---
title: "From-to chart load-distance layout comparison"
subject: "Industrial Engineering and Management"
chapter: 2
key: "semester_6/industrial_engineering_and_management:load-distance-from-to-chart-layout"
source: https://www.bctengineer.com/q/semester_6/industrial_engineering_and_management/load-distance-from-to-chart-layout
site: BCT Engineering
---

# From-to chart load-distance layout comparison

Industrial Engineering and Management · Chapter 2: Design of Production Systems

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_engineering_and_management/load-distance-from-to-chart-layout
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_engineering_and_management/chapter-2

## Question

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.

## 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%.

## Answer

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

#### Present layout P-R-S-Q (slots 1, 2, 3, 4)
Slots: P = 1, R = 2, S = 3, Q = 4.

| Pair | Loads | Slot gap | Distance (m) | Load x distance |
|---|---|---|---|---|
| P-Q | 40 | 3 | 30 | 1,200 |
| P-R | 10 | 1 | 10 | 100 |
| P-S | 5 | 2 | 20 | 100 |
| Q-R | 30 | 2 | 20 | 600 |
| Q-S | 15 | 1 | 10 | 150 |
| R-S | 50 | 1 | 10 | 500 |
| **Total** | | | | **2,650** |

#### Proposed layout P-Q-R-S (slots 1, 2, 3, 4)
Slots: P = 1, Q = 2, R = 3, S = 4.

| Pair | Loads | Slot gap | Distance (m) | Load x distance |
|---|---|---|---|---|
| P-Q | 40 | 1 | 10 | 400 |
| P-R | 10 | 2 | 20 | 200 |
| P-S | 5 | 3 | 30 | 150 |
| Q-R | 30 | 1 | 10 | 300 |
| Q-S | 15 | 2 | 20 | 300 |
| R-S | 50 | 1 | 10 | 500 |
| **Total** | | | | **1,850** |

#### Comparison
Saving $= 2650 - 1850 = 800$ load-m per day, i.e. $800/2650 = 30.2\%$ 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%.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the load-distance score calculated?
   A. Sum of distances only
   B. Sum of loads only
   C. Sum of (loads between departments x distance between them)
   D. Product of all loads
   Answer: C. Score = sum of loads_ij x d_ij; the smaller the score, the less handling. (Key point: Load-distance score)

2. In a row of equal slots 10 m apart, how is the distance between departments found?
   A. 10 x difference in slot numbers
   B. Sum of slot numbers
   C. 10 / slot gap
   D. Always 10 m
   Answer: A. Distance = 10 m x slot gap. (Key point: Distance in row)

3. What is the load-distance total for the present layout P-R-S-Q?
   A. 1,850
   B. 2,650
   C. 800
   D. 3,000
   Answer: B. The total of 1,200 + 100 + 100 + 600 + 150 + 500 is 2,650 load-m per day. (Key point: Present layout score)

4. Why is the proposed layout P-Q-R-S better?
   A. It increases distances
   B. It reduces total score to 1,850 by placing high-flow pairs P-Q, Q-R, R-S adjacent
   C. It has no flows
   D. It reduces cost to zero
   Answer: B. Saving is 800 load-m per day, about 30%, though rearranging cost must be compared. (Key point: Comparison)

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