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90 CHAPTER 3 LINEAR PROGRAMMING: SENSITIVITY ANALYSIS AND INTERPRETATION OF SOLUTION

Thus the slope of line B is 1.5 and its intercept on the D axis is 1062.
Now that the slope of lines A and B have been calculated, we see that for extreme
point fi to remain optimal we must have:

1:5 slope of objective function 0:7 (3:1)

Let us now consider the general form of the slope of the objective function line.
Let C S show the profit of a standard bag, C D show the profit of a deluxe bag, and P
show the value of the objective function. Using this notation, the objective function
line can be written as:
P ¼ C S S þ C D D

Writing this equation in slope-intercept form, we obtain:
C D D ¼ C S S þ P
and

C S P
D ¼ S þ
C D C D
We see that the slope of the objective function line is given by C S /C D . Substituting
C S /C D into Expression (3.1), we see that extreme point fi will be optimal as long as
the following expression is satisfied:

C S
1:5 0:7 (3:2)
C D

To calculate the range of optimality for the standard-bag profit contribution, we
hold the profit contribution for the deluxe bag fixed at its initial value C D ¼ 9. Doing
so in expression (3.2), we obtain:

C S
1:5 0:7
9
From the left-hand inequality, we have:

C S C S
1:5 or 1:5
9 9
Thus,
or C S 13:5
13:5 C S
From the right-hand inequality, we have:
C S C S
0:7or 0:7
9 9
Thus,

C S 6:3
Can you calculate the
range of optimality using
the graphical solution Combining the calculated limits for C S provides the following range of optimality for
procedure? Try Problem 3. the standard-bag profit contribution:
6:3 C S 13:5

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