Graphical Method . Example
Bob, a farmer, is wondering which crops he should plant in the upcoming season.
He can grow wheat and barley on his 4000 acres of farmland.
Bob uses only organic fertilizers on his farm. He estimates that a maximum of 10 Metric Tons of organic fertilizers could be procured for the upcoming season.
Bob is confused regarding which crop (or combination of crops) to grow in order to maximize his income because although wheat yields a relatively higher contribution, growing barley requires relatively less land and fertilizers compared to wheat.
Comparison of wheat and barley is as follows:
(1 Metric Ton)
(1 Metric Ton)
|Land requirement||1 acre||0.8 acre|
|Fertilizer requirement||0.004 MT||0.001 MT|
Find the optimum production plan that will maximize Bob's income.
Variables used for constraints:
W = Quantity of wheat (in metric tons) to be grown
B = Quantity of barley (in metric tons) to be grown
● Land constraint 1W + 0.8B ≤ 4000
Only such quantities of wheat and barley can be grown that their combined coverage of land does not exceed 4000 acres.
● Fertilizer constraint 0.004W + 0.001B ≤ 10
Only such quantities of wheat and barley can be grown that their combined usage of fertilizers does not exceed 10 MT.
● Non-Negativity constraint W ≥ 0 and B ≥ 0
Only positive quantities of wheat and barley can be grown.
Objective function 200W + 100B = 1000000
Since Bob's objective is to maximize his income, we have used the contribution per unit of wheat and barley in the objective function.
1000000 is just a random number that we have used to obtain the slope of the objective function. Any other number could also be used as the gradient will remain the same.
|Constraints||Wheat||Barley||Point 1||Point 2|
|1W + 0.8B ≤ 4000||0||1(0)+0.8B =4000|
0.8B = 4000
B = 4000 ÷ 0.8
B = 5000
|1W + 0.8(0) = 4000|
W = 4000
|0.004W + 0.001B ≤ 10||0||0.004(0)+0.001B = 10|
0.001B = 10
B = 10 ÷ 0.001
B = 10000
|0.004W + 0.001(0) = 10|
0.004W = 10
W = 10 ÷ 0.004
W = 2500
Feasible region is the white area on the graph.
|Objective Function||W||B||Point 1||Point 2|
|200W + 100B = 400000||0||0 + 100B =400000|
B = 400000 ÷ 100
B = 4000
|200W + 0 = 400000|
W = 4000 ÷ 200
W = 2000
Optimum point always lies on one of the corner points of the feasible region.
Co-ordinates of the optimum point are approximately 1850 W and 2750 B (1850, 2750).