There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units The cost of transportation per unit is given below:

From/To		Cost (in Rs.)
	A	B	C
P	160	100	150
Q	100	120	100

How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?

Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 - x - y) units will be transported to depot at C.

∴ x ≥ 0, y ≥ 0 and 8 - x - y ≥ 0
i.e.    x ≥ 0, y ≥ 0 and x + y ≤ 8
The weekly requirement of the depot at A is 5 units of the commodity. Since x units are transported from the factory at P, the remaining (5 - x) units need to be transported from the factory at Q.
Clearly 5 - x ≥ 0, i.e. x ≤ 5.
Similarly, (5 - y) and 6 - (5 - x + 5 - y) = x + y - 4 units are to be transported from the factory at Q to the depots at B and C respectively.
∴ 5 - y ≥ 0, x + y - 4 ≥ 0
i.e.    y ≤ 5, x + y ≥ 4
Total transportation cost Z is given by
Z = 160 x + 100 y + 100(5 - x) + 120 (5 - y) + 100 (x + y - 4) + 150 (8 - x - y)
= 160 x + 100 y + 500 - 100 x + 600 - 120 y + 100 x + 100 y - 400 + 1200 - 150 x - 150 y
= 10 x - 70 y + 1900
= 10(x - 7 y + 190)
∴ the problem reduces to
Minimise Z = 10 (x - 7 y + 190)
subject to the constraints
x ≥ 0, y ≥ 0
x + y ≤ 8
x ≤ 5
y ≤ 5
and    x + y ≥ 4
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 8
For x = 0, y = 8
For y = 0, ,y = 8
∴ line meets OX in A(8, 0) and OY in L(0, 8).
x = 5 is a straight line BM parallel to y-axis.
y = 5 is a straight line CN parallel to x-axis.
Again we draw the graph of x + y = 4
For x = 0, y = 4
For y = 0, x = 4
∴ line meets OX in D(4, 0) and OY in P(0, 4).
Since feasible region satisfies all the constraints.

∴ DBEFCP is the feasible region.
The corner points are D(4, 0), B(5, 0), E(5, 3), F(3, 5), C(0, 5), P(0, 4).
At D(4, 0), Z = 10 (4 - 7 × 0 + 190) = 10 × 194 = 1940
At B(5, 0), Z = 10(5 - 7 × 0 + 190) = 10 × 195 = 1950
At E(5, 3), Z = 10 (5 - 7 × 3 + 190) = 10 × 174 = 1740
At F(3, 5), Z = 10 (3 - 7 × 5 + 190) = 10 × 158 = 1580
At C(0, 5), Z = 10 (0 - 7 × 5 + 190) = 10 × 155 = 1550
At P(0, 4), Z = 10 (0 - 7 × 4 + 190) = 10 × 162 = 1620
∴ minimum value = 1550 at (0, 5).
∴ the optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A. B and C respectively. Corresponding to this strategy, the transportation cost would be minimum, i.e. Rs. 1550.