Question 2 of 5: Load-Distance Analysis of a Four-Department Layout

Question 2: Load-Distance Analysis of a Four-Department Layout (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. Four departments occupy a rectangular block divided into a two-by-two grid, A and B along the top and C and D beneath them. The figure brackets the block as 20 ft across the top and 10 ft deep, so each department measures 10 ft by 5 ft and the centre-to-centre pitches are 10 ft between columns and 5 ft between rows. Department sizes are stated to be appropriate, so no department needs to be enlarged or shrunk, and trip cost is proportional to distance alone.

Find. Score the present arrangement on a load-distance basis, judge whether it is a good arrangement, and if it is not, produce the arrangement of the same four departments in the same four positions that minimises weekly travel.

Approach. Convert the block plan into centre-to-centre distances, form the load-distance score by multiplying every interdepartmental trip volume by the distance separating the two departments, then enumerate all 24 ways of assigning four departments to four positions and take the minimum.

1. Fix the geometry and the three distances a two-by-two block can offer. With a column pitch $\Delta x = 10\ \text{ft}$ and a row pitch $\Delta y = 5\ \text{ft}$, and travel along the aisles (rectangular or rectilinear distance, $d = |\Delta x| + |\Delta y|$), only three separations exist: two departments in the same row are 10 ft apart, two in the same column are 5 ft apart, and two diagonally opposite are $10 + 5 = 15$ ft apart. The set of six pairwise distances is therefore fixed at $\{5,\,5,\,10,\,10,\,15,\,15\}$ no matter how the departments are arranged — all the layout decision can do is match trip volumes to distances.
2. Score the present layout. The load-distance score is $$\text{LD}=\sum_{i<j} n_{ij}\,d_{ij}$$ where $n_{ij}$ is the weekly trip count between departments $i$ and $j$ and $d_{ij}$ their centre-to-centre distance. With A and B in the top row, C and D beneath, the pairs A–C and B–D are vertical, A–B and C–D horizontal, and A–D and B–C diagonal:

Nearly six-tenths of that total — 600 of 1,025 trip-feet — is spent on the two diagonal moves, and those are precisely the two 20-trip pairs. That is the diagnosis: the layout has put two of its three busiest relationships on the longest journey it has.

3. Recognise the structure that decides the problem. The two diagonal pairs of a two-by-two block always form a perfect matching of the four departments, and so do the two row pairs and the two column pairs. There are exactly three perfect matchings of four objects, and the six trip volumes therefore split into three sums, one of which must be paid at 15 ft, one at 10 ft and one at 5 ft: $$M_{1}=n_{AB}+n_{CD}=25+5=30,\qquad M_{2}=n_{AC}+n_{BD}=15+10=25,\qquad M_{3}=n_{AD}+n_{BC}=20+20=40$$ with $M_{1}+M_{2}+M_{3}=95$, the total weekly trips. Any layout costs $$\text{LD}=5M_{\text{col}}+10M_{\text{row}}+15M_{\text{diag}}$$ and the present layout, with $M_{\text{col}}=M_{2}$, $M_{\text{row}}=M_{1}$ and $M_{\text{diag}}=M_{3}$, gives $5(25)+10(30)+15(40)=125+300+600=1{,}025$ — the table above, recovered in one line.
4. Enumerate every arrangement. Assigning four departments to four positions gives $4! = 24$ arrangements, but the block has a four-fold mirror symmetry (left–right and top–bottom flips change nothing physical), so only $24/4 = 6$ distinct scores exist — one for each of the three choices of which matching is diagonal, times two for which of the remaining pair goes into the short 5 ft column relation. Evaluating all six:

The present arrangement is the worst of the six. So the answer to “what do you think of the present layout?” is blunt: it is the poorest of every arrangement available without moving a wall.

5. State the recommended layout. The minimum, 875 ft-trips per week, puts the two heavy 20-trip pairs A–D and B–C into the short 5 ft column relation and banishes the light A–C and B–D traffic to the diagonals. One arrangement that does it is A and B unchanged in the top row with C and D interchanged beneath, so that D sits under A and C sits under B:

Checking against the matching form, $5(40)+10(30)+15(25)=200+300+375=875$, which agrees. The saving is

6. Confirm that the conclusion does not depend on the distance metric. If trips are assumed to travel in a straight line rather than along aisles, the diagonal becomes $\sqrt{10^{2}+5^{2}} = 11.18$ ft instead of 15 ft. The present layout then scores $5(25)+10(30)+11.18(40) = 872.2$ ft-trips per week and the same proposed arrangement scores $5(40)+10(30)+11.18(25) = 779.5$, a saving of 92.7 ft-trips or 10.6 per cent. The recommendation is identical under both metrics, and it was shown above that it is also identical under either reading of the figure's dimensions, so it is robust.
7. Close with the practical judgement. The change is the cheapest one available: only two departments move, they exchange places, and the question states that the department sizes are appropriate, so the two swapped rooms are the same 10 ft by 5 ft and no partition has to be rebuilt. C and D are also the natural pair to disturb, because the C–D relationship is the lightest in the building at five trips a week, so the disruption falls where the traffic is thinnest. Against that, the load-distance model values every trip equally; before committing, one would confirm that no trip type carries an unusual burden (a heavy or awkward load, a queue of waiting customers, a confidentiality requirement) and that no department has an unavoidable relationship with the entrance or with a service core, since either would add a fixed term that this model omits.