aaoczc2252
DESCRIPTION
Keep Smiling & Mailing With Warm Regards, P.Gopi, Inspection Engr, CHEM P&S - QA&I, L&T, E&C Div, Powai Mumbai - 400072. Extn - 4279 / 3049 Phone: +9122 6705 4279 Fax: +9122 6705 1383 / 1890 Keep Smiling & Mailing With Warm Regards, P.Gopi, Inspection Engr, CHEM P&S - QA&I, L&T, E&C Div, Powai Mumbai - 400072. Extn - 4279 / 3049 Phone: +9122 6705 4279 Fax: +9122 6705 1383 / 1890TRANSCRIPT
Sensitivity Analysis
The optimal solution of a LPP is based on the conditions that prevailed at the time the LP model was formulated and solved. In the real world, the decision environment rarely remains static and it is essential to determine how the optimal solution changes when the parameters of the model are changed. That is what sensitivity analysis does. It provides efficient computational techniques to study the dynamic behavior of the optimal solution resulting from making changes in the parameters of the model.
In studying the sensitivity analysis, we should be familiar with the lingo that is being used in LPP situations. A general LPP is of the form
Maximize (or Minimize) nn xcxcxcz ...2211
subject to the constraints
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 2
...
...
.
...
, ,..., 0
n n
n n
m m mn n m
n
a x a x a x b
a x a x a x b
a x a x a x b
x x x
The RHS constants of the constraints, mbbb ,...,, 21
are referred to resources or availabilities of the problem.
The objective coefficients, nccc ,...,, 21
are referred to as unit profits (or unit costs).
The decision variables, nxxx ,...,, 21
are referred to as units of activities 1, 2, …, n.
Dual Price of a constraint
This measure actually represents the unit worth of a resource - that is it gives the contribution to the objective function resulting from a unit increase or decrease in the availability of a resource. In terms of duality theory, the dual price of a resource (=constraint) i, is precisely the value of the optimal dual variable yi associated with the constraint i. (Did you understand why it is called “dual price”?). Other non-suggestive names include shadow prices and simplex multipliers.
Reduced cost of a variable xj (=activity j)
is defined as
Cost of consumed resources per unit of activity xj
- profit per unit of activity xj
= zj - cj
Changes affecting feasibility
The feasibility of the current optimum solution may be affected only if
(1) The RHS of the constraints are changed
OR
(2) A new constraint is added to the model.
In both cases, infeasibility occurs when at least one element of the RHS of the optimal tableau becomes negative – that is, one or more of the current basic variables become negative.
is changed to
First we consider the changes in the optimal solution due to changes in the RHS bi. We note that the optimal solution is given by
1B
B bxwhere b is the old RHS and B is the basic matrix. Remember B-1 is found in the optimal tableau below the entries which initially had an identity submatrix.
The optimal value is given by bB 1BC z
b bWhen , the correspondingnew solution and new objective value are got by replacing with .bb
Example 4.3-2 (Pages 135-136)
TOYCO assembles three types of toys: trains, trucks and cars using three operations. The daily limits on the available times for the three operations are 430, 460, and 420 minutes respectively. The profits per toy train, truck and car are $3, $2, and $5 respectively. The assembly times per train at the three operations are 1, 3, and 1 minute respectively. The corresponding times per truck and per car are (2, 0, 4) and (1, 2, 0) minutes respectively. A zero indicates that the operation is not used.
Letting x1, x2, and x3 represent the daily number of units assembled of trains, trucks and cars, the LPP model is:
Maximize 321 523 xxxz
subject to
0,,
4204
46023
4302
321
21
31
321
xxx
xx
xx
xxx
The optimal tableau is given in the next slide. (Note: x4, x5, and x6 are slack variables there.)
x6 0 2 0 0 -2 1 1 20
x3 0 3/2 0 1 0 1/2 0 230
z 1 4 0 0 1 2 0 1350 Basic z x1 x2 x3 x4 x5 x6 Sol
x2 0 -1/4 1 0 1/2 -1/4 0 100
This is the optimal tableau
Suppose that TOYCO wants to change the capacities of the three operations according to the following cases:
460 500 300 450
(a) 500 (b) 400 (c) 800 (d) 700
400 600 200 350
Use Sensitivity analysis to determine the optimal solution in each case.
112
02/10
04/12/1
;052 1BCB
Solution: We note
(b) New Solution is
6
3
2
x
x
x
600
400
500
112
02/10
04/12/11bB
0
200
150
Since this is feasible, it is optimal and the new optimal value =
(c) New Solution is
6
3
2
x
x
x
200
800
300
112
02/10
04/12/11bB
400
400
50
13002005150203
This is not feasible. So we apply dual simplex method to restore feasibility. We note
1900400550203 new z =
x6 0 2 0 0 -2 1 1 400
x3 0 3/2 0 1 0 1/2 0 400
z 1 4 0 0 1 2 0 1900 Basic z x1 x2 x3 x4 x5 x6 Sol
x2 0 -1/4 1 0 1/2 -1/4 0 -50
x6 0 1 4 0 0 0 1 200
x5 0 1 -4 0 -2 1 0 200
z 1 2 8 0 5 0 0 1500
x3 0 1 2 1 1 0 0 300
This is the new optimal tableau.
Feasibility Range of the Elements of the RHS
Another way of looking at the effect of changing the availabilities of the resources, bi, is to determine the range for which the current solution remains feasible.
For example if, in the TOYCO model, b2 is changed to b2+D2= 460+D2, we want to find the range of D2 so that the current solution remains optimal.
When b2 is changed to b2+D2= 460+D2, the new solution is
6
3
2
x
x
x
420
460
430
112
02/10
04/12/1
21 DbB
2
2
2
2/1
4/1
20
230
100
D
D
D (current optimal solution + D2 times the 2nd column of B-1.)
2
2
2
20
2/1230
4/1100
D
D
D
is feasible if
020
02/1230
04/1100
2
2
2
D
D
D
20
460
400
2
2
2
Dor
Dor
Dor
40020 2 DOr
Thus current solution remains optimal if RHS of the 2nd constraint lies between 440 and 860 (the other RHSs being the same).
Problem 5 Problem Set 4.5B Page 151
HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation:
Unit Resources RequirementsResource Model 1 Model2 Maximum Availability
(units) (units) (units)
Resistor 2 3 1200
Capacitor 2 1 1000
Chips 0 4 800
Unit Profit($) 3 4
Let x1, x2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP
Maximize 21 43 xxz
subject to
0,
8004
10002
120032
21
2
21
21
xx
x
xx
xx (Resistors)
(Capacitors)
(Chips)
Taking the slack variables as s1, s2, s3, the optimal tableau is:
x2 0 0 1 1/2 -1/2 0 100
s3 0 0 0 - 2 2 1 400
z 1 0 0 5/4 1/4 0 1750 Basic z x1 x2 s1 s2 s3 Sol
x1 0 1 0 -1/4 3/4 0 450
This is the optimal tableau
(a) Determine the status of each resource
Answer: Since s1 = 0 = s2, the resistor and the capacitor resources are scarce.
Since s3 > 0, the chips resource is abundant.
(b) In terms of the optimal profit, determine the worth of one resistor, one capacitor and one chip.
Answer: They are respectively y1, y2, y3 the dual optimal solution and hence are
5/4, 1/4, 0 respectively.
(c) Determine the range of applicability of the dual prices for each resource.
Resistor: If D1 is the increase in the resource 1,
the new optimal solution is given by
2
3
1
x
s
x
800
1000
1200
02/12/1
122
04/34/1 11
D
bB
1
1
1
2/1
2
4/1
100
400
450
D
D
D
0 gives 200200 1 D
Similar calculations show that, for a change D2 in the capacitor, the range of feasibility is given by
200200 2 D
And for a change D3 in the chips, the range of feasibility is given by 4003 D
(d) If the available number of resistors is increased to 1300 units, find the new optimum solution.
The new solution is:
2
3
1
x
s
x 450 25
400 200
100 50
150
200
425 This is feasible and hence optimal.And z = 1875.
Yes, as D1 = 100
(g) A new contractor is offering to sell HiDec additional resistors at 40 cents each but only if HiDec would purchase at least 500 units. Should HiDec accept the offer?
Addition of a new constraint
The addition of a new constraint to an existing model can lead to one of two cases:
1. The new constraint is redundant, meaning that it is satisfied by the current optimal solution and hence can be dropped altogether from the model altogether.
2. The current solution violates the new constraint in which case the dual simplex method can be used to recover feasibility.
We introduce this constraint in the final simplex tableau as an additional row where the usual additional (slack or surplus) variable is taken as the basic variable for this new row. Because this new row will probably have non-zero coefficients for some of the other basic variables, the conversion to proper form for Gaussian elimination is done and then reoptimization is done in the usual way. The following example illustrates this.
Example: Consider the LPP
Maximize 3212 xxxz subject to
0,,
20
102
603
321
321
321
321
xxx
xxx
xxx
xxx
Applying the Simplex method, the optimal tableau is given in the next slide. Now a new constraint
1 2 33 2 28x x x solution.
is added. Find the new optimal
x2 0 0 1 - 3/2 0 - 1/2 1/2 5
x1 0 1 0 1/2 0 1/2 1/2 15
z 1 0 0 3/2 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol
s1 0 0 0 1 1 - 1 - 2 10
s4 0 3 -2 1 0 0 0 1 28
s4
0
0
0
0
z 1 0 0 0 0 3/7 2/7 3/7 22
x2 0 0 1 0 0 4/7 5/7 -3/7 8
x1 0 1 0 0 0 1/7 3/7 1/7 14
s1 0 0 0 0 1 - 12/7 -15/7 2/7 8
x3 0 0 0 1 0 5/7 1/7 -2/7 2
0 0 0 -7/2 0 - 5/2 -1/2 1 -7
Changes affecting Optimality
1. Changes in the objective coefficients cj
The changes in the objective coefficients, cj , affect "zj - cj" and hence the optimality. We recompute the z-Row by replacing cj with c'j . We note that CB will also change to C'B.
In the previous problem, we replace the objective function with
321 323 xxxz .
Find the new optimal solution.
Thus c1 is changed from 2 to 3; c2 is changed from -1 to -2; and c3 is changed from 1 to 3;
We note that in z-Row, the coefficients of basic variables x1 , x2 , and x4 will remain zero now also. We have only to calculate the new coefficients of the non-basic variables only and the new z.
Coefficient of x3 = z3 – c3 33
1 cABCB
02
33
2/3
2/1
1
230
55
1 cABCB
02
50
2/1
2/1
1
230
Coefficient of x5 = z5– c5
Coefficient of x6 = z6– c6 66
1 cABCB
02
10
2/1
2/1
2
230
Since all zj– cj are 0, the current solution remains optimal. Of course the new z is
350352153
If in the previous problem, we replace the objective function with 321 332 xxxz
You can verify that z6– c6 = -1/2 < 0. Thus optimality is spoiled. By applying the regular Simplex method, we can show that the new optimum solution is x1 = 10, x2 = 0, x3 = 0 with new z = 20.
Addition of a new activity
(= Changes in the coefficients of an existing activity)
Suppose a new activity n+1 is added with coefficients
1,
1,2
1,1
1
.
nm
n
n
n
a
a
a
c
To find the effect of this on the current optimal solution, we pretend this activity was present initially with all coefficients zero. Hence the new coefficients are calculated using the formulae:
In z-row, the coefficient of xn+1 is
111
11
nnBnn cABCcz
And in the constraint matrix its coefficients are
11
nABIf zn+1 – cn+1 satisfies the optimality condition, the current solution is optimal. Else we apply regular simplex method to restore optimality.
The same procedure is adopted when the coefficients of an existing variable are changed. If the variable ia basic variable, we should see that the new tableau is in proper form (i.e. coefficients of other basic variables in that column should be made zero).
In the previous problem the coefficients of the (non-basic) variable x3 are changed from
1
2
1
1
33
23
13
3
a
a
a
c
to
2
1
3
2
33
23
13
3
a
a
a
c
Using sensitivity analysis, find the new optimal solution and value.
The only change in the optimal tableau will be the x3 column. We calculate the new x3 column and the coefficient of x3 in the z-row.
We first note that
2
1
2
10
2
1
2
10
211
;120 1BCB
Hence the new x3 column is 31AB
2
1
3
2
1
2
10
2
1
2
10
211
2
32
16
New coefficient of x3 in z-row =
331
33 cABCcz B
2
2
32
16
120
02
3
Thus optimality is disturbed. We replace the x3 column in the original optimal tableau by the new values and then find the new optimal solution and optimal value by regular simplex method.
x2 0 0 1 0 - 1/2 1/2 5
x1 0 1 0 0 1/2 1/2 15
z 1 0 0 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol
s1 0 0 0 1 - 1 - 2 10
z 1 0 0 0 1/4 5/4 0 55/2
x2 0 0 1 0 1/4 -3/4 0 15/2
x1 0 1 0 0 1/12 5/12 1/3 95/6
x3 0 0 0 1 1/6 - 1/6 -1/3 5/3
-3/2
6
-1/2
-3/2
In the previous problem a new variable x7 is introduced with coefficients
2
1
2
1
37
27
17
7
a
a
a
c
Using sensitivity analysis, find the new optimal solution and value.
This is like the previous case. We assume the variable x7 was already present with all coefficients 0.
We first note that
2
1
2
10
2
1
2
10
211
;120 1BCB
Hence the new x7 column is 71AB
2
1
2
2
1
2
10
2
1
2
10
211
2
12
37
New coefficient of x7 in z-row =
771
77 cABCcz B
)1(
2
12
37
120
02
7
Hence the original solution remains optimal with the same objective value and it does not help by introducing this new variable.
In the previous problem the coefficients of the (basic) variable x1 are changed from
1
1
3
2
31
21
11
1
a
a
a
c
to
Using sensitivity analysis, find the new optimal solution and value.
The only change in the optimal tableau will be the x1 column. We calculate the new x1 column and the coefficient of x1 in the z-row.
0
2
2
1
31
21
11
1
a
a
a
c
We first note that
2
1
2
10
2
1
2
10
211
;120 1BCB
Hence the new x1 column is 11AB
0
2
2
2
1
2
10
2
1
2
10
211
1
1
0
New coefficient of x1 in z-row =
111
11 cABCcz B
1
1
1
0
120
2
We replace the x1 column in the original optimal tableau by the new values.
x2 0 1 -3/2 0 - 1/2 1/2 5
x1 0 0 1/2 0 1/2 1/2 15
z 1 0 3/2 0 3/2 1/2 25
Basic z x1 x2 x3 s1 s2 s3 Sol
s1 0 0 1 1 - 1 - 2 10
z 1 0 1/2 0 0 1/2 0 5
s3 0 0 1 -1 0 0 1 20x1 0 1 -1/2 1 0 1/2 0 5
s1 0 0 2 -1 1 - 1 0 50
2
0
1
-1
1 0 0 1/2 0 1/2 -1/2 -5
0 0 1 -1 0 0 1 20
Problem 5 Problem Set 4.5B Page 151
HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation:
Unit Resources RequirementsResource Model 1 Model2 Maximum Availability
(units) (units) (units)
Resistor 2 3 1200
Capacitor 2 1 1000
Chips 0 4 800
Unit Profit($) 3 4
Let x1, x2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP
Maximize 21 43 xxz
subject to
0,
8004
10002
120032
21
2
21
21
xx
x
xx
xx (Resistors)
(Capacitors)
(Chips)
Taking the slack variables as x3, x4, x5, the optimal tableau is:
(c) Determine the range of applicability of the dual prices for each resource.
Resistors: If D1 is the increase in the resource 1,
the new optimal solution is given by
2
3
1
x
s
x
800
1000
1200
02/12/1
122
04/34/1 11
D
bB
1
1
1
2/1
2
4/1
100
400
450
D
D
D
0 gives 200200 1 D
Similar calculations show that, for a change D2 in the capacitor, the range of feasibility is given by
200200 2 D
And for a change D3 in the chips, the range of feasibility is given by 4003 D
(d) If the available number of resistors is increased to 1300 units, find the new optimum solution.
The new solution is:
2
3
1
x
s
x
50
200
25
100
400
450
150
200
425 This is feasible and hence optimal.And z = 1875.
(e) If the available number of chips is reduced to 350 units, will you be able to determine the new optimal solution directly from the given information? Explain.
Answer: NO, as the feasibility is not spoilt only if the number of chips is 400400800
(f) If the availability of capacitors is limited by the range of applicability computed in (c), determine the corresponding range of the optimal profit and the corresponding ranges for the number of units to be produced of Models 1 and 2.
If the number of capacitors is changed from b2= 400 to b2 = 400+D2, where 200200 2 D
then 4
3450 2
1
Dx
2100 2
2
Dx
41750 2D
z
Hence600300 1 x
2000 2 x
18001700 z
(g) A new contractor is offering to sell HiDec additional resistors at 40 cents each but only if HiDec would purchase at least 500 units. Should HiDec accept the offer?
Answer: Since the new value of b1, the number of resistors is (at least) 1700, the feasibility is spoilt (as D1 > 200). The new solution is (taking D1 =500)
2
3
1
x
s
x
250
1000
125
100
400
450
350
600
325
And new z = 2375. We restore Optimality by Dual Simplex.
Problem 9 Problem Set 4.5E Page 159
Consider the HiDec model in Problem 5, Problem Set 4.5B
(a) Determine the conditions that will keep the current solution optimum if the unit profits of Models 1 and 2 are changed simultaneously.
(b) Suppose that the objective function is changed to
21 25 xxz
Determine the associated optimum solution.
Solution. (a) Let c1 = 3 be changed to c1 =3+ d1. Let c2 = 4 be changed to c2 =4+ d2.
Thus the new coefficient of the non-basic variable s1 in the z-row is z3– c3 33
1 cABCB
0
2/1
2
4/1
403 21
dd
4
5
2421
dd0 for retaining optimality
Thus the new coefficient of the non-basic variable s2 in the z-row is z4– c4 44
1 cABCB
0
2/1
2
4/3
403 21
dd
4
1
24
3 21 dd
0 for retaining optimality
These are the two conditions on d1 and d2 .
Solution (b): Here new objective function is
21 25 xxz
Thus d1 = 5 - 3 =2 and d2 = 2 - 4 = -2.
New z3– c3 = 04
1 New z4– c4 = 0
4
11
Thus optimality is spoiled. And we reoptimize using regular Simplex method.
Note also new z
100
400
450
205 =2450
Problem 3 Problem Set 4.5F Page 161
In the TOYCO model, suppose that a new toy (fire engine) requires 3, 2, and 4 minutes respectively, on operations 1, 2, and 3. Determine the new optimal solution when the profit per unit is given by
(a) $5 (b) $10 (Optimal tableau in next page)
Solution: This amounts to introducing a new variable x7 with coefficients as follows.
37
27
17
7
)(
a
a
a
c
a ;
4
2
3
5
37
27
17
7
)(
a
a
a
c
b
4
2
3
10
We first note that
112
02
10
04
1
2
1
;052 1BCB
Hence the new x7 column is7
1AB
4
2
3
112
02
10
04
1
2
1
0
1
1
New coefficient of x7 in z-row =
771
77 cABCcz B
5
0
1
1
052
2
Thus in case (a) the current solution is optimal and it does not help to introduce the new toy.
in case (a)
And = -3 in case (b)
In case (b), optimality is spoiled and we restore it using Regular simplex method. (see the next slide)
Consider the LPP
Maximize 3212 xxxz subject to
0,,
4
3
15223
321
321
321
321
xxx
xxx
xxx
xxx
The optimal tableau is given in the next slide.
We first note that
021
110
031
;201 1BCB
Hence the new x1 column is 11AB
3
2
1
021
110
031
3
1
5
New coefficient of x1 in z-row =
111
11 cABCcz B
1
3
1
5
201
2
Thus optimality is spoiled. We try to restore it.