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Sem 5 Prep
← OperationsCentre of gravity and load-distance

Formulas for this chapter

Centre of gravity

x-bar = SUM(xi x wi) / SUM(wi) y-bar = SUM(yi x wi) / SUM(wi)

Placing a single facility to serve known demand points with known volumes. Assumes cost is directly proportional to distance and volume shipped.

xi, yi
Grid coordinates of demand point i
wi
Weight or volume shipped to demand point i

Euclidean distance

d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

Straight-line travel, and the default in load-distance problems that give coordinates rather than road distances.

(x1, y1)
Coordinates of the first point
(x2, y2)
Coordinates of the second point

Rectilinear distance

d = |x2 - x1| + |y2 - y1|

Movement along aisles, streets or a shop floor. The measure the class CRAFT spreadsheet uses between department centroids.

|x2 - x1|
Absolute horizontal separation
|y2 - y1|
Absolute vertical separation

Load-distance score

Score(site i) = SUM over j of ( load at j x distance from i to j )

Ranking a shortlist of real candidate sites. Lowest score wins; a cost per unit distance scales all scores equally.

load at j
Demand or tonnage carried to demand point j
distance
Road, Euclidean or rectilinear, as the question states

Great-circle approximation

d = 69 x sqrt( dLong^2 + dLat^2 ) miles d = 111 x sqrt( dLong^2 + dLat^2 ) km Decimal degrees = degrees + minutes / 60

When locations are given as latitude and longitude. Convert to decimal degrees first.

dLat
Difference of latitudes, in decimal degrees
dLong
Difference of longitudes, in decimal degrees
Step 1 of 18
The ideaTheory

The balance point of a map

Put four weights on a flat board, one for each customer, each as heavy as the tonnage they buy. Now find the single point where the board balances on your finger.

That point is the centre of gravity. It is the spot that minimises the total weight-times-distance of hauling to everybody.

The arithmetic is a weighted average, done once for the x coordinate and once for the y coordinate. Nothing harder than that.

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