TRANSPORTATION PROBLEM TRANSPORTATION PROBLEM The cheapest way of
TRANSPORTATION PROBLEM
TRANSPORTATION PROBLEM The cheapest way of transporting material from one place to another place is determined in Transportation Problem
As the cost of transport is to be minimized, the objective function will be Min type The objective function and constraints are obtained by drawing the flow diagram
There is a factory located at each of the two places P and Q. A certain commodity is delivered to each of the three depots situated at A, B and C. The weekly requirements are, 5, 5 and 4 units. The production capacity of the factories P and Q are 8 and 6 units respectively. The cost of transportation per unit is given below: TO COST IN RUPEES FROM P A B C 16 10 15 Q 10 12 10 How many units should be transported from each factory to each depot in order that the transportation cost is minimum? Formulate the above linear programming problem mathematically.
There are 02 factories and 03 depots A P B C Q
P has maximum of 8 units Let x units are transported to A, y units to B and 8 – (x + y) to C A x P y 8 -( x+ B y) C Q
Q has maximum of 6 units The requirements of the depots A, B and C are 5, 5 and 4 A P x 5 - y 5 -y x 8 -( x+ B y) C y) 5 + x 5 4 ( y 6 x+ Q
COST IN RUPEES A B C P 16 10 15 Q 10 12 10 A x P 10 16 y 10 8 -( x+ 15 y) 5 - B x 5 -y 12 -4 y x+ 0 1 C Q
The total transportation cost is 16 x + 10 y + 15(8 – x – y)+ 10(5 – x) + 12(5 – y) + 10(x + y – 4) Total cost = x – 7 y + 190 A x P 10 16 y 10 8 x 15 y 5 - B x 5 -y 12 -4 y x+ 0 1 C Q
OBJECTIVE FUNCTION Min z = x – 7 y + 190 A x P 10 16 y 10 8 x 15 y 5 - B x 5 -y 12 -4 y x+ 0 1 C Q
To get the constraints ALL COMMODITIES TRANSPORTED ARE GREATER THAN OR EQUAL TO ZERO
Constraints x ≥ 0 5 -x ≥ 0 y≥ 0 5 -y ≥ 0 8 -x-y ≥ 0 x+y-4 ≥ 0
Constraints x ≥ 0 y≥ 0 x+y≤ 8 x≤ 5 y≤ 5 x+y ≥ 4
THE LPP FOR THE TRANSPORTATION PROBLEM IS Min z = x – 7 y + 190 SUBJECT TO THE CONSTRAINTS x+y≤ 8 x+y ≥ 4 x≤ 5 y≤ 5 x ≥ 0 y≥ 0
y axis x=5 y=5 x x + y = 8 4 X axis
FEASIBLE REGION OF LPP
THE VALUE OF THE OBEJCTIVE FUNCTION AT THE CORNER POINTS ARE POINT (x , y) Value of z (4, 0) 194 (5, 0) 195 (5, 3) 174 (3, 5) 158 (0, 5) 155 (0, 4) 162 Z is minimum at x = 0 and y = 5
Z is minimum at x = 0 and y = 5 The optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory P and 5, 0 and 1 units from the factory Q to the depots A, B and C. The minimum transportation charge is Rs. 155
A brick manufacturer has two depots A and B which stocks of 30000 and 20000 bricks respectively. He receives orders from three builders P, Q and R for 15000, 20000 and 15000 bricks respectively. The cost in Rupees of transporting 1000 bricks to the builders from the depots are given below: FROM TO COST IN RUPEES P Q R A 40 20 30 B 20 60 40 How should the manufacturer fulfill the orders so as to keep the cost of transportation minimum?
OBJECTIVE FUNCTION Min z = 3 x – 30 y + 1800 P x A 20 -x 40 y 20 30 - 30 x- 15 Q y R 20 -y 60 5 1 y x+ 40 B
THE LPP FOR THE TRANSPORTATION PROBLEM IS Min z = 30 x – 30 y + 1800 SUBJECT TO THE CONSTRAINTS x + y ≤ 30 x + y ≥ 15 x ≤ 15 y ≤ 20 x ≥ 0 y≥ 0
THE VALUE OF THE OBEJCTIVE FUNCTION AT THE CORNER POINTS ARE POINT (x , y) Value of z (15, 0) 2250 (15, 15) 1800 (10 , 20) 1500 (0, 20) 1200 (0, 15) 1350 Z is minimum at x = 0 and y = 20
Z is minimum at x = 0 and y = 20 The optimal transportation strategy will be to deliver 0, 20 and 10 thousand bricks to builders P, Q, R from depot A and 15, 0 and 5 thousand of bricks to builders P, Q and R from the depot B The minimum transportation charge is Rs. 1200
An oil company has two depots A and B with capacities 7000 and 4000 liters respectively. The company has to supply oil to three petrol pumps D, E and F whose requirements are 4500, 3000 and 3500 liters. The distance in kms between the depots and the petrol pumps is given below: From / to D E F A 7 6 3 B 1 4 2 Assuming the transportation cost of 10 liters of oil if Rs. 1 per km, how should the delivery be scheduled in order that the transportation cost is mimimum? What is the minimum cost?
D x A 70 3 7 y 6 E 00 - 3 x- 45 00 -x 3000 -y 60 0 0 5 -3 y x+ 40 y F B
THE LPP FOR THE TRANSPORTATION PROBLEM IS Min z = 3 x + y + 39500 SUBJECT TO THE CONSTRAINTS x + y ≤ 7000 x + y ≥ 3500 x ≤ 4500 y ≤ 3000 x ≥ 0 y≥ 0 Z is minimum at x = 500 and y = 3000
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