Question 7 of 7: Priority Dispatching at a Single Work Centre

Question 7: Priority Dispatching at a Single Work Centre (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Six job orders waiting at one work centre, with the due dates, arrival dates and times, operation times and counts of remaining operations tabulated above. Total work content 37 hours. The queue is complete at April 21, 5 pm, when the last order arrives, and that instant is taken as $t = 0$.

Find. The priority number and processing sequence under each of the five dispatching rules, a comparison of what the sequences achieve, and a reasoned preference.

Approach. Each rule is a formula that turns the table into one number per job; sort ascending (or by arrival, for FCFS) and the sequence follows. The rules are then compared on the two measures a work centre actually cares about: mean flow time, which drives work in process, and tardiness against the due dates.

1. Compute the static slack and the slack per remaining operation. Static slack is the calendar allowance the order arrived with, $$SS = \text{due date} - \text{date received at the centre}$$ and the fifth rule divides that allowance among the operations still to be done, $$SS/RO = \frac{SS}{\text{number of remaining operations}}$$ Order 1, for example, is due May 1 and arrived April 18, so $SS = 13$ days and, with three operations left, $SS/RO = 13/3 = 4.33$ days per operation. Order 2 is the striking case: it is due April 20 but did not reach the centre until April 21, so its slack is already $-1$ day — it is late before any work starts.
2. Tabulate all five priority numbers together. Setting them side by side is what makes the comparison possible.

   Order	Arrival at centre	Operation time (h)	Due date	Static slack (days)	Remaining ops	SS/RO (days per op)
   1	Apr 18, 9 am	6	May 1	13	3	4.33
   2	Apr 21, 10 am	3	Apr 20	−1	1	−1.00
   3	Apr 19, 5 pm	7	Jun 1	43	2	21.50
   4	Apr 21, 3 pm	9	Jun 15	55	4	13.75
   5	Apr 20, 5 pm	4	May 15	25	5	5.00
   6	Apr 21, 5 pm	8	May 20	29	7	4.14

3. Rule (a), FCFS — sort by arrival at the centre. The arrival order is April 18, 19, 20, then the three April 21 arrivals at 10 am, 3 pm and 5 pm, giving $$\boxed{\ \text{FCFS: } 1 \to 3 \to 5 \to 2 \to 4 \to 6\ }$$ This rule uses no information beyond the time stamp, which is its virtue (it is seen as fair and needs no data) and its defect (it ignores both urgency and duration).
4. Rule (b), SOT — sort by ascending operation time. The operation times are 3, 4, 6, 7, 8 and 9 hours for orders 2, 5, 1, 3, 6 and 4, $$\boxed{\ \text{SOT: } 2 \to 5 \to 1 \to 3 \to 6 \to 4\ }$$
5. Rules (c) and (d), SS and FISFS — sort by slack and by due date. Ranking the static slacks $-1, 13, 25, 29, 43, 55$ gives $2 \to 1 \to 5 \to 6 \to 3 \to 4$, and ranking the due dates April 20, May 1, May 15, May 20, June 1, June 15 gives the same order: $$\boxed{\ \text{SS: } 2 \to 1 \to 5 \to 6 \to 3 \to 4 \quad\text{and}\quad \text{FISFS: } 2 \to 1 \to 5 \to 6 \to 3 \to 4\ }$$ The two coincide on this data set, which is worth remarking on rather than passing over: they agree only because the orders arrived at the centre in almost the same order as their due dates. Had a late-arriving order carried an early due date, the two rules would diverge.
6. Rule (e), SS/RO — sort by slack per remaining operation. Ranking $-1.00, 4.14, 4.33, 5.00, 13.75, 21.50$, $$\boxed{\ \text{SS/RO: } 2 \to 6 \to 1 \to 5 \to 4 \to 3\ }$$ The division by remaining operations promotes order 6, which has 29 days of slack but seven operations still to go, and demotes order 3, which has 43 days and only two. This is the only rule of the five that looks beyond this work centre.
7. Compare what the sequences actually achieve. All five clear the queue in the same 37 hours, so the comparison is about the average experience of a job. Taking the completion times in sequence and averaging:

   Rule	Sequence	Mean flow time (h)	Mean tardiness (h)	Orders tardy
   FCFS	1, 3, 5, 2, 4, 6	20.33	4.67	1
   SOT	2, 5, 1, 3, 6, 4	18.00	1.83	1
   SS = FISFS	2, 1, 5, 6, 3, 4	18.50	1.83	1
   SS/RO	2, 6, 1, 5, 4, 3	19.83	1.83	1

   Due allowances are converted to work-centre hours at eight hours per working day from $t = 0$; only order 2, which arrived already overdue, can be late under any rule, and the three slack-based rules and SOT all put it first, which is why they share the same tardiness. Under FCFS order 2 waits until hour 20 and finishes 28 hours late.
8. State a preference and justify it. On this data SOT gives the lowest mean flow time, 18.00 hours against 20.33 for FCFS, and that is not an accident of the numbers: shortest operation time is provably the rule that minimises mean flow time, and therefore mean work in process and mean job lateness, for a single machine with all jobs available. It is the rule to prefer when the objective is throughput and low work in process. Its known weakness is that a long job can be starved indefinitely as short jobs keep arriving, so in practice it is run with a truncation rule that promotes any job waiting beyond a set age. If instead the objective is delivery performance, the slack-based rules are preferable, and SS/RO is the better of them because it is the only one that accounts for the work an order still faces downstream — an order with 29 days of slack and seven operations left is genuinely more urgent than one with 43 days and two. My preference for this centre is SOT with a truncation rule if it feeds a stock point, and SS/RO if it feeds customer orders with firm promise dates; in both cases order 2 should be expedited immediately, since it is already past due on arrival and no sequencing rule can recover a negative slack.