time. The available production time per day and the required output give the pace at which a unit must leave the line: $$C=\frac{\text{production time available per day}}{\text{units required per day}} =\frac{8\ \text{h}\times 3600\ \text{s/h}}{400\ \text{units}} =\frac{28{,}800}{400}=72\ \text{s per unit}$$ No station may be loaded beyond 72 s, or the line cannot meet demand.
3. Sum the work content and divide by the cycle time. The total task time is $$\textstyle\sum t_i=50+40+20+45+20+25+10+35=245\ \text{s}$$ so the theoretical minimum number of stations is $$N_{\min}=\left\lceil \frac{\sum t_i}{C}\right\rceil =\left\lceil \frac{245}{72}\right\rceil=\lceil 3.40\rceil \ \Rightarrow\ \boxed{\ N_{\min}=4\ \text{stations}\ }$$ The ceiling is essential: 3.40 stations cannot be staffed, and three stations would provide only $3\times 72=216$ s of capacity against 245 s of work. Note also that the longest single task, A at 50 s, is below the cycle time, so no task has to be split — the theoretical minimum is at least feasible on that count.
4. Part (c) — choose a balancing rule. The longest-task-time heuristic is selected: at each step, from the tasks whose predecessors are all already assigned and whose time fits in the station's remaining capacity, assign the one with the longest time; when nothing fits, open a new station. The rule is chosen because it places the awkward, large tasks early while stations are still empty, leaving the small tasks as the flexible filler that closes the remaining gaps — the opposite order tends to strand a large task in a station of its own.
5. Assign tasks to stations. Working through the rule with $C=72$ s: Station 1 — available tasks are A (50) and B (40); take A, leaving 22 s. C (20) now becomes available and fits, leaving 2 s, which accommodates nothing. Station 1 = {A, C}, load 70 s. Station 2 — available are B (40), D (45), E (20); take D, leaving 27 s; B does not fit but E does, leaving 7 s. Station 2 = {D, E}, load 65 s. Station 3 — available are B (40), F (25), G (10); take B, leaving 32 s; take F, leaving 7 s; G does not fit alongside. Station 3 = {B, F}, load 65 s. Station 4 — the remaining tasks G (10) and H (35) both have their predecessors assigned and together load 45 s. Station 4 = {G, H}.

   Station	Tasks assigned	Station time (s)	Idle time (s)
   1	A, C	70	2
   2	D, E	65	7
   3	B, F	65	7
   4	G, H	45	27
   Total	8 tasks	245	43

   Every precedence relation is respected — C follows A in the same station, D and E follow C from Station 1, F follows D from Station 2, and G and H follow E and F respectively — and no station exceeds 72 s, so the balance is feasible in $\boxed{\ 4\ \text{stations, the theoretical minimum}\ }$
6. Evaluate the balance. Line efficiency compares the work actually done with the labour time purchased: $$E=\frac{\sum t_i}{N_{\text{act}}\,C}=\frac{245}{4\times 72}=\frac{245}{288}=0.8507$$ $$\boxed{\ E=85.07\ \%\ \text{efficiency},\qquad d=1-E=14.93\ \%\ \text{balance delay}\ }$$ The 43 s of idle time per cycle is the price of indivisible tasks; over a day it is $43\times 400=17{,}200$ s, or about 4.8 worker-hours. Actual daily output is $28{,}800/72=400$ units, exactly the requirement, and the line's bottleneck is Station 1 at 70 s, which would cap output at $28{,}800/70=411$ units if the pace were allowed to float.