Assembly line balancing with cycle time

(a) Cycle time

$$C = \frac{\text{available time}}{\text{required output}} = \frac{480}{60} = 8\ \text{min}$$

(b) Minimum number of stations

Total task time $= 4+3+5+4+3+6+2+3 = 30$ min.

$$N_{min} = \frac{\sum t}{C} = \frac{30}{8} = 3.75 \Rightarrow 4\ \text{stations}$$

(c) Ranked positional weight (RPW)

RPW of a task = its own time plus the times of all tasks that follow it.

Order of rank: A, C, B, D, E, F, G, H. Assign to each station the highest-ranked available task that fits in the remaining cycle time.

Explanation: after A (4 min) only 4 min remain in the station. C (RPW 19) needs 5 min, which does not fit, so B (3 min) is taken. Task D cannot join station 2 because C+E already use 8 min, and F cannot start until D is finished, so D takes a station alone. F (6) and G (2) fill station 4 exactly.

(d) Efficiency and balance delay

$$\text{Efficiency} = \frac{\sum t}{N\,C} \times 100 = \frac{30}{5 \times 8} \times 100 = 75\%$$ $$\text{Balance delay} = 100 - 75 = 25\%$$

The four-station minimum is not reachable because of the precedence limits, so five stations are needed.

Answer: Cycle time 8 min; minimum stations 4 (theoretical); 5 stations used; efficiency 75%; balance delay 25%.