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From-to chart load-distance layout comparison

Short answer

Load-distance score = sum (loads × distance). Present layout P-R-S-Q yields 2,650 load-m/day; proposed P-Q-R-S yields 1,850 load-m/day. The proposed layout reduces material handling by about 30%.

A plant has four departments P, Q, R and S placed in a straight row in four equal slots with centre-to-centre spacing of 10 m. The daily material flow (loads per day, either direction) is: P-Q 40, P-R 10, P-S 5, Q-R 30, Q-S 15, R-S 50. The present order is P, R, S, Q and the proposed order is P, Q, R, S. Using the load-distance score (loads x distance), compare the two layouts.

Method

The load-distance score is ∑(loadsij×dij)\sum (\text{loads}_{ij} \times d_{ij}), where dijd_{ij} is the distance between departments along the row. The layout with the smaller score has less material handling. In a row, distance =10×= 10 \times (difference in slot numbers).

Present layout P-R-S-Q (slots 1, 2, 3, 4)

Slots: P = 1, R = 2, S = 3, Q = 4.

PairLoadsSlot gapDistance (m)Load x distance
P-Q403301,200
P-R10110100
P-S5220100
Q-R30220600
Q-S15110150
R-S50110500
Total2,650

Proposed layout P-Q-R-S (slots 1, 2, 3, 4)

Slots: P = 1, Q = 2, R = 3, S = 4.

PairLoadsSlot gapDistance (m)Load x distance
P-Q40110400
P-R10220200
P-S5330150
Q-R30110300
Q-S15220300
R-S50110500
Total1,850

Comparison

Saving =2650−1850=800= 2650 - 1850 = 800 load-m per day, i.e. 800/2650=30.2%800/2650 = 30.2\% reduction. The proposed layout puts the high-flow pairs P-Q, Q-R and R-S next to each other. It is therefore better, though rearranging costs must be compared with the saving.

Answer: Present = 2,650 load-m/day; proposed = 1,850 load-m/day; proposed layout reduces handling by about 30%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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