22-Mec-B4 Integrated Manufacturing Systems · December 2016

Question 1 of 6: Operation Sequence Analysis and Block Layout

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Operation Sequence Analysis and Block 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. A four-department shop whose inter-departmental traffic is recorded as a from-to chart of moves per period, together with the floor area each department needs. Blank cells in the chart carry no flow.

Find. (a) the relative arrangement of the four departments that minimises total travel, obtained by operation sequence analysis, and (b) a dimensioned block diagram that realises that arrangement within the required areas.

Approach. Operation sequence analysis collapses the directional from-to chart into undirected pair flows, ranks the pairs by intensity, places the heaviest pairs on a trial grid so that they share a boundary, and then scores the trial by the flow-distance product $\sum f_{ij}d_{ij}$; the trial is improved by exchanging departments until no exchange lowers the score.

1. Part (a) — total the chart to fix the size of the problem. Adding the eight non-blank entries, $N=2+2+2+4+3+1+2+1=17$ moves per period must be made somewhere in the plant. This total is the yardstick: any layout can only redistribute these 17 moves over longer or shorter distances, never reduce their number.
2. Collapse the directional chart into undirected pair flows. Travel cost does not care which way a load moves, so for each pair of departments the two directions are added, $$f_{ij}=n_{ij}+n_{ji}.$$ Working through the six pairs gives $f_{AB}=2+2=4$, $f_{AC}=0+0=0$, $f_{AD}=2+2=4$, $f_{BC}=4+3=7$, $f_{BD}=0+0=0$ and $f_{CD}=1+1=2$. The six pair flows sum to 17, which reconciles with Step 1.
3. Rank the pairs and read the pattern. In descending order the pairs are B–C (7), A–B (4), A–D (4), C–D (2), and then A–C and B–D with nothing at all. The four non-zero pairs are exactly the links of the closed chain A → B → C → D → A, and the two pairs that carry nothing are the two chords of that chain. That is the whole structure of the problem, and it is what Figure 1 shows.
4. Place the departments on a trial grid so the chain closes. A closed chain of four cannot be laid out in a single row without breaking one link, but it fits a two-by-two block exactly: in a two-by-two grid each cell touches two others along a wall and faces the fourth across a corner. Put the two zero-flow pairs on the corner-to-corner diagonals and every link of the chain lands on a shared wall. The trial arrangement is therefore A and B along the top, D and C along the bottom, so that A faces B and D, and C faces B and D.
5. Score the trial layout. Measure distance in grid steps, so departments sharing a wall are one step apart and departments meeting only at a corner are two. The score is then $$Z=\sum_{i<j} f_{ij}\,d_{ij}=4(1)+7(1)+2(1)+4(1)+0(2)+0(2)$$ which evaluates to $\boxed{Z=17\ \text{flow-distance units}}$, the theoretical floor, because it equals the raw move count of Step 1. Every one of the 17 moves travels the shortest distance the grid allows.
6. Confirm by exchange that no rival arrangement is better. On a two-by-two grid there are only three genuinely different arrangements, distinguished by which pair of departments is placed on the diagonals. Putting A–B and C–D on the diagonals scores $11+2(6)=23$, and putting A–D and B–C on the diagonals scores $6+2(11)=28$. The best single-row arrangement, D–A–B–C, scores $4(1)+4(1)+7(1)+2(3)=21$ because C and D then sit three bays apart. The two-by-two block therefore wins by 4 units over the best row and by 6 over the next-best block, and no exchange improves it.
7. Part (b) — size the building envelope. The four departments require $3600+2400+2400+1600=10{,}000$ square feet in total. A square envelope of $100\ \text{ft}\times100\ \text{ft}$ delivers exactly that with no circulation allowance, and a square is the natural shape for a layout whose flow pattern is a closed loop rather than a line.
8. Split the envelope into two depth bands matching the row areas. The top row of the arrangement carries A and B, needing $3600+2400=6000$ square feet, and the bottom row carries D and C, needing $1600+2400=4000$. Across a 100 ft frontage those become bands of depth $$d_{\text{top}}=\frac{6000}{100}=60\ \text{ft},\qquad d_{\text{bot}}=\frac{4000}{100}=40\ \text{ft},$$ and the two depths add to the 100 ft building depth as they must.
9. Size each department within its band. Dividing each area by its band depth gives the frontage: A occupies $3600/60=60$ ft and B $2400/60=40$ ft along the top band, while D occupies $1600/40=40$ ft and C $2400/40=60$ ft along the bottom band. Both bands close on the 100 ft frontage, so the blocks tile the envelope with no residual space.
10. Check that the party walls reproduce the required adjacencies. A and B share a 60 ft wall, A and D share a 40 ft wall, B and C share a 40 ft wall, and D and C share a 40 ft wall, so all four links of the flow chain have a real, usable opening between the departments; B and D share no boundary at all, which costs nothing because they exchange no material. The layout therefore achieves $\boxed{\text{zero non-adjacent flow, all 17 moves between touching departments}}$.

The block diagram is deliberately drawn with the two largest departments, A and C, on opposite corners. That keeps the two heavy pairs, B–C and the two four-move links out of A, on short walls near the centre of the building, which is where an aisle and a handling spine would naturally run. If a receiving dock is placed on the left wall and shipping on the right, the dominant path A → B → C → D runs clockwise around that spine without any material crossing the building twice.