OpenSpecEngineering

Estimates how long parts wait in front of a production workstation, and how much work piles up there, from its utilization, process time and variability. The workstation can be one machine or several identical machines fed from one queue. Industrial and production engineers use it to see why a busy station builds long queues and how much cutting variability or adding a machine would help [3].

The Kingman (VUT) equation

CT_q ≈ ((c_a² + c_e²) / 2) · (u^{√(2(m+1)) − 1} / (m(1 − u))) · t_e

CT_q is the average time a part waits in the queue. The equation is the product of three factors: variability (V), utilization (U) and time (T) [3]. c_a is the coefficient of variation (standard deviation over mean) of the times between arrivals, and c_e that of the effective process time, which includes setups, breakdowns and rework. u is the utilization of the station, m the number of parallel machines and t_e the mean effective process time on one machine. For one machine the U factor is u/(1 − u), Kingman’s heavy-traffic result [1]. The exponent for m machines is Sakasegawa’s fit [2].

CT = CT_q + t_e    r = u·m / t_e    WIP = r·CT

CT is the total time at the station and r the throughput. The WIP follows from Little’s Law.

Limits and common mistakes

• It is an approximation, most accurate at high utilization [1], [3]. At low utilization or with very high variability it can be well off.
• The queue grows without limit as u approaches 1. On one machine the queue at 95 % is more than three times the queue at 85 %, so check that the utilization is real, not planned.
• Use the effective process time. Leaving out setups and downtime makes both t_e and c_e too small.
• A c of 1.0 means exponential, random-looking times. Regular, paced arrivals give c_a well below 1. Batched arrivals from an upstream batch process give much more than 1.

Worked example

Two identical CNC lathes share one queue. Each part takes 12 minutes on average, setups and stoppages included, and the lathes run at 88 % utilization. The arrival CV is 0.8 and the process-time CV is 0.6. How long do parts wait?

1. Enter Utilization (u, 0–1) 0.88 and Mean Effective Process Time (te, min) 12.
2. Enter Number of Parallel Machines (m) 2.
3. Enter Arrival CV (ca — 1.0 = Poisson) 0.8 and Effective Process-Time CV (c_e) 0.6.
4. Press Solve. The Solution shows: Queue time CTq: 20.7715 min
   Total cycle time CT (queue + process): 32.7715 min
   Throughput rate: 8.8 units/hr
   WIP (Little’s Law, L = rate × CT): 4.80649 units
   Queue WIP: 3.04649 units

Check: V = (0.64 + 0.36)/2 = 0.5. U = 0.88^1.449 / (2 × 0.12) = 0.8309 / 0.24 = 3.462. CT_q = 0.5 × 3.462 × 12 = 20.8 min. The throughput is 0.88 × 2 / 12 min = 8.8 parts an hour.