Solutions for

Problems on Parametric Analysis

1.The modified objective is Min (17+xxxxrewritten as Min 17x1 + 29x2 + x4 + (x1 – 2x2 – 2x3). If we define a new variable Z =x1 – 2x2 – 2x3 (and add this relationship to the original set of constraints), we get the following model:

Min 17x1+29x2+ x4+Z

ST 2x1+ 3x2+2x3+3x4<=40

4x1+4x2+4x4>=10

3x3 –1x4 = 0

1x1–2x2+2x3–1Z = 0

The sensitivity analysis on  yields the following results:

When  is… The Range for isThe optimal solution is

0 [-9.6, infinity]x1=1; x2=0; x3=0.5

-9.7 [-17, -9.6]x1=2.5; x2=0; x3=0

-17.1 [-infinity, -17]x1=20; x2=0; x3=0

X11 / X12 / X13 / X14 / X15 / X22 / X23 / X24 / X25 / X33 / X34 / X35 / X44 / X45 / X55
0 / 0 / 0 / 0 / 15 / 0 / 0 / 0 / 0 / 5 / 0 / 0 / 0 / 0 / 10

2.The optimal solution as obtained from Excel Solver is summarized below:

a.Make a lease contract for the months of January through May for 15,000 sq-ft;
Make a lease contract for the month of March for 5,000 sq-ft;

Make a lease contract for the month of May for 10,000 sq-ft.

  1. An increase of 1,500 sq-ft in the required space for the month of January falls within the range of feasibility of constraint #1 (the maximum increase is 5,000). The cost increases by (260)(1,500) = $39,000 where 260, is the shadow price.
  2. From the 100% rule and the sensitivity analysis results we have:
    (1500/5000) +(1500/infinity) +(2000/5000 = 3500/5000 < 1. So, the shadow prices remain unchanged, and therefore we can calculate the new total cost by:
    New Objective = 16,500+260(1500) + 0(1500) +280(-2000)
  3. To answer this problem we run parametric analysis on four constraints:
    The new model is:

X11 / X12 / X13 / X14 / X15 / X22 / X23 / X24 / X25 / X33 / X34 / X35 / X44 / X45 / X55
0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0
COST / 280 / 450 / 600 / 730 / 820 / 280 / 450 / 600 / 730 / 280 / 450 / 600 / 280 / 450 / 280
Jan / 1 / 1 / 1 / 1 / 1 / >= / 15 / +
Feb / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / >= / 10 / +
Mar / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / >= / 20 / 
Apr / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / >= / 5
May / 1 / 1 / 1 / 1 / 1 / >= / 25 / 

We treat  as a variable and remove it from the right hand side of each constraint where it appears. Then we add the constraint whereis a parameter whose value is going to change based on the ranges of feasibility associated with it (such an approach was taken when performing the parametric analysis for the objective coefficients). Since represents a change in constraints’ right hand side, we shall identify ranges of for which the shadow prices don’t change, and then use them appropriately to calculate new values of the objective function. Observe the modified model

X11 / X12 / X13 / X14 / X15 / X22 / X23 / X24 / X25 / X33 / X34 / X35 / X44 / X45 / X55 
0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0
COST / 280 / 450 / 600 / 730 / 820 / 280 / 450 / 600 / 730 / 280 / 450 / 600 / 280 / 450 / 280
Jan / 1 / 1 / 1 / 1 / 1 / -1 / >= / 15
Feb / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / -1 / >= / 10
Mar / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / +1 / >= / 20
Apr / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / >= / 5
May / 1 / 1 / 1 / 1 / 1 +1 / >= / 25

Theta 1= b

‘b’ is a changing parameter

Now we perform the parametric analysis on b to find ranges for where the shadow prices remain unchanged. We start with which gives the original model ranges of feasibility. A summary of this procedure appears in the table below.

When  is:The Range for is: The shadow prices are:

Constraints:JanFebMarAprMayTheta

0 [-10, 2.5] 260 0 280 0 280-300

2.6 [2.5, 5] 280 0260 0280-260

5.1 [5, 15] 280170 90 0280+80

15.1 [15, 20]280170 0 90280170

20.1 [20, 25]280170 0280 90360

25.1 [25, infinity]280170 0280 0450

-10.1 [-15, -10]220 0280 40280-340

-15.1 [-infinity, -15] 0 0280 40280-560

d.For a required space of 8 sq-ft in January = -7. You can use the known solution for the range of feasibility of that is [-10, 2.5]: Cost = 16500+(-300)(-7) = 18600

Comment: We could use the shadow prices of the original constraints to get the same results:16500+(-7)(260)+(-7)(0)+(+7)(280)+(+7)(280) = 18600.

Let us add another question here. Suppose the required space in Jan. increase from 15 to 25. That is Observing the results of our parametric analysis we realize that we cross the borders of two ranges(at = 2.5 and at = 5). The Cost for this set of changes is calculated as follows:

  • Cost (when = 2.5) = 16500 +(-300)(2.5) = 15750
  • Cost (when =5) = 15750 + (-260)(2.5) = 15100
  • Cost (when =10) = 15100 + (80)(5) = 15500

The following graph will demonstrate the situation

3.To formulate the model we define the variables Xij that represent how many units of solid tubes type ‘i’ are produced in mill ‘j’, and Yij that representhow many units of hollow tubes type ‘i’ are produced in mill ‘j’ (types of tubes are defined by their diameters).

The model:

Min .10X11+.1X12+.15X14+.15X21+.18X22+.20X24+…+ .55X41+.50X42+
.20Y12+.13Y13+.25Y14+…+1.0Y42+.60Y43

Subject to:

Capacity constraints:

.5X11+.6X21+.8X31+.10X41 360,000 [(3 shifts)(40 hours)(50 weeks)(60 minutes) = 360,000]

.50X12+.60X22+.60X32+1.0X42+1.0Y12+1.2Y22+1.6Y31+2.0Y42360,000

.50Y13+.60Y23+.80Y33+1.0Y43360,000

.10X14+.30X24+.60X34+.50Y14+.60Y24+.80Y34360,000

Demand constraints:

X11+X12+X14250,000Y12+Y13+Y14190,000

X21+X22+X24150,000Y22+Y23+Y24190,000

X31+X32+X34150,000Y32+Y33+Y34160,000

X41+X4280,000Y42+Y43150,000

All the Xij arenon-negative.

  1. There are four objective coefficients changing simultaneously by the same amount. The new objective is:

Min .10X11 +…+(.15 – )X14+(.20 – )X24+(.30 – )X34+...+(.25 – )Y14+(.35 – )Y24+
+(.55 – )Y34+… = Min [Original objective] – whereX14+X24+X34+Y14+Y24+Y34 (to be added as a new constraint).

Run SOLVER when  = 0 and observe its range of optimality. Notice the allowable decrease. Atmax there will be a change in the optimal solution. Specifically,for the range of optimality is [-.045, 0.009]. Thus, when there is around $.05 cost reduction on .5 inch solid, 1 inch solid, and 2 inch solid tubes; as well as $.055 cost reduction on .5 inch hollow, 1 inch hollow, and 2 inch hollow tubes, the optimal production plan will change.

  1. We need to add the variable X44 and Y44 to the model. In the objective function they will appear with the coefficients 1000. These large coefficients are selected arbitrarily, just to ensure that in our minimization problem X44 = 0 and Y44 = 0 at the outset (before the cost reduction is considered). So the Initial objective function (to be altered soon) isMin[.10X11+…+1000X44+.2Y14 + …+1000Y44].
    Now the cost reduction is implemented. The modified objective function isMin [.10X11+…+(1000-)X44+(1000--.9)Y44] =
    Min[Original function - .9Y44 –], where Z = X44 + Y44 (to be added as a new constraint). Three corrections in the current constraints are needed to reflect the inclusion of X44 and Y44 in our model. Here are the new constraints:

.10X14+.30X24+.60X34+1.0X44+.50Y14+.60Y24+.80Y34+1.0Y44360,000

X41+X42+X4480,000

Y42+Y43+Y44150,000,

and of course –X44 – Y44 +Z = 0

With these changes we run solver when  = 0. Observing the range of optimality the maximum increase for  is.5 (note that the cost reduction is defined as
1000 -  at the modified objective function, therefore to lower the cost  need to grow larger). Thus, the optimal solution changes when  is just above .5. To make sure that at  = .5 the changes in the optimal solution include at least one of the new products run in Mill 4 run SOLVER with .