Line balancing when demand rises by 20%
A final-assembly line for a pump housing has ten work elements with precedence and times: 1 (40 s) precedes 2 (30 s) and 3 (25 s); 2 and 3 precede 4 (35 s); 4 precedes 5 (20 s) and 6 (45 s); 5 and 6 precede 7 (30 s); 7 precedes 8 (25 s); 8 precedes 9 (30 s); 9 precedes 10 (20 s). The line works 390 min per day after breaks and must supply 360 housings per day. A proposed five-station balance assigns {1,3}, {2,4}, {5,6}, {7,8}, {9,10}. The plant is negotiating a contract that would raise demand by 20% without additional shift time. The industrial engineer claims the existing five stations can absorb the increase by merely working faster, because total work content is only 300 s.
1. Compute the present cycle time, the theoretical minimum number of stations, and the efficiency of the proposed five-station balance.
Read the worked answer
Available time = 390×60 = 23,400 s/day. Cycle time = 23,400/360 = 65 s/unit.
Total work content = 40+30+25+35+20+45+30+25+30+20 = 300 s. Theoretical minimum stations = ceil(300/65) = ceil(4.62) = 5.
Proposed station times: {1,3} = 65 s; {2,4} = 65 s; {5,6} = 65 s; {7,8} = 55 s; {9,10} = 50 s. All are ≤65 s and precedence is respected, so the balance is feasible. Idle time = 5×65 − 300 = 25 s. Efficiency = 300/(5×65) = 92.3%.
2. The contract raises demand by 20%. Compute the new cycle time and check whether five stations can meet it. Determine the actual minimum number of stations under the precedence and task times.
Read the worked answer
New demand = 360×1.20 = 432 units/day. Required cycle time = 23,400/432 = 54.17 s/unit.
The theoretical minimum is ceil(300/54.17) = ceil(5.54) = 6 stations, but that ignores precedence and indivisible task times. Check the pairings: 1+3 = 65 s; 2+3 = 55 s; 2+4 = 65 s; 4+5 = 55 s; 5+6 = 65 s; 7+8 = 55 s; 8+9 = 55 s. Every pair that mattered in the old balance now exceeds 54.17 s. In fact task 1 cannot be combined with 2 or 3, tasks 2 and 3 cannot combine with each other, task 4 cannot combine with 5, and only 9+10 = 50 s is feasible as a pair.
A feasible minimum grouping is {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9,10}: 9 stations. Efficiency would fall to 300/(9×54.17) = 61.5%. The engineer's claim is wrong: the line cannot simply work faster, and the theoretical six-station figure is unattainable without splitting or re-sequencing tasks.
3. What practical options should the industrial engineer evaluate before accepting a nine-station line?
Read the worked answer
First, question the 54.17 s cycle: overtime, a second shift, or a small finished-goods buffer could decouple the new contract from the existing line. Second, attack the task times that destroy the balance: task 1 at 40 s is isolated because 1+3 = 65 s, 2+3 = 55 s and 4+5 = 55 s. Reducing task 3 from 25 s to 14 s would allow 1+3 = 54 s; reducing task 5 from 20 s to 19 s would allow 4+5 = 54 s. Third, consider parallel stations at the bottleneck rather than nine serial stations, or duplicate the short elements 9 and 10.
The judgement is not to accept the nine-station arithmetic as inevitable. The real problem is the combination of a tight cycle and poor task indivisibility. A small method improvement at task 3 or task 5 is likely cheaper than adding four stations and accepting 61.5% efficiency.