Badr Alghamdi
العربية
Lessons

Transfer lots: the next station starts before this one finishes

Split one batch into transfer lots and the next station starts before this one finishes. The overlap is real time back, and it has a price the schedule never shows.

Your prediction before the experiment

A second station has to wait until the first station finishes the whole batch. The reference batch is unsplit. Move the transfer lot size to reveal the comparison.
A second station has to wait until the first station finishes the whole batch. The reference batch is unsplit. Move the transfer lot size to reveal the comparison.

The schedule held still

Total quantity
100 units
Stage one
10 s setup, then 1 s per unit
Stage two
5 s setup, then 1.5 s per unit
Order
Two serial stages, transfer lots processed in order
Excluded
No transport time, blocking, variability, or lot dependent setup

The one change

units

Enter a whole number from 10 through 100.

Units finished at stage two, over timeStage two finishes 20 units at 60 seconds, 40 at 90, 60 at 120 and all 100 at 180 seconds. The unsplit batch finishes all 100 at 260 seconds, so the split lands 80 seconds sooner.50Move the lot size to draw this

Open the recurrence

How the ledger advances

Each row waits for two facts: its upstream completion and the previous downstream completion. The later of those becomes the downstream start. Its own downstream processing time is then added.

C1,j=S1+qcum,j p1,S2,j=max⁡(C1,j, C2,j−1),C2,j=S2,j+qj p2C_{1,j} = S_1 + q_{\mathrm{cum},j}\,p_1,\qquad S_{2,j} = \max\bigl(C_{1,j},\,C_{2,j-1}\bigr),\qquad C_{2,j} = S_{2,j} + q_j\,p_2C1,j=S1+qcum,j p1,S2,j=max⁡(C1,j, C2,j−1),C2,j=S2,j+qj p2\begin{gathered}C_{1,j} = S_1 + q_{\mathrm{cum},j}\,p_1, \\ S_{2,j} = \max\bigl(C_{1,j},\,C_{2,j-1}\bigr), \\ C_{2,j} = S_{2,j} + q_j\,p_2\end{gathered}

where C1,jC_{1,j} is the upstream completion of a transfer lot, S1S_1 the upstream start, qcum,jq_{\mathrm{cum},j} the units released up to that lot, p1p_1 the upstream time a unit, S2,jS_{2,j} the downstream start, C2,jC_{2,j} the downstream completion, qjq_j the units in the lot, and p2p_2 the downstream time a unit.

Worked example

With 20 unit transfer lots, stage one releases at 30, 50, 70, 90, and 110 seconds. Stage two finishes those lots at 60, 90, 120, 150, and 180 seconds. The unsplit batch finishes at 260 seconds. The processing work itself, setup and run time together, stays the same in both cases and is not the completion time, in seconds:

10+5+100×1+100×1.5=26510 + 5 + 100 \times 1 + 100 \times 1.5 = 265

Across the whole admitted range the finish time is 160 seconds plus the transfer lot size, so every ten units removed from the lot returns ten seconds.

T=160+qT = 160 + q

where TT is the finish time in seconds and qq the transfer lot size in units.

This model does not optimize a transfer lot, promise completion, infer capacity, recommend staffing or sequencing, represent variability, or establish production, quality, commercial, or safety adequacy.

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