a) Identi Identify the two binding constraints. Hence, find the optimal solution and optimal value of the above linear programming problem.
Plot an isoprofit line as 2X + 3Y =18 on the graph.
Then move it towards the maximization direction (away from the origin). The last point at which it leaves the feasibility region is (4,9).
So, the optimal solution is X = 4; Y = 9 Max. Objective = 2*4+3*9 = 35
The first and the third constraints are taking part in the optimal solution whereas the constraint 3X + Y <= 27 is non-binding.
3X + 2y <= 30 - binding
3X + Y <= 27 - non-binding
2X + 5Y <= 53 - binding
b) Find the slack/surplus associated to the non-binding constraint and state clearly whether it is slack or surplus.
LHS of the non-binding constraint = 3*4 + 9 = 21
RHS = 27
So, there is a slack = 27 - 21 = 6
c) Find the range of optimality of the coefficient of X of the objective function. Hence, find the optimal solution if the coefficient of X is to be increased from 2 to 4.
The two extremes of the slope of the objective line for which (4, 9) is the optimal is when:
The slope is parallel to the 2X + 5Y = 53 line i.e. -2/5
The slope is parallel to the 3X + 2Y = 30 line i.e. -3/2
The slope of the objective line is -C1/C2. Keeping C2 = 3 to be constant,
The limits of the slope can be written as:
-3/2 <= -C1/3 <= -2/5
or, -4.5 <= -C1 <= -1.2
or, 1.2 <= C1 <= 4.5
So, the optimality range for the objective coefficient of X is [1.2, 4.5]
d) Suppose the value of the right-hand side of the Constraint III is increased by 1 (i.e. from 53 to 54). Find the new optimal solution and the dual price for the Constraint III.
The new optimal solution will be found at the intersection of the following two lines:
2X + 5Y = 53+1
3X + 2Y = 30
which is Y = 102/11 and X = 42/11
Objective = 2X + 3Y = 2*42/11 + 3*102/11 = 35.45
The earlier objective value was 35. So, the shadow price = 35.45 - 35 = 0.45