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The information that can be obtained from the final
simplex tableau are:
1. The optimum solution.
2. The status of the resources.
3. The unit worth of each resource (Dual prices)
4. The sensitivity of the optimum solution to change
in:
Range of objective function coefficients
(Range of Optimality)
Range of right-hand-side values (Range of
feasibility)
2
Example
50 x1  40 x2
Max
s.t.
3x1  5x2  150 Assembly time
1x2  20 Portable display
8 x1  5 x2  300 Warehouse capacity
x1 , x2  0
where
x1 = number of units of the Deskpro
x 2 = number of units of the Portable
The final simplex tableau for the problem is:
x1
x2
s1
s2
s3
0
0
0
1
3
25
0
3
25
0
5
Basis
cB
50
40
x2
s2
x1
40
0
1
8
0
0
0
8
50
1
0
5
zj
50
40
14
cj  zj
0
0
25
5
 14
5
x1 =30, x 2 =12, and Z=1980
0
0
25
25
25
26
5
 26
5
12
8
30
1980
3
Resource
Slack Variable
1. Assembly time
S1=0
2. Portable display
S2=8
3. Warehouse capacity
S3=0
Status of Resource
Scarce (binding)
Abundant (non-binding)
Scarce(binding)
The range of optimality for an objective function coefficient for any
basic variable is determined by
cj  zj  0
for all values of j.
For Example: The range of optimality for c1
x1
x2
s1
s2
s3
0
0
0
25
1
3
25
0
3
25
64  c1
5
0
5
Basis
cB
c1
40
x2
s2
x1
40
0
1
8
0
0
0
8
c1
1
0
5
zj
c1
40
cj  zj
0
0
c1  64
5
0
25
c1  64  0
24  c1
5
24  c1  0
24  c1  64
12
8
30
25
c1  24
480+30 c1
5
24  c1
0
5
c1  64
0
5
c1  64
0
25
c1  24
4
Hints :
c
1. If there are two or more upper bounds on k , the smallest of
these is the upper bound on the range of optimality.
c
2. If there are two or more lower bounds on k , the largest of
these is the lower bound on the range of optimality.
3. The range of optimality for the objective function coefficient of
any nonbasic variable is given by
Problems
1. Suppose an increase in material costs reduces
the profit per unit for the Deskpro to \$30,
what is the optimal solution?
2. Suppose an increase in material costs reduces
the profit per unit for the Deskpro to \$15,
what is the optimal solution?
Steps to compute the range of optimality for objective
function. (See Textbook page 50 (82)
5
The right-hand-side values (the bi s) are the resource
available.
Dual Prices (Shadow prices) provide information on
Dual price is the improvement in value of the optimal
solution per-unit increase in a constraint's right-handside value.
The zj values for the slack variables in the final
simplex tableau are the dual prices
x1
x2
s1
s2
s3
0
0
0
1
3
25
0
3
25
0
5
Basis
cB
50
40
x2
s2
x1
40
0
1
8
0
0
0
8
50
1
0
5
25
25
12
25
8
25
30
26
0
5
5 1980
cj  zj
0
0  14 5
0  26 5
The dual prices for the assembly time constraint, Portable
display constraint, and warehouse capacity constraint are
respectively 14 = \$2.8, 0.0, and 26 =\$5.2.
5
5
More warehouse space will have the biggest positive impact
on profit.
zj
50
40
14
6
Table 1 summarizes the rules for determining the dual prices
for the various constraint types.
Constraint Type Dual Price Given by
zj value for the slack variable associated
≤
with the constraint
Negative of the zj value for the surplus
≥
variable associated with the constraint
zj value for the artificial variable
=
associated with the constraint
Dual prices for a minimization problem
Table 2 DUAL PRICES
7
The range of values over which a particular bi can vary
without any of the current basic variables becoming
infeasible (i.e., less than zero) will be referred to as the
range of feasibility.
Calculate the range of values for bi that satisfy the following
inequalities.
Current solution 
Column of the final simplex
(last column of 
 tableau correspond ing to the

  bi 
 the final simplex 
slack vari able associated



 tableau)

 with constraint i
 0 
 0 
 
 .
  
 0 
The range of feasibility can be established by the
maximum of the lower limits and the minimum of the
upper limits.
8
Example
The final simplex tableau for the problem is:
x1
x2
s1
s2
s3
0
0
0
1
3
25
0
3
25
0
5
Basis
cB
50
40
x2
s2
x1
40
0
1
8
0
0
0
8
50
1
0
5
zj
50
40
14
cj  zj
0
0
25
5
 14
5
0
0
25
25
25
26
5
 26
5
12
8
30
1980
To compute upper and lower bounds for the maximum
amount that b1 can be changed before the current optimal
basis becomes infeasible.
 8  0 
12 
25
 8   b  8   0
1
25  
 

5  0
30

 25
12  8
b  0
25 1
b1  37.5
8 8
b  0
30  5 b1  0
25 1
25
b1  25
b1  150
 37.5  b1  25
150  37.5  150  b1  150  25
112.5  b1  175
The range of feasibility for b1 indicates that as long as the
available assembly time is between 112.5 and 175 hours, the
current optimal basis will remain feasible.
9
Problems
1. Suppose that we increase b1 by 5, what is the optimal
solution?
2. Suppose that we increase b1 by 25, what is the optimal
solution? (See textbook page 56)
3. Suppose that we increase b1 by 40, what is the optimal
solution?
1. Because the dual price for b1 (assembly time) is 14 5 , we know
profit can be increased by \$2.80 by obtaining an additional hour of
assembly time.
The profit = \$1980+(\$2.8)5=\$1994
The values of the optimal basic variables
 25  13.6
s  8  5 8   6.4
25
x  30  5 5   29
25
x2  12  5 8
2
1
To determine the range of feasibility for the right-hand-side value of
≥ constraint See
textbook page 57
Changes that force bi outside its range of feasibility will force us to
resolve the problem to find the new optimal solution consisting of a
different set of basic variables.
(More advanced linear programming texts show how it can be done
without completely resolving the problem.)
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