Transcript Sensitivity analysis and duality

Sensitivity analysis
LI Xiao-lei
A graphical introduction to sensitivity
analysis

Sensitivity analysis is concerned
with how changes in an LP’s
parameters affect the LP’s optimal
solution.
A graphical introduction to sensitivity
analysis

The Giapetto problem
max z=3x1+2x2
s.t.
2x1 + x2≤100 (finishing constraint)
x1 + x2≤80 (carpentry constraint)
x1
≤40 (demand constraint)
x1 , x2 ≥0
Where
x1=number of soldiers produced per week
x2=number of trains produced per week
A graphical introduction to sensitivity
analysis
The optimal solution is z=180, x1=20, x2=60
3 x+2 y-180 = 0
100
80
y
60
40
z=60
20
z=100
z=180
0
0
20
40
60
x
80
100
Graphical analysis of the effect of a change in
an objective function coefficient
Let c1 be the contribution to profit by
each soldier. For what values of c1
does the current basis remain optimal?
 At present, c1=3 and the profit line has
the form

3x1+2x2=constant,
or x2=-3x1/2 + constant/2.
Figure 1
3 x 1+2 x 2-120 = 0
100
90
finishing constraint, slope=-2
A
80
demand constraint
70
B
x
2
60
profit line z=120, slope=-3/2
50
D
40
30
carpentry constraint, slope =-1
C
20
10
E
0
0
10
20
30
40
50
x1
60
70
80
90
100
Graphical analysis of the effect of a
change in an objective function coefficient

If a change in c1 cause the profit lines
to be flatter than the carpentry
constraint, the optimal solution will
change from point B to a new optimal
solution (point A).
The slope of each profit line is –c1/2
The profit line will be flatter than the
carpentry constraint if –c1/2>-1, or
c1<2, and the new optimal solution will
be (0,80), point A.
Graphical analysis of the effect of a
change in an objective function coefficient

If the profit line s are steeper than the
finishing constraint, the optimal
solution will change from point B to
point C.
The slope of the finishing constraint is 2.
If –c1/2<-2, or c1>4, and the new optimal
solution will be (40,20).
Graphical analysis of the effect of a
change in an objective function coefficient
In summary, if all other parameters
remain unchanged, the current basis
remains optimal for 2≤c1≤4. and
Giapetto should still manufacture 20
soldiers and 60 trains.
 But Giapetto’s profit will change. For
instance, if c1=4, Giapetto’s profit will
now be 4(20)+2(60)=$200 instead of
$180.

Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution
Whether a change in the right-hand
side of a constraint will make the
current basis no longer optimal?
 Let b1 the number of available
finishing hours. Currently, b1=100.

Figure 2
2 x 1+x 2-120 = 0
100
90
A
80
finishing constraint, slope=-2
finishing constraint b1=80
finishing constraint b1=120
demand constraint
70
B
x
2
60
profit line z=120, slope=-3/2
50
D
40
30
carpentry constraint, slope =-1
C
20
10
0
0
10
20
30
40
50
x1
60
70
80
90
100
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution
A change in b1 shifts the finishing
constraint parallel to its current
position.
 The current optimal solution is where
the carpentry and finishing constraints
are binding.
Then as long as the point where the
finishing and carpentry constraints are
binding remains feasible, the optimal
solution will still occur where these
constraints intersect.

Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution



From figure 2, for 80≤b1≤120, the current
basis remains optimal.
With the changing of b1, the values of the
decision variables and the objective function
value change.
For example, if 80≤b1≤100, the optimal
solution will change from point B to some
point on the line segment AB. Similarly, if
100≤b1≤120, the optimal solution will
change from point B to some point on the
line segment BD.
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution

To determine how a change in the righthand side of a constraint changes the
values of the decision variables.
Let b1=number of available finishing hours.
If we change b1 to 100+Δ, the current basis
remains optimal for -20≤Δ≤20.
Note: as b1 changes, the optimal solution to
the LP is still the point where the finishinghour and carpentry-hour constraints are
binding.
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution

Thus, we can find the new values of the
decision variables by solving
2x1+x2=100+Δ
and x1+x2=80
This yields x1=20+Δ and x2=60-Δ
Thus, a increase in the number of available
finishing hours results in an increase in the
number of soldiers produced and a decrease
in the number of trains produced.
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution

Let b2=the number of available
carpentry hours
If we change b2 to 80+Δ, the current
basis remains optimal for -20≤Δ≤20.
Note: as b2 changes, the optimal
solution to the LP is still the point
where the finishing-hour and
carpentry-hour constraints are binding.
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution
Thus, if b2=80+ Δ, the optimal solution to the
LP is the solution to
2x1+x2=100
and x1+x2=80+ Δ
This yields x1=20-Δ and x2=60+2Δ.
It shows that an increase in the amount of
available carpentry hours decrease the
number of soldiers produced and increases
the number of trains produced.
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution

Let b3=the demand for soldiers.
If b3 is changed to 40+Δ, the current basis
remains optimal for Δ≥-20. for Δ in this range,
the optimal solution to the LP will still occur
where the finishing and carpentry constrains are
binding.
Thus, the optimal solution to the LP is the solution
to
2x1+x2=100
and x1+x2=80
This yields x1=20 and x2=60
Graphical analysis of the effect of a
change in a right-hand side on the LP’s
optimal solution
Note:
In a constraint with positive slack( or
positive excess) in an LP’s optimal
solution, if we change the righthand side of the constraint to a
value in the range where the
current basis remains optimal, the
optimal solution to the LP is
unchanged.
Shadow prices
We define the shadow price for the
ith constraint of an LP to be the
amount by which the optimal z-value
is improved if the right-hand side of
the ith constraint is increased by 1.
 This definition applies only if the
change in the right-hand side of
constraint i leaves the current basis
optimal.

Shadow prices

For example, for the Δchanging in the
finishing hours, the optimal solution is
x1=20+Δ and x2=60-Δ. Then the optimal zvalue will equal 3x1+2x2=3(20 +Δ) +2(60-Δ)
=180+ Δ. Thus, as long as the current basis
remains optimal, a unit increase in the
number of available finishing hours will
increase the optimal z-value by $1.so the
shadow price of the first constraint is $1.
Shadow prices

For the second constraint, if Δchanging in
the finishing hours, the optimal solution is
x1=20-Δ and x2=60+2Δ. Then the optimal zvalue will equal 3x1+2x2=3(20 -Δ) +2(60+2Δ)
=180+ Δ. Thus, as long as the current basis
remains optimal, a unit increase in the
number of available finishing hours will
increase the optimal z-value by $1.so the
shadow price of the second constraint is $1.
Shadow prices


For the third constraint, the optimal values
of the decision variables remain unchanged,
as long as the current basis remains optimal.
Then the optimal z-value will also remain
unchanged, which shows that the shadow
price of the third constraint is $0.
It turns out that whenever the slack variable
or excess variable for a constraint is
positive in an LP’s optimal solution, the
constraint will have a zero shadow price.
Shadow prices


Each unit by which constraint i’s right-hand
side is increased will increase the optimal zvalue( for a max problem) by the shadow price.
Suppose we increase the rhs of the ith
constraint of an LP by Δbi , thus, the new
optimal z-value is given by
(new optimal z-value)=(old optimal z-value)
+(constraint i’s shadow price)Δbi
For a minimization problem,
(new optimal z-value)=(old optimal z-value)
-(constraint i’s shadow price)Δbi
Importance of sensitivity
analysis
If a parameter changes, sensitivity
analysis often makes it unnecessary
to solve the problem again.
 A knowledge of sensitivity analysis
often enables the analyst to determine
from the original solution how
changes in an LP’s parameters
change the optimal solution.

Some important formulas

Assume a max LP problem with m constraints
and n variables. Although some of these
variables may be slack, excess, or artificial
ones, we choose to label them x1,x2,…,xn.
max z=c1x1+c2x2+…+cnxn
s.t. a11x1+a12x2+…+a1nxn=b1
a21x1+a22x2+…+a2nxn=b2
:
:
:
:
am1x1+am2x2+…+amnxn=bm
xi≥0 (i=1,2,…,n)
(1)
Some important formulas

For the Dakota problem,
max z=60x1+30x2+20x3+0s1+0s2+0s3
s.t. 8x1+ 6x2+ x3+ s1
=48(lumber)
4x1+ 2x2+1.5x3+
s2
=20 (finishing)
2x1+1.5x2+0.5x3+
s3=8 (carpentry)
x1,x2,x3,s1,s2,s3≥0
(1’)
Some important formulas


Let BVi be the basic variable for row i of the
optimal tableau. Also define BV={BV1,
BV2 , … , BVm} to be the set of basic
variables in the optimal tableau, and define
the m × 1 vector xBv=[xBv1, xBv2 , …, xBvm]T
We also define
NBV=the set of nonbasic variables in the
optimal tableau
xNBv=(n-m) ×1 vector listing the nonbasic
variables.
Some important formulas

The optimal tableau for the LP 1’
z
+ 5x2
+10s2+10s3=280
- 2x2
+ s1 +2 s2 - 8s3=24
- 2x2+ x3+
2s2 - 4s3 =8
x1+1.25x2
-0.5s2+1.5s3=2
For this optimal tableau, BV1=s1, BV2=x3,
and BV3=x1, then xBV=[s1,x3,x1]’
We may choose NBV={x2,s2,s3}, then
xNBV=[x2,s2,s3]’
Some important formulas

DEFINATION
cBV is the 1×m row vector [cBV1 cBV2 …
cBVm].
Thus, the elements of cBV are the
objective function coefficients for the
optimal tableau’s basic variables. For
the Dakota problem, BV={s1,x3,x1},
then form (1’) we find that cBV=[0 20
60].
Some important formulas

DEFINATION
cNBV is the 1×(n-m) row vector whose
elements are the coefficients of the
nonbasic variables( in the order of
NBV).
If we choose to list the nonbasic
variables for the Dakota problem in
the order NBV={x2,s2,s3}, then form (1’)
we find that cNBV=[30 0 0].
Some important formulas

DEFINATION
The m×m matrix B is the matrix
whose jth column is the column for
BVj in (1).
For the Dakota problem, the first
column of B is the s1 column in (1’);
the second, the x3 column; and the
third, the x1 column.
8
1 1Thus,
B  0 1.5 4
0 0.5 2
Some important formulas

DEFINATION
aj is the column (in the constraints) for
the variable xj in (1).
For example, in the Dakota problem,
6 
a2  2 
1.5
Some important formulas

DEFINATION
N is the m×(n-m) matrix whose
columns are the columns for the
nonbasic variables (in the NBV
order) in (1).
For the Dakota problem, we write
NBV={x2,s2,s3}, then
 6 0 0
N   2 1 0
1.5 0 1
Some important formulas

DEFINATION
The m×1 column vector b is the
right-hand side of the
constraints in (1).
 48 
For the Dakota problem, b  20
8 
Matrix notation

(1) may be written as
z  cBV xBV  cNBV xNBV
s.t. Bx BV  NxNBV  b
xBV , xNBV  0
(3)
Matrix notation

The Dakota problem can be written as
 s1 
max z  0 20 60 x3   30
 x1 
1 1 8   s1   6 0
s.t. 0 1.5 4  x3    2 1
0 0.5 2  x1  1.5 0
 s1  0  x2  0
 x   0  ,  s   0 
 3    2   
 x1  0  s3  0
 x2 
0 0 s2 
 s3 
0  x2  48
0  s2   20
1  s3  8 
Matrix notation

Multiplying the constraints in (3) through by
B-1, we obtain
B 1 Bx BV  B 1 NxNBV  B 1b
or
xBV  B 1 NxNBV  B 1b
(4)
In (4),BVi occurs with a coefficient of 1 in
the ith constraint and a zero coefficient
in each other constraint. Thus, BV is the
set of the basic variables for (4), and (4)
yields the constraints for the optimal
tableau.
Matrix notation

For the Dakota problem,
1 2  8
B 1  0 2  4
0  0.5 1.5 
Then,
2  8  x2  24
s1   - 2
x    - 2
 s   8 
2

4
 3 
 2   
 x1  1.25  0.5 1.5  s3  2 
Of course, these are the constraints for the
optimal tableau.
Matrix notation

From (4), we see that the column of a nonbasic
variable xj in the constraints of the optimal
tableau is given by
B-1(column for xj in (1) )=B-1aj
The following two equations summarize the
preceding discussion:
Column for xj in optimal tableau’s constraints=B-1aj
(5)
Right-hand side of optimal tableau’s constraints=B-1b
(6)
Matrix notation

How to express row 0 of the optimal tableau in terms
of BV?
We multiply the constraints through by the vector cBVB-1:
cBV xBV  cBV B 1 NxNBV  cBV B 1b
(7)
And rewrite the original objective function as,
z  cBV xBV  cNBV xNBV  0
(8)
Adding (7) to (8), we can eliminate the optimal
tableau’s basic variables,
z  (cBV B 1 N  cNBV ) x NBV  cBV B 1b
(9)
Matrix notation

From (9), the coefficient of xj in row 0 is
cBVB-1(column of N for xj)(coefficient for xj in cNBV)=cBVB-1aj-cj
the right hand side of row 0 is cBVB-1b

To summarize, we let c j be the coefficient of xj in
the optimal tableau’s row 0. then we have shown
that
1
c j  cBV B a j  c j
(10)
and
right-hand side of optimal tableau’s row 0=cBVB-1b (11)
Matrix notation

To illustrate, we determine row 0 of the Dakota
problem’s optimal tableau.
cBV
1 2  8
 [0 20 60] and B 1  0 2  4
0  0.5 1.5 
Then cBVB-1=[0 10 10], from (10) we find that the
coefficients of the nonbasic variables in row 0 are
6
c2  cBV B 1a2  c2  0 10 10 2   30  5
1.5
Matrix notation
and
0 
Coefficient of s2 in optimal row0  cBV B 1 1  0  10
0
0
Coefficient of s3 in optimal row0  cBV B 1 0  0  10
1
Of course, the optimal tableau’s basic variables
(x1,x3,and s1) will have zero coefficients in row 0.
Matrix notation

From (11), the right-hand side of row 0 is
48


1
cBV B b  0 10 1020  280
 8 

Putting it all together, we see that row 0 is
z+5x2+10s2+10s3=280
Matrix notation

Simplifying formula (10) for slack, excess,
and artificial variables
If xj is the slack variable si, the coefficient of
si in the objective function is 0, and the
column for si in the original tableau has 1
in row I and 0 in all other rows. Then (10)
yields
Coefficient of si in optimal row 0
= ith element of cBVB-1-0
=ith element of cBVB-1
(10’)
Matrix notation

Similarly, if xj is the excess variable ei,
the coefficient of ei in the objective
function is 0 and the column for ei in the
original tableau has -1 in row i and 0 in
all other rows. Then (10) reduces to
Coefficient of ei in optimal row 0
=-( ith element of cBVB-1)-0
=-(ith element of cBVB-1)
(10’’)
Matrix notation

Finally, if xj is an artificial variable ai,
the objective function coefficient of
ai( for a max problem) is –M and the
original column for ai has 1 in row i
and 0 in all other rows. Then (10)
reduces to
Coefficient of ai in optimal row 0
=( ith element of cBVB-1)-(-M)
=(ith element of cBVB-1)+(M)
(10’’’)
Summary of formulas for computing
the optimal tableau from the initial LP
Column for xj in optimal tableau’s constraints=B-1aj
(5)
Right-hand side of optimal tableau’s constraints=B-1b
(6)
c j  cBV B1a j  c j
Coefficient of si in optimal row 0=ith element of cBVB-1
(10’)
Coefficient of ei in optimal row 0=-(ith element of cBVB-1) (10’’)
Coefficient of ai in optimal row 0=-(ith element of cBVB-1)+(M)
(10’’’)
right-hand side of optimal tableau’s row 0=cBVB-1b
(11)
We must first find B-1 because it is necessary in order to compute
all parts of the optimal tableau.
Example 1
Solution
 After adding slack variables s1 and s2,
we obtain:
max z=x1+4x2
s.t.
x1+2x2+s1
=6
2x1+ x2 +s2=8
Example 1

For the following LP, the optimal basis
is BV={x2,s2}, computer the optimal
tableau.
max z=x1+4x2
s.t.
x1+2x2≤6
2x1+ x2≤8
x1,x2≥0
Example 1
First we compute B-1,
1

0
 2

 2 0
1
B
and B  


1
1 1

1
 2 
Example 1
Use (5) and (6) to determine the optimal tableau’s
constraints.
The column for x1 in the optimal tableau is
 1
 1   1 
0
     2
B 1a1   2
 1 2 1 2  3 
 2
The other nonbasic variable is s1,
 1
 1   1 
0
    2 
Column for s1 in optimal tableau  B 1a1   2
 1 2 1 0  1 
 2
Example 1

From (6)
 1
0 6 3
1
    
Right - hand side of optimal tableau  B b   2
 1 2 1 8  5

Since BV is listed as {x2,s2},
 1

0
  2 0
c BV B 1  4 0 2
 1 2 1
Example 1

Then (10) yields
coefficien t of x1 in row 0 of optimal tableau
1 
 cBV B a1  c1  2 0   1  1
 2
1

From (10’)
coefficien t of s1 in optimal tableau
 first element of cBV B 1  2
Example 1

(11) yields
the right - hand side of the optimal tableau' s row 0
6 
 cBV B b  2 0   12
6 
1


Of course, the basic variables x2 and s2 will
have zero coefficients in row 0.
Thus, the optimal tableau’s row 0 is
z+x1+2s1=12
Example 1

The complete optimal tableau is
z +x1
+2s1
=12
½x1+x2+ ½ s1
=3
3/2x1
- ½ s1 +s2=5
Sensitivity analysis



The study of how an LP’s optimal solution
depends on its parameters is called
sensitivity analysis.
We let BV be the set of basic variables in the
optimal tableau. Given a change in an LP, we
want to determine whether BV remains optimal.
Important observation:
A simplex tableau( for a max problem) for a set of
BV is optimal if and only if each constraint has a
nonnegative right-hand side and each variable
has a nonnegative coefficient in row 0.
Sensitivity analysis

Suppose we have solved an LP and have found
that BV is an optimal basis. We can use the
following procedure to determine if any change
in the LP will cause BV to be no longer optimal.
Step 1 Using the preceding formulas, determine
how changes in the LP’s parameters change the
right-hand side and row 0 of the optimal tableau.
Step 2 If each variable in row 0 has a nonnegative
coefficient and each constraint has a
nonnegative right-hand side, BV is still optimal.
Otherwise , BV is no longer optimal.
Sensitivity analysis

Two reasons why a change in an LP’s
parameters causes BV to be no longer optimal.


A variable in row 0 may have a negative
coefficient. In this case , a better bfs can be
obtained by pivoting in a nonbasic variable with a
negative coefficient in row 0. we say that the BV
is now a suboptimal basis.
A constraint may now have a negative right-hand
side. In this case, at least one member of BV will
now be negative and BV will no longer yield a bfs.
We say that BV is now an infeasible basis.
Illustrate in example

The Dakota furniture example
x1=number of desks manufactured
x2=number of tables manufactured
x3=number of chairs manufactured
The objective function was
max z=60x1+30x2+20x3
Illustrate in example

The initial tableau was
z -60x1 -30x2 - 20x3
=0
8x1+ 6x2+ x3 +s1
=48 (lumber constraint)
4x1+ 2x2+1.5x3
+s2 =20 (finishing constraint)
2x1+1.5x2+0.5x3
+s3=8 (carpentry constraint)
The optimal tableau was
z
+5x2
+10s2 +10s3 =280
-2x2 + s1+ 2s2 -8s3=24
-2x2+x3
+2s2 -4s3=8
x1+1.25x2
-0.5s2 +1.5s3=2
BV={s1,x3,x1} and NBV={x2,s2,s3}. The optimal bfs is
z=280,s1=24,x3=8,x1=2,x2=0,s2=0,s3=0
Illustrate in example

Six types of changes in an LP’s parameters:






Change 1 changing the objective function
coefficient of a nonbasic variable
Change 2 changing the objective function
coefficient of a basic variable
Change 3 changing the right-hand side of a
constraint
Change 4 changing the column of a nonbasic
variable
Change 5 adding a new variable or activity
Change 6 adding a new constraint
Illustrate in example
Changing the objective function
coefficient of a nonbasic variable
 The only nonbasic decision variable is
x2. the objective function coefficient of
x2 is c2=30.
 How would a change in c2 affect the
optimal solution?
Illustrate in example




Suppose we change c2 from 30 to 30+Δ.
From (6), since the B-1 and b are unchanged, the
right-hand side of BV’s tableau (B-1b) has not
changed, so BV is still feasible.
From (10), BV will remain optimal if c2≥0, and BV
will be suboptimal if c2<0.
From preceding, we know that cBVB-1=[0 10 10].
6
c2  0 10 10 2   (30  )  5  
1.5
Illustrate in example

Thus, c2≥0 holds if 5-Δ≥0,or Δ≤5.
This means that if the price of tables is
decreased or increased by $5 or less,
BV remains optimal.

If BV remains optimal after a change in
a nonbasic variable’s objective function
coefficient, the values of the decision
variables and the optimal z-value
remain unchanged.
Illustrate in example



For example, if the price of tables increase to
$33(c2=33), the optimal solution to the Dakota
problem remains unchanged.
On the other hand, if c2>35, BV will no longer
be optimal, because c2<0.
For example, if c2=40,
6
c2  0 10 10 2   40  5
1.5
Illustrate in example

Now the BV “final” tableau is,
Basic
variables ratio
z
-5x2
+10s2+10s3=280 z=280
-2x2 +s1+ 2s2 -8s3=24 s1=24
none
-2x2+x3 +2s2 -4s3=8
x3=8
none
x1+1.25x2
-0.5s2+1.5s3=2
x1=2
1.6*
This is not an optimal tableau, and we can increase z by
making x2 a basic variable in row 3.
Illustrate in example
Changing the objective function coefficient
of a basic variable
 In the Dakota problem, x1 and x3 are basic
variables.


Since we are not changing B or b, (6) shows
that the right-hand side of each constraint
will remain unchanged, and BV will remain
feasible.
Since we are changing cBV. From (10), this
may change more than one coefficient in row
0.
Illustrate in example

The Dakota problem,
Suppose that c1 is changed from 60 to 60+Δ.
2
 8
1
B 1  0 2  4
0  0.5 1.5 
 8
1 2
cBV B 1  0 20 60   0 2  4
0  0.5 1.5 
 0 10  0.5 10  1.5 
1 1 8
B  0 1.5 4
0 0.5 2
Illustrate in example

Noting that,
8
6 
1 
a1  4, a2  2 , a3  1.5 , c1  60  , c2  30, c3  20
2
1.5
0.5

The coefficient of each nonbasic variable in the new
row 0 is,
6
c2  cBV B 1a2  c2  0 10  0.5 10  1.5  2   30  5  1.25
1.5
coefficien t of s2 in row 0  second element of cBV B 1  10  0.5
coefficien t of s3 in row 0  second element of cBV B 1  10  1.5
Illustrate in example

Thus, row 0 of the optimal tableau is now
z+(5+1.25Δ)x2+(10-0.5Δ)s2+(10+1.5Δ)s3=?
BV will remain optimal if anf only if the
following hold:
5+1.25Δ≥0 Δ≥-4
10-0.5Δ≥0 Δ≤20
10+1.5Δ≥0 Δ≥-20/3
the current basis will remain optimal if and
only if -4≤Δ≤20.
Illustrate in example
Thus, as long as 56=604≤c1≤60+20=80, the current basis
remains optimal.
 Suppose c1=70, z=70x1+30x2+20x3.
revenue should increase by 2(10)=$20,
and new revenue=280+20=$300.

Illustrate in example


If c1<56 or c1>80, the current basis is no longer
optimal.
Suppose c1=100(Δ=40), how can we determine
the new optimal solution?
c1  0, c2  5  1.25  55, c3  0
s1 coefficien t in row 0  0
s2 coefficien t in row 0  10  0.5  10
s3 coefficien t in row 0  10  1.5  70
48
right  hand side of row 0  cBV B 1b  0  10 7020  360
 8 
Illustrate in example

The BV tableau is ,
Basic
variables ratio
z
+55x2
-10s2+70s3=360 z=360
-2x2 +s1+ 2s2 -8s3=24 s1=24
12
-2x2+x3
+2s2 -4s3=8
x3=8
4*
x1+1.25x2
-0.5s2+1.5s3=2
x1=2
none
Illustrate in example
Changing the right-hand side of a constraint

Effect on the current basis
Since b does not appear in (10), changing
the right-hand side of a constraint will leave
row 0 of the optimal tableau unchanged;
 changing a right-hand side cannot cause
the current basis to become suboptimal.
 Will affect the right-hand side of the
constraints in the optimal tableau.

Illustrate in example
Note
As long as the right-hand side of each
constraint in the optimal tableau
remains nonnegative, the current
basis remains feasible and optimal.
If at least one right-hand side in the
optimal tableau becomes negative,
the current basis is no longer feasible
and therefore no longer optimal.
Illustrate in example


Suppose we change the amount of finishing hours (b2)
in the Dakota problem.
Currently,b2=20, we change b2 to 20+Δ, from (6)
48  1 2  8 48   24  2 
B 1 20    0 2  4 20     8  2 
8
 0  0.5 1.5  8
 2  0.5

The current basis will remain optimal if and only if the
following hold:
24+2Δ≥0 Δ≥-12
8+20Δ≥0 Δ≥-4
2-0.50Δ≥0 Δ≤4
Illustrate in example

For -4≤Δ≤4, 20-4≤b2≤20+4, or
16≤b2≤24, the current basis remains
feasible and therefore optimal.
Illustrate in example

Effect on decision variables and z
Even if the current basis remains optimal, the values of
the decision variables and z change.
The new values of the basic variables are as follows
 8 48 28
s1 
1 2
 x   B 1b  0 2  4 22  12 
 3

   
 x1 
0  0.5 1.5  8  1 
48
New z  value  cBV B 1 (new b)  0 10 1022  300
8 
From (11),
Illustrate in example

When the current basis is no longer optimal
Suppose we change b2 to 30. from (6), the righthand side of the constraints in the tableau for
BV={s1,x3,x1} is
1 2  8 48 44 
B 1b  0 2  4 30  28 
0  0.5 1.5   8   3
From (11), the right-hand side of row 0 is
48
cBV B 1b  0 10 1030  380
 8 
Illustrate in example

The tableau is,
z
+5x2
+10s2+10s3=380
-2x2 +s1+ 2s2 -8s3=44
-2x2+x3
+2s2 -4s3=28
x1+1.25x2
-0.5s2+1.5s3=-3
Basic
variables
z=380
s1=44
x3=28
x1=-3
Since x1=-3, BV is no longer feasible or optimal.
Illustrate in example
Changing the column of a nonbasic variable
 x2 is a nonbasic variable in the optimal solution,
this means that Dakota should not manufacture
any tables at present.
 Suppose, the price of tables increased to $43
and, because of changes in production
technology, a table required 5 board feet of
lumber, 2 finishing hours and 2 carpentry hours.
Illustrate in example
Here we are changing elements of the
column for x2 in the original problem.
The change leaves B and b unchanged.
Thus, the right-hand side of the optimal
tableau remains unchanged.
 From (10),the only part of row 0 that is
changed is c2; the current basis will
remain optimal if and only if c2≥0 holds.

Illustrate in example

Use (10) to compute the new coefficient of x2 in row 0.
cBV B 1  0 10 10
5 
c2  43 and a2  2
2

5 
c2  0 10 102  43  3  0
2
Since c2<0, the current basis is no longer optimal. c2=-3
means that each table now increases revenues by $3.
Illustrate in example

From (5),
2
 8 5   7
1
B 1a2  0
2
 4 2   4
0  0.5 1.5  2 2 
The tableau is now,
z
Basic
variables
-3x2
+10s2+10s3=280 z=280
-7x2 +s1+ 2s2 -8s3=24 s1=24
-4x2+x3
+2s2 -4s3=8
x3=8
x1+2x2
-0.5s2+1.5s3=2
x1=2*
Illustrate in example
Adding a new activity
 Suppose that Dakota is considering
making footstools. A stool sells for $15
and requires 1 board foot of lumber, 1
finishing hour, and 1 carpentry hour.
Should the company manufacture any
stools?
Illustrate in example

Our new initial tableau is
z -60x1 -30x2 -20x3-15x4
=0
8x1 +6x2 +x3 +x4+s1
=48
4x1 +2x2+1.5x3 +x4 +s2 =20
2x1+1.5x2+0.5x3 +x4
+s3=8
We call the addition of the x4 column to
the problem adding a new activity.
Illustrate in example
From (6), we see that the right-hand
sides of all constraints in the optimal
tableau will remain unchanged.
 From (10), we see that the coefficient
of each of the old variables in row 0
will remain unchanged.
 Compute c4, the current basis will
remain optimal if c4≥0.

Illustrate in example

1

Since


c4  15 and a4  1
1
1


c4  0 10 101  15  5
1
The current basis is still optimal.
This means that each stool manufactured will
decrease revenues by $5. for this reason, we
choose not to manufacture any stools.
Summary of sensitivity
analysis
Change in initial
problem
Effect on optimal
tableau
Current basis is still
optimal if
Changing nonbasic
objective function
coefficient cj
Coefficient of xj in
optimal row 0 is
changed
Coefficient of cj in row 0
for current basis is still
nonnegative
Changing basic
objective function
coefficient cj
Entire row 0 may
change
Each variable still has a
nonnegative coefficient
in row 0
Changing right-hand
side of a constraint
Right-hand side of
constraints and row 0
are changed
Right-hand side of each
constraint is still
nonnegative
Changing the column of
a nonbasic variable xj or
adding a new variable xj
Changes the coefficient The coefficient of xj in
for xj in row 0 and xj’s
row 0 is still
constraint column in
nonnegative
optimal tableau