Shop scheduling is one of the most important and most challenging things we do. The best master plans can be for naught if the execution side of the business cannot get things done on time.
In this example, we’ll apply the Theory of Constraints thinking to optimize the throughput and utilization of a manufacturing train.
Consider a train with 3 work centers A, B and C
| Work Center | A | B | C |
| Rate /Hr | 50 | 80 | 100 |
| Time/Unit | 0.02 | 0.0125 | 0.01 |
Lot Size = 100 Units
Scenario 1 – Complete the lot in one work center before moving to the next work center
| Work Center | A | B | C | ||
| Rate /Hr | 50 | 80 | 100 | ||
| Time/Unit | 0.02 | 0.0125 | 0.01 | ||
| Start Time | 0 | 2 | 3.25 | ||
| End Time | 2 | 3.25 | 4.25 | 4.25 | Lapsed Time |
| Lot Process Time (Hr) | 2 | 1.25 | 1 | 4.25 | Processing Time |
Here there is no overlap of operations. Downstream work centers have the longest wait time for upstream activities.
Scenario 2 – Transfer to Downstream WC when ½ lot is completed.
| Work Center | A | B | C | ||
| Rate /Hr | 50 | 80 | 100 | ||
| Time/Unit | 0.02 | 0.0125 | 0.01 | ||
| Start Time Sublot 1 | 0 | 1 | 1.625 | ||
| End Time Sublot 1 | 1 | 1.625 | 2.125 | ||
| Start Time Sublot 2 | 1 | 2 | 2.625 | ||
| End Time Sublot 2 | 2 | 2.625 | 3.125 | 3.125 | Lapsed Time |
| Idle Time | 0.375 | 0.5 | 0.875 | Total Split Lot Idle Time | |
| Lot Process Time (Hr) | 2 | 1.25 | 1 | 4.25 | Processing Time |
This technique saves 1.125 total elapsed time. This gives an increase of (1.125/4.25) 29% effective capacity.
Scenario 3 – Maximize the overlap of the work center operations. Here we build a queue in front of W/C B to a level where W/C B will finish 1 unit processing time after W/C A completes. We do the same from W/C B to W/C C.
| Work Center | A | B | C | ||
| Rate /Hr | 50 | 80 | 100 | ||
| Time/Unit | 0.02 | 0.0125 | 0.01 | ||
| Start Time Build Queue | 0 | ||||
| End Time Build Queue | |||||
| Start Time | 0 | 0.7625 | 1.0225 | ||
| End Time | 2 | 2.0125 | 2.0225 | 2.0225 | Lapsed Time |
| Idle Time | 0 | Total Split Lot Idle Time | |||
| Lot Process Time (Hr) | 2 | 1.25 | 1 | 4.25 | Processing Time |
Here there is a minimum of wait time before downstream W/C start. There is only enough to build a small queue so that the downstream work center does not wait for the upstream W/C to complete an operation.
By maximizing the overlap (concurrent running of equipment) this technique doubles the effective capacity of the line.
The time to build a queue is shown as Tq and the queue size is in units.
| Queue and Start Time for W/C B | Queue and Start Time for W/C C | |||
| Tq | 0.7625 | Tq | 0.26 | |
| Queue Size | 38.125 | Queue Size | 20.8 | |
| Remaining To Process A (units) | 61.875 | Remaining To Process B (units) | 79.2 | |
| Remaining Time A | 1.2375 | Remaining Time B | 0.99 | |
| Total Time A | 2 | Total Time B | 1.25 | |
| Time Start B | 0.7625 | Start Time C | 1.0225 | |
Logic to Find Start times and queue size for downstream W/C in a maximum overlap of processes scenario
Consider two work centers (W/C) A followed by B.
Production Rate of W/C A = Ra units/hr Rate of W/C B = Rb units/hr Let Lot size = L
Find the starting point of B such that B finishes 1 make unit (takt time) after A completes. This prevents B from pausing its processing to wait for A.
(1) Time for A to complete a full lot = TFa = L / Ra
(2) Start time for A = TSa = 0 we start the process on W/C A at time 0 (zero)
(3) The time for W/C B to process 1 unit = 1 / Rb
(4) The time for W/C B to process a full lot = TFb= L / Rb
(5) The objective is to find the start time of W/C B = TSb
(6) If the End Time for W/C B = TFb
then the start time for W/C A = TFb – L/Rb
Start time = Finish time – processing time.
(7) TSb = TFb – L / Rb
Remember we want the W/C B to finish 1 processing cycle time for B after A finishes.
Combining (1) and (3) above
(8) TFb = TFa + 1/Rb = L/Ra + 1/Rb
Now Start time of W/C B = End Time of W/C B – Processing time of W/C B. Equation (7)
Substitute (8) into (7)
(9) TSb = TFb – L/Rb = L/Ra + 1/Rb – L/Rb
The size of the queue that is built before W/C B before it starts is
(10) QueueSize B = TSb * Ra
The Wait time for the start of B times the rate of production of W/C A.
