OpenSpecEngineering

Sizes a paced assembly line before you assign tasks to stations. From the total work content and the output you need, it gives the cycle time, the fewest stations that could possibly do the work, and the line’s top output rate. Enter the number of stations in your proposed layout and it also gives the line efficiency, the balance delay and the idle time per unit. Manufacturing and industrial engineers use it to judge a line layout [1], [3].

Equations

C = T_op / D     N_min = ⌈Σt / C⌉

E = Σt / (N · C)     d = 1 − E     idle = N · C − Σt

T_op is the operating time available per shift and D the units needed per shift, so C is the cycle time: the time each station has to finish its share of one unit. You can also enter C directly. Σt is the total work content, the sum of all task times for one unit. N_min is Σt/C rounded up, since stations come in whole numbers. N is the number of stations in your layout, E the line efficiency and d the balance delay, the share of paid station time spent idle [1], [2]. The top output rate is 60/C units per hour.

Limits and common mistakes

• N_min is a lower bound. Precedence rules and task times that don’t pack neatly usually force one or more extra stations. Only a real task assignment, such as the largest-candidate rule [1], gives the true count.
• No task can be longer than C unless it is split or the station is doubled. The page stops with an error when the longest task exceeds C.
• Use the operating time after breaks and planned downtime, not the clock length of the shift.
• Fewer stations than N_min would need an efficiency above 100 %, so the page rejects it.

Worked example

A line assembles a small appliance with 22.4 min of work per unit. The longest task takes 1.0 min. It must make 400 units in a shift with 450 min of operating time. The proposed layout has 22 stations. How good is it?

1. Enter Total Work Content Σt 22.4.
2. Choose Cycle Time Basis From required output rate, then enter Available Operating Time per Shift 450 and Required Output per Shift 400.
3. Enter Longest Single Task Time 1.0 and Actual / Proposed Number of Stations 22.
4. Press Solve. The Solution shows: Cycle time C: 1.125 min/unit
   Theoretical minimum stations: 20
   Max theoretical output rate: 53.3333 units/hr
   Line efficiency (Σt / (N·C)): 90.5051 %
   Balance delay (idle fraction): 9.49495 %
   Total idle time per unit: 2.35 min

Check by hand. 450 / 400 = 1.125 min, and 22.4 / 1.125 = 19.9, so at least 20 stations. 22 stations hold 22 × 1.125 = 24.75 min of station time per unit, 2.35 min more than the work, and 22.4 / 24.75 = 90.5 %. Twenty stations would mean 99.6 %, rarely reached in practice.