There are two factories located at place P and the other at place Q. From these factories, a certain commodity is to be delivered to each of the depots situated at A, B and C. Weekly production of commodity by P and Q are 120 kg and 150 kg respectively. Weekly requirement of commodity by A, B and C are 80 kg, 90 kg and 100 kg respectively. P delivers 60 kg to A, 40 kg to B and 20 kg to C. How much amount of the commodity should Q deliver to A, B and C to meet their requirement? If the rate of the commodity is ? 20 per kg, find the total amount to be paid to P and Q.
Answers
Answer:
Let the factory at P transports x units of commodity to depot at A and y units to depot at B. Then , 8 − (x + y) units of commodity would be transported to depot C .
The requirement at depot A is 5 units since, x units are transported from factory P.
Therefore, the remaining (5 − x) units are transported from factory Q.
Similarly, (5 − y) units and 4 − [8 − (x + y)] i.e. (x + y − 4) units will be transported from factory Q to depot B and C respectively.
Units of commodity cannot be negative.Therefore,
x≥0, y≥0 and 8−x−y≥0⇒x≥0, y≥0 and x+y≤85−x≥0, 5−y≥0 and x+y−4≥0⇒x≤ 5, y≤5 and x+y≥4The mathematical formulation of the given problem is
Minimize Z=160x+100y+150(8−x−y)+100(5−x)+120(5−y)+100(x+y−4) =10x−70y+1900
subject to
x+y≤8x≤5y≤5x+y≥4x, y≥0
First we will convert inequations into equations as follows:
x + y = 8, x = 5, y = 5, x + y = 4, x = 0 and y = 0
Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A1(8, 0) and B1(0, 8) respectively. By joining these points we obtain the line x + y = 8. Clearly (0,0) satisfies the x + y = 8. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 8.
Region represented by x ≤ 5:
x = 5 is the line that passes C1(5, 0) and is parallel to the Y axis.The region to the left of the line x = 5 will satisfy the inequation x ≤ 5.
Region represented by y ≤ 5:
y = 5 is the line that passes D1(0, 5) and is parallel to the X axis.The region below the line y = 5 will satisfy the inequation y ≤ 5.
Region represented by x + y ≥ 4:
The line x + y = 4 meets the coordinate axes at E1(4, 0) and F1(0, 4) respectively. By joining these points we obtain the line x + y = 4. Clearly (0,0) does not satisfies the inequation x + y ≥ 4. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 4.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region of the LPP is the shaded portion in the graph
The corner points of the feasible region are E1(4, 0), C1(5, 0), G1(5, 3), H1(3, 5), D1(0, 5) and F1(0, 4).
The values of the objective function at the corner points are
Corner points Z = 10x−70y+1900
E1(4, 0) 1940
C1(5, 0) 1950
G1(5, 3) 1740
H1(3, 5) 1580
D1(0, 5) 1550
F1(0, 4) 1620
Thus, the minimum value of Z is 1550 which is attained at D1(0, 5).
Therefore, from P to A: 0 units, P to B : 5 units , P to C: 3 units
from Q to A : 5 units, Q to B : 0 units, Q to C: 1 unit