1

Parametric analysis extends the single parameter sensitivity analysis by allowing wide range changes in the model parameters. Such an analysis can deal also with multiple changes (when several parameters are changing simultaneously), if these changes are related to one another in a linear fashion.
Two types of changes are discussed next: Changes in objective function coefficients and in constraints’ right hand side. These changes were discussed before in the context of single change sensitivity analysis. Although the 100% rule extended this discussion to cover also multiple changes, its scope was limited. The parametric analysis studied here will shed more light on the effects of wide range changes of input parameters, for both the single and multiple change cases.
The following example will support our discussion.
Max 7X
1
+ 10X
2
ST 3X
1
+ 4X
2

12000
X
1
+ X
2

3500
2X
1
+ 3X
2

7500
10X
1
+12X
2

36000
X
1
and X
2
are non-negative
Study the objective function value as a function of Q when the coefficient ‘7’ in the objective function changes to ‘7 + Q’.
The objective function is now re-written as OBJ = (7+Q)X
1
+ 10X
2
= 7X
1
+ 10X
2
+ QX
1
.
To perform the analysis more easily we’ll define a new variable X
3
formulating a new constraint X
3
= X
1
, and write the objective function as OBJ = 7X
1
+ 10X
2
+ QX
3
. The parametric analysis is performed by changing Q. Note that 7X
1
+10X
2
is actually the objective value before any change was made, that is OBJ(Q=0). Thus OBJ(Q) =
OBJ(0)+X
3
Q.
Step 1:
When running the model in SOLVER for Q=0 (which results in the original model), the range of optimality for Q is [– .333, 3], while the optimal solution is X
1
= 3000 and X
2
=
500 and X
3
=3000. Comment: The range of optimality for C
1
is [7-.333, 7+3].
The objective value for -.333

Q

3 is OBJ(Q) = OBJ(0)+3000Q = 26000 + 3000Q. The graph that describes OBJ is a straight line in Q, with a slope of 3000. For Q= -.333, OBJ
= 26000 + 3000(-.333) = 25000, and for Q = 3, OBJ = 26000+3000(3) = 35000 (see the graph below). The slope of this function is 3000.
2

Now we extend the study to ranges above 7 + 3 = 10 and below 7 – .333 = 6.667.
Step 2: Studying the objective function for values of C
1 above 7+3=10.
Select Q = 3.1 (a value slightly greater than 3). The optimal solution changes because
7+Q falls above the upper bound of the range of optimality. Observation: Note that X
1
is expected to increase because its coefficient increases, making it more attractive when
OBJ is maximized. Note that in the modified objective function OBJ(Q) = 7X
1
+ 10X
2
+X
3
Q, X
3
is the slope and Q is the variable. Now, if X
1
(and thus X
3
) increases the slope of OBJ increases. When running SOLVER again with Q = 3.1 we get the following solution:
X
1
= 3500; X
2
= 0; X
3
= 3500; the allowable decrease for Q is .1, and the allowable increase is infinity. S o for 3

Q

OBJ(Q) = OBJ(3) + 3500(Q – 3) = 35000 + 3500(Q – 3) =
24500 + 3500Q

The slope now is 3500 (larger than before as expected).

This result tells us that for any value of Q greater than 3 (3.1-.1 = 3) the optimal solution will not change because the range of optimality upper bound for Q is infinity. Therefore there is no need to increase Q any more in order to find a new equation for OBJ.
Step 3: Studying the objective function for values of C
1 below 7-.333=6.667.
Select Q = -0.4 (a value slightly below -.333). When running SOLVER we get:
X
1
= 0; X
2
= 2500; X
3
= 0; the range of optimality for Q is [-

,-.333]. Thus OBJ(Q) =
OBJ(-.333) + 0(Q-(-.333)) = 25000+0 = 25000. Again, no new value of Q needs to be selected in order to find a new line, because the line OBJ = 25000 exists for all Q from
-.333 down to -

We can now describe the optimal value of the objective function as a function of Q.
OBJ = (7X
1
+10X
2
) + X
1
Q
OBJ=25000
OBJ=26000+3000Q
OBJ=24500+3500Q
Q
-.333 3
From this graph one can read the objective value for any change Q in the original value
(7) of the coefficient C
1
.
Parametric analysis can be performed on a constraint right hand side in a similar manner to the one performed on an objective coefficient. In the example presented above, we
3

allow the first constraint change from 12000 to 12000+Q. As you certainly remember, for changes in the right hand side of a constraint within the range of feasibility the shadow prices don’t change; so for all Q between the allowable decrease and increase one can recalculate the objective value for different values of Q using the current objective value, and the known shadow price. For right hand side changes wider than the range of feasibility, new shadow prices need to be found, and the new objective value can then be determined.
Let us follow the procedure that produces a graph describing the objective value as a function of Q. To do this conveniently we make a few changes in the original model. The constraint of interest, 3X
1
+ 4X
2

12000+Q, is written first as 3X
1
+ 4X
2
– Q

12000, then Q is replaced by a new variable X
3
, so 3X
1
+ 4X
2
– X
3

12000, and a new constraint is added, X
3
= Q. The change ‘Q’ in the original constraint appears now as a right hand side of the new constraint (X
3
= Q), and so has a range of feasibility and a shadow price of its own. With these observations we’ll be able to more easily describe
OBJ as a function of Q. One more observation is needed before we start the procedure.
The new variable X
3
is of the urs type, because Q can be either positive or negative (the constraint right hand side, 12000, can either increase or decrease). So we’ll use the transformation X
3
= X
3
+ - X
3
. For clarity, here is the modified model.
Max 7X
1
+ 10X
2
ST 3X
1
+ 4X
2
– (X
3
+
- X
3
-
)

12000
X
1
+ X
2

3500
2X
1
+ 3X
2

7500
10X
1
+12X
2

36000
X
3
+ - X
3
= Q
All variables are non-negative
Generally, we’ll change the value of Q in the constraint X
3
+
- X
3
-
= Q over a wide range, create consecutive ranges of feasibility, use Solver to observe the shadow prices, and formulate the objective function for each range as a function of Q.
Step 1:
The process begins with the original model where Q = 0. The new constraint becomes
X
3
+
- X
3
-
= 0. The optimal objective value is 26000; the range of feasibility for the new constraint is [0-1000]

Q

0+

OBJ remains 26000 for all Q within this range of feasibility of (-1,000,

), because the shadow price is zero for this range (which by the way should not surprise you since at the current optimal solution there is a slack of 1000 units in this constraint). Comment: Note the range of feasibility for constraint #1 that corresponds to the current range of feasibility of Q: [12000-1000, 12000+
 
Step 2:
Let Q = -1001, that is slightly below -1000. For this run, the allowable increase of Q is 1 and the allowable decrease is 999 (from the Solver output). This means the range of feasibility for Q is: Lower bound = -1001-999 = -2000; Upper bound = -1001+ 1 = -1000; so the range of feasibility for Q is -2000

Q

-1000; the shadow price is 1.
Then, OBJ(Q) = OBJ(-1000) + [Shadow price][

in right hand side] =
4

26,000+1(Q-(-1000)) = 27,000 + 1Q for -2000

Q

-1000.

-2000 Q -1000 0
The objective value drops in this range of feasibility from OBJ(-1000) = 26,000 to
OBJ(-2000) = 25,000.
Step 3:
Let Q = -2001, that is slightly below -2000. Following the same guidelines for calculating the range of feasibility as in step 2, it is found that -12,000

Q

-2000, while the shadow price is 2.5. Then OBJ(Q) = OBJ(-2000)+ [Shadow price][

in right hand side]= =
25,000 + 2.5[Q-(-2000)] = 30,000 + 2.5Q for -12,000

Q

-2000.
Step 4:
Q cannot take on values below -12,000, because the right hand side of constraint #1 becomes negative (12,000 – Q), which lead to an infeasible solution.
The objective function plot follows (in the graph below ‘b’ represents the right hand side of 3X
1
+ 4X
2
 b, where b = 12000+Q):
OBJ
OBJ = 27,000+1Q
OBJ = 26,000
OBJ = 30,000+2.5Q
Q = -12000 b = 0
Q = -2000 b = 10,000
Q= -1000 b = 11000
Q = 0 b = 12000
Q
5

Wide range multiple changes in the input parameters of the linear model may be required for various reasons. For example, in a multi-period production problem a 10% increase of demand in all periods might occur, a change that alters the right hand sides of several constraints simultaneously. Such scenarios are the topic of this section.
As before, we focus on changes in the objective coefficients first, and turn to the constraints later.
Two types of changes will be considered:
(i) An increase (addition) or decrease (subtraction) in the value of a parameter, that is “new parameter value = old parameter value   ”
(ii) A percentage increase or decrease in a parameter value, that is
“new parameter value = old parameter value(1  
".
Changes in the two objective coefficients of the linear programming model presented below are to be analyzed.
Max 7X
1
+ 10X
2
ST 3X
1
+ 4X
2

12000
X
1
+ X
2

3500
2X
1
+ 3X
2

7500
10X
1
+12X
2

36000
Assume for each one unit change in C
1
=7 there is a same-direction 2 units change in
C
2
=10. For example, the objective coefficients may take on the new values C
1
= 7+1,
C
2
=10+2; or C
1
=7–1.5, 10–3; etc.
If C
1 is changing by
 the new coefficient values are C
1
= 7 +
 and
C
2
=10 + 2

Thus, the objective function becomes Max (7+

X
1

)X
2 function can be re-written as Max 7X
1
+10X
2
+

(X
1
. This
+2X
2
). If we define a new variable called X
3
by X
3
= X
1
+2X
2
the objective function becomes Max 7X
1
+ 10X
2
+

X
3
. Recall that

is a constant, so we have an objective coefficient
 that changes while the other coefficients (‘7’ and ‘10’) remain unchanged. Single parameter sensitivity analysis can be performed in the usual manner in order to find ranges for
 for which optimal solutions remain unchanged; note that by changing the parameter
 more than one original objective coefficient is changing (in our case the parameters ‘7’ and ‘10’).
This provides a tool for determining controlled ranges of optimality for both C
1
and C
2
. Let us proceed with the above model, by repeatedly running it with different values of

The modified model is:
Max 7X
1
+ 10X
2
+

X
3
ST 3X
X
1
1
+ 4X
+ X
2
2
2X
1
+ 3X
2

12000


3500
7500
6

10X
– X
1
+12X
2

36000
1
– 2X
2
+ X
3
= 0
 This is the additional constraint that
 defines X
3.
To find the ranges of optimality for
 we follow the procedure demonstrated next.
The procedure
Step 1: Let

= 0. With this value for

we are actually solving the original model. Below you’ll find an excerpt of the SOLVER output containing the optimal solution for this problem, including the

range of optimality, when the original value of

is ‘0’.
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$B$25 X1
$C$25 X2
3000
500
0
0
7
10
3
0.5
0.333333333
3
$D$25 X3 4000 0 0 1 3
From the output we see that for all the values of
 between -3 and 1 (0 – 3= -3 and
0 + 1=1) the optimal solution is X
1
= 3000 and X
2
= 500, and X
3
= X
1
+ 2X
2
=
3000+2(500)=4000, (but remember X
3
is not part of the original model). Consequently, for all the values of C
1
+

and C
2
+2
 this optimal solution does not change as long as
 remains within the boundaries of (-3,

). To demonstrate, assume we know that there is a decrease of 2 units in C
1
. So

= -2, and as a result C
1
= 7 – 2 = 5, and C
2
= 10 – 2(2) = 6.
Since

= -2 falls inside the range of optimality the optimal solution remains unchanged
(X
1
= 3000 and X
2
= 500). Comment: The 100% rule does not lead to a clear conclusion here because the sum of ratios for the mentioned changes is greater than 1:
2
.
333
 4
3

1 .
The objective function for the range of -3
  
1
 is OBJ(

= OBJ(0) +

(4000) = 26000 +
4000

Note that OBJ(

-3) = 26000+4000(-3) = 14000, and OBJ(

26000+4000(1)
= 30000. These two values will be used to construct the graph of the objective functions.
Step 2: We extend the study to cover ranges for

above and below its current range of optimality

graphical demonstration of the current situation for the range of optimality for
 is shown below:

-3 0 +1
7

2.1

Let us select

= 1.1 (slightly beyond the upper bound of the current range-ofoptimality). We rerun the modified linear programming again, this time with the new value of

.
The Solver output follows:
Cell Name
$B$25 X1
$C$25 X2
Final
Value
0
2500
Reduced
Cost
-0.0333
0
Objective Allowable Allowable
Coefficient Increase Decrease
7
10
0.0333
1E+30
1E+30
0.05
$D$25 X3 5000 0 1.1 1E+30 0.1
From the output we see that for all the values of
 between 1 and
 
( 1.1 – 0.1=1, and
1.1 +

) the optimal solution is X
1
= 0 and X
2
= 2500. Consequently, for all the values of C
1
+

and C
2
+2
 this optimal solution does not change as long as

remains within the boundaries of its range-of-optimality (1,

). For example, assume we know that there is an increase of 2 units in C
1
. That is

= 2 (C
1
= 7 + 2 = 9, and C
2
=
10 + 2(2) = 14). Since

= 2 falls inside the range of optimality (1,

) the optimal solution remains unchanged (X
1
= 0, X
2
= 2500, and X
3
= 5000).
The optimal objective function for the range 1
     is OBJ(

) = OBJ(1) +
5000(

= 30000+5000(

2.2
Since the upper bound of the new range of optimality is
 there is no need to check other values of range boundaries on the right hand side since the optimal solution remains X
1
= 0 and X
2
= 2500 for every value of

above 1.
2.3
We change direction, and check boundaries on the left hand side. Since we started with a range of optimality of (-3, 1), we let

= -3.1. From the rerun of the linear programming modified model with

= -3.1 we find a range of optimality for

of
(-7, -3). The optimal solution for this range is X
1
= 3500, X
2
= 0 and X
3
= 3500. The optimal objective function for the range is OBJ(Q) = OBJ(-3) + 3500(

=
Recall, OBJ(-3) = 14000, so OBJ(Q) = 24500 + 3500

.
2.4
We continue our search by selecting the value

= -7.1, and find the range
(-7, -

The optimal solution is

X
1
= 0, X
2
= 0 and X
3
= 0. The optimal objective function for the range is OBJ(Q) = 0

2.5
The search can be stopped because the current range spans down to -

8

A graphical demonstration of the optimal objective function is provided next:
OBJ
30000
25000+5000
 
-7
24500+3500
 
26000+4000
 
-3
14000
1
 
Additional example
Let us re-visit example 2 of the notes on special LP models. In this example we presented a multi-period production problem for a four quarter time horizon. The model formulated there was:
Cost
Prod1<40
Prod2<50
Prod3<40
Prod4<60
Q1
Q2
Q3
Q4
X1 X2 X3 X4 Y1 Y2 Y3 Y4 Inv1 Inv2 Inv3
400 420 390 375 450 450 450 450 40 42 39
1
1
1
1
1 1 -1
1 1 1 -1
1 1 1
<= 40
<= 50
<= 40
<= 60
=
=
-1 =
30
60
75
1 1 1 = 25
X j
represents the regular production in quarter j; Y j represents the overtime production in quarter j; Inv j
represents the inventory at the end of quarter j, and the holding cost is 10% of the regular time production cost. The current optimal production plan is given below:
9

Final Reduced Objective Allowable Allowable
Cell Name Value
$B$2 X1 40
$C$2 X2
$D$2 X3
50
40
$E$2 X4
$F$2 Y1
$G$2 Y2
$H$2 Y3
$I$2
$J$2
Y4
Inv1
$K$2 Inv2
0
10
0
25
0
0
35
Cost
40
0
0
75
0
42
0
0
0
0
Coefficient Increase Decrease
400 10 1E+30
420
390
375
450
450
450
450
40
42
30
60
75
1E+30
40
42
1E+30
10
1E+30
1E+30
1E+30
1E+30
40
10
60
75
40
42
$L$2 Inv3 0 114 39 1E+30 114
Management would like to know the minimum change in regular production cost needed in quarter 1, 2 and 3 (same change in each quarter) before inventory in these quarters will be considered.
Answer: A minimum change will be achieved if the production cost in quarter 1 is increased, while the production cost in quarter 2 and 3 is decreased. Such changes will cause an increase in holding cost in quarter 1 and a decrease in holding cost in quarters 2 and 3. Let’s assume the change is 
. Then the production costs are 400 +

, 420 -

, and 390 -

. The holding costs are 40+.1

, 42-.1

, and 39-.1

.
For comparison we’ll answer by the 100% rule as well as by parametric analysis.

Using the 100% rule:
 infinity) + (

Infinity

+ (

)/(

42) + (

)/(

114) = 1
Quart1 Quart2 Quart3 Quart1 Quart2 Quart3
Production Production Production Inventory Inventory Inventory
Solving the equation we find the minimum production cost change is

= 8.829. This does not guarantee the solution will change if the cost change is larger (note the parametric analysis below), but certainly a smaller production cost change will not force any change in the optimal production plan.
10

Using parametric analysis
The model is changing to:
Cost
Inv Inv Inv
X1 X2 X3 X4 Y1 Y2 Y3 Y4
40
40
0
50
42
0
40
39
0
25
37
5
0
45
0
1 2 3
 
0 35 0 10
45 45 45
0 0
0 0
-
49
0 40 42 39 0
Prod1<40
Prod2<50
Prod3<40
Prod4<60
1
1
1
1
Q1 1 1 -1 =
Q2 1 1 1 -1 =
Q3 1 1 1 -1 =
Q4 1 1 1 =
New
Const. 1 -1 -1 0.1 -0.1 -0.1 -1 =
As explained above

= X1 – X2 – X3 + .1Inv1 – .1Inv2 – .1Inv3.

is a new variable added to the objective function to reflect the changing coefficients, along with a new constraint that defines

. Theta must be defined as an URS, Note that the current value of the Theta coefficient is set to “zero”. From the sensitivity analysis
<
=
<
<
=
<
=
=
0
7
5
2
5
0
3
0
6
4
0
6
4
0
5
0
0 results for the modified model, the range of optimality of the Theta coefficient is [0 +
9.0909, 0 – 30]. We conclude that the production cost change should be at least $9.09 before the solution changes. This agrees with the results obtained by the 100% rule and demonstrates that the result there is a lower bound on the required change.
Suppose an original objective coefficient ‘C’ changes by a fraction of

Then, its new values is C(1 +

)
 or C(1–

). This concept can be applied to multiple changes of the objective function. For example if 1% increase of the coefficient 7 in our example occurs simultaneously with 2 percent decrease in the coefficient 10, the objective can be written as 7(1+

)X
1
+ 10(1-2

)X
2
= 7X
1
+ 10X
2
+

(7X
1
– 20X
2
) = 7X
1
+ 10X
2
+

X
3
, where
X
3
= 7X
1
– 20X
2
(which needs to be added to the original model as a new constraint, as before). Here is the modified model:
Max 7X
1
+ 10X
2
+

X
3
ST 3X
X
2X
1
1
1
+ 4X
+ X
+ 3X
2
2
2

12000


3500
7500
11

10X
1
+12X
2
+7X
1
– 20X
2
– X
3

36000
= 0
 This is the additional constraint that defines X
3.
The rest of the analysis is similar to the one performed above.
Important comment: X
3
must be defined as urs , because 7X
1
– 20X
2
might take on negative values.
For the original model presented before, assume for each unit change in the right hand side of constraint 1 there is an opposite unit change in the right hand side of constraint 2.
The resulting constraints can be expressed as:
Constraint 1: 3X
1
+ 4X
2

12000 +
 or

3X
1
+ 4X
2
-
 
12000

Constraint 2: 1X
1
+ 1X
2

3500 -
 or 1X
1
+ 1X
2
+
 
3500.
Let us add a constraint of the form X
3
=
 and change the two constraints to:
+ 1X
2
+ X
3

3500. The modified model is: 3X
1
+ 4X
2
– X
3

12000, and 1X
1
Max 7X
1
+ 10X
2

ST 3X
1
+ 4X
2
- X
3
 
12000
1X
1
+ 1X
2
2X
1
10X
+ 3X
2
1
+ 12X
+X
3
2
 
3500

7500

36000
X
3

=

Changing ‘  ’ actually changes X
3
 which in turn affects the right hand sides of the two constraints simultaneously. Since only one right hand side is formally changing (

), we can argue that as long as ‘  ’ falls inside its range of feasibility, all the shadow prices remain unchanged, thus can be used to calculate the new value of the objective function.
We then ‘cross the borders’ of the range of feasibility and generate new ranges and new shadow prices. The process stops when no feasible solution can be obtained. Observe the results and read the explanations below.
Important comment: X
3
must be defined as an urs variable, because both an increase and a decrease in the two constraints of interest, constraint 1 and 2, must be considered.

Range of feasibility
0 -1000 to 0
Shadow price (for X
0
3
=

0.1 0 to 1000 - 1
1001 1000 to 3500 -10
At this point we can stop, because
 cannot take on values greater than 3500 otherwise constraint 2 right hand side becomes negative: 1X
1
+ 1X
2

3500 -
 
However, the parametric analysis is not completed yet. Negative values of

need to be applied to the right hand sides in order to complete the analysis.
12

Parametric analysis – continued.

Range of feasibility
0 -1000 to 0
-1001 -2000 to -1000
-2001 -12000 to -2000
Shadow price (for X
3
=

0
1
2.5
First run results repeated
+2.5
OBJ
-10
-12K -2K -1K 1K 3.5K

For the original model presented before, assume for each 1% increase in constraint 1 there is 1% decrease in constraint 2. Let
 represent the change expressed as a fraction
(i.e. for 1% change

=.01), then for
 units increase in constraint 1 the value of 12000 on the right hand side of the constraint becomes 12000(1+

, while the value 3500 on the right hand side of constraint 2 becomes 3500(1-

). The modified model follows:
Max 7X
1
+ 10X
2
ST 3X
1
+ 4X
2

12000(1+

)
1X
1
+ 1X
2

3500(1-

)
2X
1
+ 3X
2

7500
10X
1
+ 12X
2

36000
The two changing constraints become 3X
1
+ 4X
2
- 12000
  
12000, and
1X
1
+ 1X
2
+ 3500
  
3500. A new variable X
3
 and a new constraint X
3
=

are added.
The modified model is now:
Max 7X
1
+ 10X
2
+ 0X
3

ST 3X
1
+ 4X
2
- 12000X
3
 
12000
1X
1
+ 1X
2
+ 3500X
3
 
3500
2X
1
+ 3X
2

7500
10X
1
+ 12X
2

36000
X
3

=

(‘0’ is the initial value of ‘
 ’. See the results below)
13

Analysis results:

Range of feasibility
0 0 to .285
.286 .285 to 1
Shadow price (for X
3
=

- 3500
-25000
For negative values of
 we have:

Range of feasibility
-.01 -0.083 to 0
-.085 -0.167 to -0.083
-.168 -1 to -0.167
Shadow price (for X
3
=

0
12000
30000
Comment:

can only range from -1 through +1. This is so because

represents the fractional change in the right hand side of constraints 1 and 2. Since the right hand side of the original constraints cannot be negative for a feasible solution to exists, and since 1% change in constraint 1 draws a 1% opposite change in constraint 2,

cannot be either greater that 1 or less than -1.
Interpretation: . The ranges of feasibility allow

to change without altering the shadow prices. These shadow prices can be used to determine the objective value for a given percent change in constraints 1 and 2. The shadow price used is the one present at the range of feasibility where

falls. For example, let us allow

= -.25 (25% decrease in constraint 1 and 25% increase in constraint 2). This

falls in the range [-1, -.167], so the relevant shadow price is 30000. Since the objective value was 26000 at the original model before any change was made, the new value can be calculated ( range by range) as follows:
OBJ(-.25) = 26000+[Shadow price 1][Range 1]+[Shadow price 2][Range 2]+Shadow price 3][Partial range 3] =26000+0[-.083-0]+12000[-.167-(-.083)]+30000[-.25-(-.167)] =
22500.
Additional Example
A farming company has two farms that grow wheat and corn. Yield and costs of growing the crops differ between the farms because of differing soil conditions. Particularly,
Bushels per acre
Corn
Wheat
Farm 1 Farm 2
500 650
400 350
Cost per acre
Corn
Farm 1 Farm 2
$100 $120
Wheat $90 $80
Each farm has 100 acres available for cultivation; 11,000 bushels of wheat and 7000 bushels of corn must be grown. (a) Determine the plan that minimizes the cost of meeting
14

the requirements. (b) How the total cost changes if the company transfers a certain percentage of the land between the farms.
Solution
The model:
CORN1 WHEAT1 CORN2 WHEAT2
MIN COST
WHEAT=7K
CORN=11K
100
500
90
400
120
650
80 =
350 =
<=
11000
7000
100
ACRES 1
ACRES 2
1 1
1
<=
1 =
100
0
To answer part (b) we run parametric analysis on the last two constraints. This calls for the following modified model:
CORN1 WHEAT1 CORN2 WHEAT2
 
MIN
WHEAT=7K
100 90
400
120 80
350 = 11000
CORN=11K
ACRES 1
500
1 1
650
100
=
<=
7000
100
ACRES 2 1 1 -100 <= 100
New cont. 1 = 0
The range of feasibility for the new constraint is [-.892,

]. This means that farm 1 can grow smaller up to 89.2% or grow larger up to 72.5%, while the shadow price remains unchanged (0), thus having no effect on the total cost. Re-running the model with

=
.726, the range of feasibility is [.725, 1], while the shadow price is 142. 85. This means the total cost increases at a rate of $142.85 for every 1% of the land transferred from farm
1 to farm 2. For this case the cost as a function of theta is: Cost(

= .725) + 142.85

for
.725
<  ≤ 1. Now we turn to the negative values of Theta. Re-running the model with

.893, the range of feasibility is [-1, -.892], while the shadow price is -1000. This means the total cost drops at a rate of $1000 for every 1% of the land transferred from farm 1 to farm 2. For this case the cost as a function of theta is: Cost(

= -.892) - 1000
 for
-1 <  ≤ −.892
.
Total
Cost

-.892
15
.752

Solutions
1.
For the following model perform parametric analyses as requested next: a.
On the objective function coefficients, assuming for each unit change in C
1
, there are opposite direction 2 units change in C
2 and C
3
. b.
On the right hand side of the constraints, assuming for each unit increase in constraint 1, there are 2 units decrease in constraint 2
.
c.
Repeat part ‘a’, this time for the two cases of simultaneous changes: (i) +1; -2; +2 and (ii) -1; +2; -2 for C
1
, C
2
, C
3
respectively.
Min 17x
1
+29x
2
+ x
4
2x
1
+ 3x
2
+ 2x
3
+3x
4

40
4x
1
+ 4x
2
+ x
4

10
3x
3
- x
4
= 0 x
1
, x
2
, x
3
, x
4

0
2.
Cox Cable Company (COX) needs to lease warehouse storage space for five months at the start of the year. COX can purchase variety of lease contracts to meet the requirements for space per each month. There are many ways COX can meet its lease space requirements per month. Here are a few examples: (i) COX can renew leases monthly five times (January through May); (ii) Lease for two months (January and
February), then for another three months (March through May); (iii) Lease for
January through March, and from February through April and from March through
May. As you see, more than one lease contract can the same month. The space requirements (in square feet) and the leasing costs (in dollar per thousand square feet) are given in the two separate tables below:
Space
Month Required
Jan 15,000
Feb
Mar
Apr
May
10,000
20,000
5,000
25,000
Lease
Length
Lease
Cost
1 month $280
2
3
4
5
450
600
730
820.
Define the following variables: X ij
– the space leased in thousand square feet for the period starting in month ‘i’ and ending in month ‘j’. For example, X
24
indicates how many thousand square feet are leased for a three months contract from February through April. The objective function of this model is
Min 280X
11
+450X
12
+600X
13
+730X
14
++820X
15
+280X
22
+450X
23
+….(complete the objective function). The constraints are:
January: X
11
+X
12
+X
13
+X
14
+X
15

15
February: X
12
+ X
13
+X
14
+X
15
+X
22
+X
23
+X
24
+X
25

10
March: X
13
+X
14
+X
15
+ X
23
+X
24
+X
25
+X
33
+X
34
+X
35

20
April: X
14
+X
15
+ X
24
+X
25
X
34
+X
35
+X
44
+X
45

5

May: X
15
+X
25
+X
35
+X
45
+X
55

25
Now answer the following questions (both simple sensitivity analysis and parametric analysis questions are presented):
16

a.
Determine the optimal leasing schedule, and the optimal leasing cost. b.
What is the effect of a 10% increase in the space requirement in January on the total cost? c.
What is the effect of 10% increase in the space requirement in January as well as a 10% reduction in the space needed February and March? Use the 100% rule first. d.
Suppose for each
 sq-ft change in space required in January and February each, there is an expected
 sq-ft change in March and May each in the opposite direction. Show how the total cost is affected when
 changes

A graph will make the presentation very clear. e.
Using the model defined in part ‘d’, determine the total cost if the January space required is 8.000 sq-ft. Use both Solver and a manual calculation.
3. Tube Steel Incorporated (TSI) is optimizing production at its 4 hot mills. TSI make
8 types of tubular products which are either solid or hollow and come in 4 diameters.
The following table show unit cost (in $s) of each product at each mill and the extrusion times (in minutes) for each allowed combination (product - mill).
Product
.5 inch Solid
1 inch Solid
2 inch Solid
4 inch Solid
.5 inch Hollow
1 inch Hollow
2 inch Hollow
4 inch Hollow
Unit Cost
Mill 1 Mill 2 Mill 3 Mill 4
.10 .10 .15
.15 .18 .20
.25 .15 .30
.55 .50
.20 .13 .25
.30 .18 .35
.50 .28 .55
1.0 .60
Unit Time
Mill 1 Mill 2 Mill 3 Mill 4
.50 .50 .10
.60 .60 .30
.80 .60 .60
.10 1.0
1.0 .50 .50
1.2 .60 .60
1.6 .80 .80
2.0 1.0
Yearly minimum requirements for the solid sizes (in thousands) are 250, 150, 150, and 80. For the hollow sizes they are 190, 190, 160, and 150. The mills can operate up to three 40 – hour shifts per week, 50 weeks a year. Several ideas are raised by management aiming at cost savings. To answer some questions about these ideas use parametric analysis after formulating a linear model that meets demand and uses available capacity restrictions at minimum cost. Here are the ideas to study: a.
Reduce unit costs of production at high-cost Mill 4. To what levels unit cost of each of the six products that can be produced in Mill 4 would have to be reduced before there could be any change in the optimal solution, if the changes are simultaneous and hollow tubes cost reduction is 10% larger than these of the solid tubes? b.
Install equipment to produce 4-inch solid and 4-inch hollow tubes at Mill 4. The new equipment would produce either product in 1 minute per unit. Taking both products together, determine the unit production cost that would have to be achieved to make it economical to use the new facilities, if it is estimated that 4 –
17

inch hollow tube will be $.90 more expensive than solid tube when produced in
Mill 4. c.
Changing the production capacities of Mills 3 and 4 in one direction (either increase or decrease), while those of Mills 1 and 2 changes in the opposite direction. If the percent changes in capacity are the same, what percent yields
$10,000 dollars of savings, and in which direction for each mill?
18