In a clear logical manner, show your steps to solve for substitution for the opt
ID: 422473 • Letter: I
Question
In a clear logical manner, show your steps to solve for substitution for the optimal answer to the linear programming problem above. The linear program and the associated graph are provided below (see Figure 1). Again assume you are a textbook author. Layout your solution such that a student would be glad they made the investment in your textbook. Do not skip or abbreviate algebra steps. You are strongly encouraged to use fractions throughout your solution to eliminate round off error 1. 1000 Max z- 1x, + 1.25x s.t. 900 800 3x1+ 1x, $ 2080 2x1+ 2x2s$ 1600 700 600 500 400 300 z=250 200 z-250 100 300 00 500 600 700 800 900 1000 100 200 C1C2C3-- obj 1(z-250)--obj 2(z-500)Explanation / Answer
Answer 1:First of all we plot the inequalities and find the region bounded by all the inequalities in the graph.
Then find the vertices and corners and find the value for which the function is max or min. This can be done by plugging in the values of the corner points and vertices to the objective function.
For the given problem we apply the above steps:
We plot the inequalities and find their bounded region as shown as the shaded portion. Now we find the vertices and corners.
Some of the vertices and corners are very much evident . These are
(X1,X2) = (0,640) -----> intercept of C1 on X2 axis
(X1,X2) = (0,0) -----> the origin
(X1,X2) = (693.33,0) -----> the intercept of C2 on X1
For the rest of the vertices we find the intersection points of the inequality lines (C1 & C3 ) and (C2 &C3)
Turning the inequalities to equalities for solving the equations for C1 and C3:
X1 +7*X2 =4480............(1)
2*X1 +2*X2 =1600----------(2)
Multiplying (1) by 2 and subtracting from (2) we get
-12*X2 = -7360
ie X2 =613.33
Therefore plugging the value of X2 in any of the equations (1) or (2) we get the value of X1 as 186.67
Therefore the corner point is
(X1,X2) = (186.67,613.33) .......... intersection of C1 and C3
We will find the remaining cornet point by finding the intersection of C2 and C3:
Using similar approach
3*X1 +X2=2080..........(3)
28X1 +2*X2 =1600.......(4)
Multiplying (3) by 2 and subtracting from (4) we get
-4*X1 = -2560
X1=640
plugging the value of X1 in any of the equations (3) and (4) we get X2 =160
Therefore the remaining corner point is
(X1,X2) =(640,160)........intersection of C2 and C3
now we plug in the values of each of the vertices and corner points in the objective function to find the max and min values.
We see that the value of the objective function is maximum when X1=186.67 and X2=613.33.
Hence this is the optimal solution
Answer 2:First of all we plot the inequalities and find the region bounded by all the inequalities in the graph.
Then find the vertices and corners and find the value for which the function is max or min. This can be done by plugging in the values of the corner points and vertices to the objective function.
For the given problem we apply the above steps:
We plot the inequalities and find their bounded region as shown as the shaded portion. Now we find the vertices and corners.
Some of the vertices and corners are very much evident . These are
(X1,X2) = (0,640) -----> intercept of C1 on X2 axis
(X1,X2) = (0,0) -----> the origin
(X1,X2) = (693.33,0) -----> the intercept of C2 on X1
For the rest of the vertices we find the intersection points of the inequality lines (C1 & C3 ) and (C2 &C3)
Turning the inequalities to equalities for solving the equations for C1 and C3:
X1 +7*X2 =4480............(1)
2*X1 +2*X2 =1620----------(2)
Multiplying (1) by 2 and subtracting from (2) we get
-12*X2 = -7340
ie X2 =611.67
Therefore plugging the value of X2 in any of the equations (1) or (2) we get the value of X1 as 198.31
Therefore the corner point is
(X1,X2) = (198.31,611.67) .......... intersection of C1 and C3
We will find the remaining cornet point by finding the intersection of C2 and C3:
Using similar approach
3*X1 +X2=2080..........(3)
28X1 +2*X2 =1620.......(4)
Multiplying (3) by 2 and subtracting from (4) we get
-4*X1 = -2540
X1=635
plugging the value of X1 in any of the equations (3) and (4) we get X2 =165
Therefore the remaining corner point is
(X1,X2) =(635,165)........intersection of C2 and C3
now we plug in the values of each of the vertices and corner points in the objective function to find the max and min values.
We see that the value of the objective function is maximum when X1=198.31 and X2=611.67
Hence this is the optimal solution
Answer 3: In order to calculate the dual price we need to increase the RHS of the constraint and see the impact on the objective function value. From steps 1 and 2 we see that for change in the value of RHS of the constraint 3 by 20 units the objective function changes by (962.9 -953.33) =9.57 units.
To solve ad find the dual price we need to increase the value of the constraint by one unit ie make 2*X1 +2*X2 <=1601
We solve the values for intersection of C1 and C3 :
We get X2=613.25
X1 =187.25
Now finding the value of the objective function using the new set of optimal solution (187.25,613.25)
The value of objective function is =953.81
Therefore for increase in the value of the resource constraint 3 by one unit the value of the objective function increases by (953.81-953.33) =0.48 which is the shadow price for constraint 3.
Answer 4: The meaning of dual price or shadow price is what is the change in the value of the objective function if the RHS ie the inequality value of the binding constraint is increased by one unit.If the constraints are non binding meaning that they have slack>0 and hence their shadow price will also be zero. The binding constraints have slack as zero and hence their dual or shadow price is non zero.
Objective function value X1 X2 Objective function value 0 640 800 0 0 0 693.33 0 693.33 186.67 613.33 953.33 640 160 840