Question

Two godowns A and B have grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transporation per quintal from the godowns to the shop are given in the following table :

To/From	A	B
D	6	4
E	3	2
F	2.50	3

How should the supplies be transported in order that the transportation cost is minimum ?

Answer 1

Let store A supplies x quintals to shop D and y quintals to shop E.

∴ remaining ration (100 – x – y) quintal will be supplied to shop F.

∴ Transporation charges from store A to shop D = Rs 6x
and transportation charges from store A to shop E = Rs 3y
and transportation charges from store A to shop E
= Rs 5/2(100 – x – y)
∴ Total transportation charges from store A to shops D,
E and F = 6x + 3y + 5/2(100 – x – y).
The remaining requirement of shop D = (60 -x) quintal
The remaining requirements of shop E = (50 – y) quintal
and the remaining requirement of shop F
= [40 – (100 – x – y)] quintal
which will be supplied from store B.
∴ Transportation charges from store B to shop D = RS 4 (60 – x).
Transportation charges from store B to shop E = Rs 2 (50 – y)
and transportation charges from store B to shop F = Rs 3(x + y – 60)
∴ Total trapsortation charges from store B to shops D, E and F = 4(60 – x) + 2(50 – y) + 3(x + y – 60).
∴ Total transportation charges from both stores A and B to the shops D, E and F.
Z = 6x + 3y + 5/2(100 – x – y) + 4(60 – x) + 2(50 – y) + 3(x + y – 60)
= 2.5x + 1.5y + 410
Total capacity of stored = 100 quintal.
∴ x + y ≤ 100
Ration availed by shop D is x quintal from store A and remaining from store B.
∴ x ≤ 60
Similarly shop E avails y quintal from store A and remaining from store B.
∴ y ≤ 50
Similarly shop F avails (100 – x – y) quintals from store A and remaining from store B.
∴ x + y ≥ 60
∵ x and y are quantity of ration in quintal.
∴ x ≥ 0 and y ≥ 0
So mathematical formulation of Linear Programming Probelm is as following :
Z = 2.5x + 1.5y + 410
x + y ≤ 100
x ≤ 60
y ≤ 50
x + y ≥ 60
x ≥ 0, y ≥ 0
Converting these inequations into the equations
x + y = 100 …..(1)
x = 60 …..(2)
y = 50 …..(3)
x + y = 60 …..(4)
x = 0 …..(5)
y = 0 …..(6)
Region represented by x + y ≤ 100 : Line x + y = 100 meets the coordinate axis on points A(100,0) and B(0,100).
x + y = 100

X	100	0
Y	0	100

A( 100, 0); B(0, 100)
Join points A to B to obtain the line. Clearly (0,0) satisfies the inequation 0 + 0 = 0 ≤ 100. So the region containing the origin represents the solution set of the inequation.
Region represented by x ≤ 60 : Line x = 60 is parallal toy-axis and its each point will satisfy the inequation in first quadrent so its solution region will be towards origin.
Region represented by y ≤ 50 : Line y = 50 is parallel to x-axis and its each point will satisfy the inequation in first quadrant. So. its solution region will be towards origin.
Region represented by x + y ≥ 60: Line x + y = 60 meets the coordinate axis on the points G(60, 0) and H(0, 60).
x + y = 60

X	60	0
Y	0	60

G(60, 0); H(0, 60)
Join G and H to obtain the line. Clearly the origin (0,0) does not satisfying the inequation so the region of solution set is opposite to the origin.

Region represented by x ≥ 0 and y ≥ 0 : Since every point in first quadrant satisfies the inequations x ≥ 0 and y ≥ 0 so the solution set of these inequations is in the first quadrant.

The shaded region GJFM is the common region representing the inequations. This region is the feasible solution region of the inequations.

Coordinates of comer points of this region are G(60,0), J(60,40), F(50,50) and M(10, 50) where point J is the point of intersection of lines x + y = 100 and x = 60, point F is the point of intersection of x + y = 100 and y = 50 and M is the point of intersection of x + y = 60 and y = 50.
The value of objective function on these points is given in following table :

Point	x-coordinate	y-coordinate	Objective function Z = 2.5x + 1.5y + 410
G	60	0	ZG = 2.5(60) + 1.5(0) + 410 = 560
J	60	40	Zj = 2.5(40) + 1.5(40) + 410 = 620
F	50	50	F = 2.5(50) + 1.5(50) + 410 = 610
M	10	50	ZM = 2.5(10) + 1.5(50) + 410 = 510

It is clear from the table the objective function has minimum value Z = RS 510 at the point M(10,50).

Hence for minimum transportation charges merchant should supply the ration from store A to shops D, E and F as 10,50 and 40 quintals respectively and from store B to shops D, E and F as 50,0,0 quintal respectively.