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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 5 of 18
Quick checkPractical

The class's seven-plant problem. Coordinates in km: (100, 300), (200, 500), (300, 600), (600, 400), (500, 300), (300, 200), (200, 100). Tonnages transported to them: 1,000, 2,500, 1,000, 2,700, 1,500, 1,200 and 1,800.

What is the x coordinate of the optimal location for the new plant, to one decimal?

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