OR6205: Assignment 4 Solution
Question 1
For the LP given below:
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 𝑥1 + 𝑥2
subject to:
𝑥1 + 2𝑥2 = 10
2𝑥1 + 𝑥2 ≥ 2
𝑥1 𝑖𝑠 𝑢𝑛𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑡𝑒𝑑, 𝑥2 ≥ 0.
a. Convert the given problem to a normal LP.
b. Construct the dual problem for the normal LP obtained in part (a).
Solution
Let 𝑥1 = 𝑥1′ − 𝑥1′′ , where 𝑥1′ ≥ 0, 𝑥1′′ ≥ 0
Then the LP normal is :
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 𝑥1′ − 𝑥1′′ + 𝑥2
subject to:
𝑥1′ − 𝑥1′′ + 2𝑥2 ≤ 10
−𝑥1′ + 𝑥1′′ − 2𝑥2 ≤ −10
−2𝑥1′ + 2𝑥1′′ − 𝑥2 ≤ −2
𝑥1′ , 𝑥1′′ , 𝑥2 ≥ 0.
The dual LP is :
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 10𝑦1 − 10𝑦2 − 2𝑦3
subject to:
𝑦1 − 𝑦2 − 2𝑦3 ≥ 1
−𝑦1 + 𝑦2 + 2𝑦3 ≥ −1
2𝑦1 − 2𝑦2 − 𝑦3 ≥ 1
𝑦1 , 𝑦2 , 𝑦3 ≥ 0.
1
Question 2
For the following LP
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 3𝑥1 + 7𝑥2 + 5𝑥3
𝑥1 + 𝑥2 + 𝑥3 ≤ 50
subject to:
2𝑥1 + 3𝑥2 + 𝑥3 ≤ 100
𝑥1 ,
𝑥2 , 𝑥3 ≥ 0.
The final simplex tableau is given below, where 𝑠1 𝑎𝑛𝑑 𝑠2 are the slack variables for constraint 1
and 2. The optimal solution given by this tableau is:
𝑥1 = 0,
Basic
variable
Z
𝑥3
𝑥2
Eq.
(0)
(1)
(2)
𝑥2 = 25,
Z
1
0
0
𝑥3 = 25,
Coefficient of:
𝑥2
𝑥3
𝑥1
3
1/2
1/2
𝑠1 = 0,
0
0
1
0
1
0
𝑠2 = 0,
𝑠1
4
3/2
-1/2
𝑍 = 300.
𝑠2
1
-1/2
1/2
Right
Side
300
25
25
Using the information from the optimal simplex tableau:
a. Determine the range of 𝑐1 (coefficient of 𝑥1 in the objective function) within which current
basis remains optimal.
b. Determine the range of 𝑐2 (coefficient of 𝑥2 in the objective function) within which current
basis remains optimal.
c. Determine the range of 𝑏1 (right hand side of the first constraint) within which current basis
remains optimal.
d. If the RHS of the first constraint (𝑏1 ) is changed from 50 to 60, find out the new objective
function value for 𝑍 from the information given in the optimal simplex tableau (without resolving).
2
Solution
(a) 𝑥1 is non-basic so changing the coefficient of 𝑥1 in the objective function will only change the
coefficient of 𝑥1 in the Row 0 of the optimal tableau.
Let the change in the coefficient of 𝑥1 in the objective function is 𝛥𝑐1 . So, the new coefficient of 𝑥1
in the objective function be (3 + 𝛥𝑐1 ). The new coefficient of 𝑥1 in Row-0 of the optimal tableau
will be (3 − 𝛥𝑐1 ).
Basic
variable
Z
𝑥3
𝑥2
Eq.
(0)
(1)
(2)
Coefficient of:
𝑥2
𝑥3
Z
𝑥1
1
0
0
3 − 𝛥𝑐1
½
½
0
0
1
0
1
0
𝑠1
𝑠2
4
3/2
-1/2
1
-1/2
1/2
Right
Side
300
25
25
Thus if 3 − 𝛥𝑐1 ≥ 0 𝑜𝑟 𝛥𝑐1 ≤ 3, the current basis remains optimal.
Therefore, if coefficient of 𝑥1 in the objective function: 𝑐1 ≤ 6, the current basis remains optimal.
(b) 𝑥2 is a basic variable, so changing the coefficient of 𝑥2 in the objective function will change the
coefficient of other variables along with the coefficient of 𝑥2 in the Row 0 of the optimal tableau.
Let the change in the coefficient of 𝑥2 in the objective function is 𝛥𝑐2. So, the new coefficient of 𝑥2
in the objective function be (7 + 𝛥𝑐2 ). The new coefficient of 𝑥2 in Row-0 of the optimal tableau
will be (0 − 𝛥𝑐2 ).
Basic
variable
Z
𝑥3
𝑥2
Eq.
(0)
(1)
(2)
Coefficient of:
𝑥2
𝑥3
𝑥1
Z
1
0
0
− 𝛥𝑐2
0
1
3
½
½
0
1
0
𝑠1
𝑠2
4
3/2
-1/2
1
-1/2
1/2
Right
Side
300
25
25
Since 𝑥2 is a basic variable, we have to do elementary row operations to turn the coefficient of 𝑥2
in Row-0 of the optimal tableau into “0’. We multiply Row-2 by 𝛥𝑐2 and add it to Row-0.
Basic
variable
Z
𝑥3
𝑥2
Eq. Z
(0)
(1)
(2)
1
0
0
𝑥1
𝑥2
3 + 𝛥𝑐2 /2 0
½
0
½
1
Coefficient of:
𝑥3
𝑠1
0
1
0
4 − 𝛥𝑐2 /2
3/2
-1/2
𝑠2
Right Side
1 + 𝛥𝑐2 /2
-1/2
1/2
300+25𝛥𝑐2
25
25
3
To keep the current basis optimal, all the coefficient of in Row-0 of the optimal tableau must be
non-negative. Therefore, the current basis remains optimal if following restrictions (1)-(3) are met:
(1)
3+
(2)
(3)
Thus, if
𝛥𝑐2
≥0
2
4−
1+
𝑜𝑟, 𝛥𝑐2 ≥ −6
𝛥𝑐2
≥0
2
𝛥𝑐2
≥0
2
𝑜𝑟,
𝑜𝑟, 𝛥𝑐2 ≤ 8
𝛥𝑐2 ≥ −2
−2 ≤ 𝛥𝑐2 ≤ 8 or 7 − 2 ≤ 𝑐2 ≤ 7 + 8, the current basis remains optimal.
So, if 𝑐2 remains between [5,15], the current basis remains optimal.
(c) Let the change in the RHS of the first constraint is 𝛥𝑏1 . So, the new RHS of the first constraint
is (50 + 𝛥𝑏1 ). So,
∆𝒃 = [
∆𝑏1
]
0
From the optimal Simplex tableau:
𝑺∗ = [
3/2 −1/2
] , 𝒚∗ = [4 1]
−1/2 1/2
25
𝑺∗ 𝒃 = [ ] , 𝒚∗ 𝒃 = 300
25
RHS adjustment in the final tableau:
Row-0 of the RHS:
𝒚∗ 𝒃 + 𝒚 ∗ ∆𝒃 = 300 + [4
∆𝑏
1] [ 1 ] = 300 + 4∆𝑏1
0
Other rows of the RHS:
𝑺∗ 𝒃 + 𝑺∗ ∆𝒃
[
3/2
25
]+[
−1/2
25
3∆𝑏1
25
+
−1/2 ∆𝑏1
2 ]
][
]=[
1/2
∆𝑏
0
1
25 −
2
4
So, the revised tableau:
Coefficient of:
𝑥2
𝑥3
𝑠1
Basic
variable
Z
𝑥3
(0)
(1)
1
0
3
½
0
0
0
1
4
3/2
1
-1/2
𝑥2
(2)
0
½
1
0
-1/2
1/2
Eq. Z
𝑥1
Right Side
𝑠2
300 + 4∆𝑏1
3∆𝑏1
25 +
2
∆𝑏1
25 −
2
To stay feasible, we must satisfy:
[
25 +
25 −
3∆𝑏1
2
∆𝑏1
0
] ≥ [ ] , which provides the following ranges for ∆𝑏1:
0
2
25 +
3∆𝑏1
≥0
2
25 −
50
So,
−
Or,
50 −
Or, if
3
50
3
𝑜𝑟 ∆𝑏1 ≤ 50
≤ ∆𝑏1 ≤ 50
50
3
100
3
∆𝑏1
≥0
2
𝑜𝑟 ∆𝑏1 ≥ −
≤ 𝑏1 ≤ 50 + 50
≤ 𝑏1 ≤ 100, the current basis remains optimal.
5
(d) From the previous question (c), we found that if we change 𝑏1 from 50 to 60, the current optimal
basis will not change.
So, we can use shadow price of the first constraint to calculate the increase in the objective function
value 𝑍 for this change.
Shadow price for the first constraint is : 4
Basic
variable
Z
𝑥3
𝑥2
Eq.
(0)
(1)
(2)
1
0
0
Coefficient of:
𝑥2
𝑥3
𝑥1
Z
3
½
½
0
0
1
0
1
0
𝑠1
4
3/2
-1/2
𝑠2
1
-1/2
1/2
Right
Side
300
25
25
So, increase in the objective function value 𝑍 for if we increase 𝑏1 by the amount ∆𝑏1
∆𝑍 = (𝑠ℎ𝑎𝑑𝑜𝑤 𝑝𝑟𝑖𝑐𝑒 𝑓𝑜𝑟 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 1) ∗ ∆𝑏1
∆𝑍 = 4 ∗ 10 = 40
So, the new objective function value: 𝑍𝑛𝑒𝑤 = 𝑍𝑜𝑙𝑑 + ∆𝑍 = 300 + 40 = 340
6
Question 3
For the LP given below:
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 3𝑥1 − 2𝑥2
2𝑥1 − 𝑥2 ≤ 30
subject to:
𝑥1 − 𝑥2 ≤ 10
𝑥1 , 𝑥2 ≥ 0.
The final simplex tableau is shown below, where 𝑠1 𝑎𝑛𝑑 𝑠2 are the slack variables for constraint
1 and 2. The optimal solution given by this tableau is:
𝑥1 = 20,
Basic
variable
Z
𝑥2
𝑥1
Eq.
(0)
(1)
(2)
𝑥2 = 10,
𝑠2 = 0,
Coefficient of:
𝑥1
𝑥2
Z
1
0
0
𝑠1 = 0,
0
0
1
0
1
0
𝑍 = 40.
𝑠1
1
1
1
𝑠2
1
-2
-1
Right
Side
40
10
20
Using the information from the optimal simplex tableau:
a. Determine the range of 𝑐1 (coefficient of 𝑥1 in the objective function) within which current
basis remains optimal.
b. Determine the range of 𝑐2 (coefficient of 𝑥2 in the objective function) within which current
basis remains optimal.
c. Determine the range of 𝑏1 (right hand side of the first constraint) within which current basis
remains optimal.
d. If the RHS of the second constraint (𝑏2 ) is changed from 10 to 13, find out the new objective
function value for 𝑍 from the information given in the optimal simplex tableau (without resolving).
Solution
(a) 𝑥1 is a basic variable, so changing the coefficient of 𝑥1 in the objective function will change the
coefficient of other variables along with 𝑥1 in the Row 0 of the optimal tableau.
Let the change in the coefficient of 𝑥1 in the objective function is 𝛥𝑐1 . The new coefficient of 𝑥1 in
Row-0 of the optimal tableau will be (0 − 𝛥𝑐1 ).
7
Basic
variable
Z
𝑥2
𝑥1
Eq.
(0)
(1)
(2)
Coefficient of:
𝑥1
𝑥2
Z
− 𝛥𝑐1
0
1
1
0
0
0
1
0
𝑠1
1
1
1
𝑠2
1
-2
-1
Right
Side
40
10
20
Since 𝑥1 is a basic variable, we have to do elementary row operations to turn the coefficient of 𝑥1
in Row-0 of the optimal tableau into “0’. We multiply Row-2 by 𝛥𝑐1 and add it to Row-0.
Basic
variable
Z
𝑥2
𝑥1
Eq.
(0)
(1)
(2)
Z
1
0
0
0
0
1
Coefficient of:
𝑥1
𝑥2
𝑠1
𝑠2
Right Side
0
1
0
1+𝛥𝑐1
1
1
1−𝛥𝑐1
-2
-1
40+20𝛥𝑐1
10
20
To keep the current basis optimal, all the coefficient of in Row-0 of the optimal tableau must be nonnegative. Therefore, the current basis remains optimal if following restrictions (1)-(2) are met:
(1)
1 + 𝛥𝑐1 ≥ 0
(2)
Thus, if
𝑜𝑟, 𝛥𝑐1 ≥ −1
1 − 𝛥𝑐1 ≥ 0
𝑜𝑟, 𝛥𝑐1 ≤ 1
−1 ≤ 𝛥𝑐1 ≤ 1 or 3 − 1 ≤ 𝑐1 ≤ 3 + 1, the current basis remains optimal.
So, if 𝑐1 remains between [2,4], the current basis remains optimal.
(b) 𝑥2 is a basic variable, so changing the coefficient of 𝑥2 in the objective function will change the
coefficient of other variables along with 𝑥2 in the Row 0 of the optimal tableau.
Let the change in the coefficient of 𝑥2 in the objective function is 𝛥𝑐2. The new coefficient of 𝑥2 in
Row-0 of the optimal tableau will be (0 − 𝛥𝑐2 ).
Basic
variable
Z
𝑥2
𝑥1
Eq.
(0)
(1)
(2)
Coefficient of:
𝑥1
𝑥2
Z
1
0
0
0
0
1
− 𝛥𝑐2
1
0
𝑠1
1
1
1
𝑠2
1
-2
-1
Right
Side
40
10
20
Since 𝑥2 is a basic variable, we have to do elementary row operations to turn the coefficient of 𝑥2
in Row-0 of the optimal tableau into “0’. We multiply Row-1 by 𝛥𝑐2 and add it to Row-0.
8
Basic
variable
Z
𝑥2
𝑥1
Eq.
(0)
(1)
(2)
Z
1
0
0
Coefficient of:
𝑥1
𝑥2
𝑠1
0
1
0
1+𝛥𝑐2
1
1
0
0
1
𝑠2
Right Side
1−2𝛥𝑐2 40+10𝛥𝑐2
-2
10
-1
20
To keep the current basis optimal, all the coefficient of in Row-0 of the optimal tableau must be nonnegative. Therefore, the current basis remains optimal if following restrictions (1)-(2) are met:
(1)
1 + 𝛥𝑐2 ≥ 0
(2)
Thus, if
𝑜𝑟, 𝛥𝑐2 ≥ −1
1 − 2𝛥𝑐2 ≥ 0
1
𝑜𝑟, 𝛥𝑐1 ≤
1
2
1
−1 ≤ 𝛥𝑐2 ≤ 2 or −2 − 1 ≤ 𝑐2 ≤ −2 + 2, the current basis remains optimal.
3
So, if 𝑐2 remains between [−3, − 2], the current basis remains optimal.
(c) Let the change in the RHS of the first constraint is 𝛥𝑏1 . So, the new RHS of the first constraint
is (30 + 𝛥𝑏1 ). So,
∆𝒃 = [
∆𝑏1
]
0
From the optimal Simplex tableau:
1
𝑺∗ = [
1
−2 ∗ [
] , 𝒚 = 1 1]
−1
𝑺∗ 𝒃 = [
10
] , 𝒚∗ 𝒃 = 40
20
RHS adjustment in the final tableau:
Row-0 of the RHS:
𝒚∗ 𝒃 + 𝒚 ∗ ∆𝒃 = 40 + [1 1] [
∆𝑏1
] = 40 + ∆𝑏1
0
Other rows of the RHS:
𝑺∗ 𝒃 + 𝑺∗ ∆𝒃
9
[
10 + ∆𝑏1
10
1 −2 ∆𝑏1
]+[
][
]=[
]
20 + ∆𝑏1
20
1 −1 0
So, the revised tableau:
Basic
variable
Z
𝑥2
𝑥1
Eq.
(0)
(1)
(2)
Coefficient of:
𝑥1
𝑥2
Z
1
0
0
0
0
1
0
1
0
𝑠1
1
1
1
𝑠2
1
-2
-1
Right Side
40+∆𝑏1
10 +∆𝑏1
20+∆𝑏1
To stay feasible:
[
So,
∆𝑏1 ≤ −10
Or,
𝑏1 ≥ 30 − 10
So, if
10 + ∆𝑏1
0
]≥[ ]
20 + ∆𝑏1
0
10 + ∆𝑏1 ≥ 0
𝑜𝑟 ∆𝑏1 ≥ −10
20 + ∆𝑏1 ≥ 0
𝑜𝑟 ∆𝑏1 ≥ −20
𝑏1 ≥ 20, the current basis remains optimal.
(d) Following the procedure in part (c) for range of 𝑏2 , we find that :
if
𝑏2 ≤ 15, the current basis remains optimal.
So, if we increase 𝑏2 from 10 to 13, the current basis remains optimal.
We can use shadow price to calculate the change in the objective function value.
Or, alternatively,
Row-0 of the RHS after changing 𝑏2 from 10 to 13
𝒚∗ 𝒃 + 𝒚 ∗ ∆𝒃 = 40 + [1
3
1] [ ] = 40 + 3 = 43
0
10
Question 4
Consider the following LP model:
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 2𝑥1 − 𝑥2 + 𝑥3
3𝑥1 − 2𝑥2 + 2𝑥3 ≤ 15
subject to:
−𝑥1 + 𝑥2 + 𝑥3 ≤ 3
𝑥1 − 𝑥2 + 𝑥3 ≤ 4
𝑥1 , 𝑥2 , 𝑥3 ≥ 0.
The final simplex tableau is given below, where 𝑥4 , 𝑥5 𝑎𝑛𝑑 𝑥6 are the slack variables for constraint
1, 2, and 3. The optimal solution given by this tableau is:
𝑥1 = 21,
Basic
Variable
𝑥2 = 24,
𝑥3 = 0,
𝑥4 = 0,
𝑥5 = 0,
𝑥6 = 7, 𝑍 = 18.
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
2
1
1
0
18
x2
(1)
0
0
1
5
1
3
0
24
x6
(2)
0
0
0
2
0
1
1
7
x1
(3)
0
1
0
4
1
2
0
21
Perform sensitivity analysis by separately investigating each of the following seven changes in the
original model. For each change, revise the final simplex tableau for identifying and evaluating
the current basic solution. Then test this solution for feasibility and for optimality. If either test
fails, reoptimize to find a new optimal solution.
(a) Change the right-hand sides of the constraints:
𝑓𝑟𝑜𝑚
𝑏1
15
𝒃 = [𝑏2 ] = [ 3 ]
𝑏3
4
𝑡𝑜
𝑏1
10
𝒃 = [𝑏2 ] = [ 4 ]
𝑏3
2
Solution
11
(a) Only change will be in the RHS column and we will need to check the feasibility.
−5
∆𝒃 = [ 1 ]
−2
Given that
And from the optimal Simplex tableau:
1
𝑺 = [0
1
∗
3 0
1 1] , 𝒚∗ = [1 1 0]
2 0
24
𝑺∗ 𝒃 = [ 7 ] , 𝒚∗ 𝒃 = 18
21
So, RHS adjustment in the final tableau will be the following:
Row-0 of the RHS:
−5
[
]
𝒚 𝒃 + 𝒚 ∗ ∆𝒃 = 18 + 1 1 0 [ 1 ] = 18 − 5 + 1 = 14
−2
∗
Other rows of the RHS:
𝑺∗ 𝒃 + 𝑺∗ ∆𝒃
24
1 3
[ 7 ] + [0 1
21
1 2
0 −5
22
1] [ 1 ] = [ 6 ]
0 −2
18
So, the revised tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
2
1
1
0
14
x2
(1)
0
0
1
5
1
3
0
22
x6
(2)
0
0
0
2
0
1
1
6
x1
(3)
0
1
0
4
1
2
0
18
We see that the current basic solution is feasible and optimal.
𝑥1 = 18,
𝑥2 = 22,
𝑥3 = 0,
𝑥4 = 0,
𝑥5 = 0,
𝑥6 = 6, 𝑍 = 14.
12
(b) Change the coefficient of 𝑥3 in the objective function:
𝑓𝑟𝑜𝑚 𝑐3 = 1
𝑡𝑜
𝑐3 = 2 .
Solution
(b) 𝑥3 is non-basic so changing the coefficient of 𝑥3 in the objective function will only change the
coefficient of 𝑥3 in the Row 0 of the optimal tableau.
Given that 𝛥𝑐3 = 2 − 1 = 1, the new coefficient of 𝑥3 in Row-0 of the optimal tableau will be
(2 − 𝛥𝑐3 ) = 2 − 1 = 1
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
1
1
1
0
18
x2
(1)
0
0
1
5
1
3
0
24
x6
(2)
0
0
0
2
0
1
1
7
x1
(3)
0
1
0
4
1
2
0
21
We see that the current basic solution remains feasible and optimal.
13
(c) Change the coefficient of 𝑥1 in the objective function:
𝑓𝑟𝑜𝑚 𝑐1 = 2
𝑡𝑜
𝑐1 = 3 .
Solution
(c) Given that 𝛥𝑐1 = 3 − 2 = 1, the new coefficient of 𝑥1 in Row-0 of the optimal tableau will be
(0 − 𝛥𝑐1 ) = 0 − 1 = −1
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
-1
0
2
1
1
0
18
x2
(1)
0
0
1
5
1
3
0
24
x6
(2)
0
0
0
2
0
1
1
7
x1
(3)
0
1
0
4
1
2
0
21
Since 𝑥1 is a basic variable, we have to do elementary row operations to turn the coefficient of 𝑥1
in Row-0 of the optimal tableau into “0’. We multiply Row-3 by 1 and add it to Row-0.
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
6
2
3
0
39
x2
(1)
0
0
1
5
1
3
0
24
x6
(2)
0
0
0
2
0
1
1
7
x1
(3)
0
1
0
4
1
2
0
21
We see that the current basic solution is feasible and optimal.
𝑥1 = 21,
𝑥2 = 24,
𝑥3 = 0,
𝑥4 = 0,
𝑥5 = 0,
𝑥6 = 7, 𝑍 = 39.
14
(d) Change the coefficients of 𝑥3 :
𝑓𝑟𝑜𝑚
𝑐3
1
𝑎13
2
[𝑎 ] = [ ]
23
1
𝑎33
1
𝑐3
4
𝑎13
3
[𝑎 ] = [ ]
23
2
𝑎33
1
𝑡𝑜
where 𝑎13 , 𝑎23 , 𝑎33 are the coefficients of 𝑥3 in constraints 1,2,3 and 𝑐3 is the coefficient of 𝑥3 in
the objective function.
Solution
(d)
Given that
𝛥𝑎13
1
𝛥𝑐3 = 4 − 1 = 3 and [𝛥𝑎23 ] = [1],
𝛥𝑎33
0
0
0 1
So, 𝜟𝒄 = [0 0 3] and 𝜟𝑨 = [0 0 1]
0
0 0
Then the adjusted Row-0 coefficient under 𝑥1 , 𝑥2 , 𝑥3 columns would be:
𝒚 ∗ 𝑨 − 𝒄 + 𝒚 ∗ 𝜟𝑨 − 𝜟𝒄
Now,
𝒚 ∗ 𝜟𝑨 − 𝜟𝒄 = [1 1
0 0 1
]
0 [0 0 1] − [0 0 3]
0 0 0
= [0 0 2] − [0 0
= [0 0
So,
𝒚 ∗ 𝑨 − 𝒄 + 𝒚 ∗ 𝜟𝑨 − 𝜟𝒄 = [0 0
3]
−1]
2] + [0
0 −1] = [0
0 1]
Other row’s adjusted coefficient under 𝑥1 , 𝑥2 , 𝑥3 columns would be:
𝑺 ∗ 𝑨 + 𝑺 ∗ 𝜟𝑨
Now,
1
𝑺 ∗ 𝜟𝑨 = [0
1
3 0 0 0
1 1 ] [0 0
2 0 0 0
1
1]
0
15
0
= [0
0
0
𝑺 ∗ 𝑨 + 𝑺 ∗ 𝜟𝑨 = [0
1
So,
0 4
0 1]
0 3
0
1 5
0 2] + [0
0
0 4
0 4
0
0 1] = [0
0 3
1
1 9
0 3]
0 7
Revised final tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
1
1
1
0
18
x2
(1)
0
0
1
9
1
3
0
24
x6
(2)
0
0
0
3
0
1
1
7
x1
(3)
0
1
0
7
1
2
0
21
We see that the current basic solution is feasible and optimal.
𝑥1 = 21,
𝑥2 = 24,
𝑥3 = 0,
𝑥4 = 0,
𝑥5 = 0,
𝑥6 = 7, 𝑍 = 18.
(e) Change the coefficients of 𝑥1 and 𝑥2 :
𝑓𝑟𝑜𝑚
𝑐1
2
𝑎11
3
[𝑎 ] = [ ]
21
−1
𝑎31
1
𝑓𝑟𝑜𝑚
𝑐2
−1
𝑎12
−2
[𝑎 ] = [ ]
22
1
𝑎32
−1
𝑡𝑜
𝑐1
1
𝑎11
1
[𝑎 ] = [ ]
21
−2
𝑎31
3
𝑡𝑜
𝑐2
−2
𝑎12
−2
[𝑎 ] = [ ]
22
3
𝑎32
2
and
where 𝑎11 , 𝑎21 , 𝑎31 are the coefficients of 𝑥1 in constraints 1,2,3 and 𝑐1 is the coefficient of 𝑥1 in
the objective function.
where 𝑎12 , 𝑎22 , 𝑎32 are the coefficients of 𝑥2 in constraints 1,2,3 and 𝑐2 is the coefficient of 𝑥2 in
the objective function.
16
Solution
(e)
𝛥𝑐1
𝛥𝑎11
[
]=
𝛥𝑎21
𝛥𝑎31
Given that
𝜟𝒄 = [−1
So,
𝛥𝑐2
−1
−1
𝛥
𝑎
−2
0
[ ] and [ 12 ] = [ ]
𝛥
𝑎
−1
2
22
𝛥𝑎32
2
3
−2 0
−1 0] and 𝜟𝑨 = [−1 2
2 3
0
0]
0
Then the adjusted Row-0 coefficient under 𝑥1 , 𝑥2 , 𝑥3 columns would be:
𝒚 ∗ 𝑨 − 𝒄 + 𝒚 ∗ 𝜟𝑨 − 𝜟𝒄
Now,
𝒚 ∗ 𝜟𝑨 − 𝜟𝒄 = [1 1
= [−3
−2 0 0
0] [−1 2 0] − [−1 −1
2 3 0
2 0] − [−1
0]
−1 0]
= [−2 3 0]
So,
𝒚 ∗ 𝑨 − 𝒄 + 𝒚 ∗ 𝜟𝑨 − 𝜟𝒄 = [0 0
2] + [−2 3
0] = [−2 3
2]
Other row’s adjusted coefficient under 𝑥1 , 𝑥2 , 𝑥3 columns would be:
𝑺 ∗ 𝑨 + 𝑺 ∗ 𝜟𝑨
Now,
1
𝑺 ∗ 𝜟𝑨 = [0
1
3 0 −2 0 0
1 1] [−1 2 0]
2 0 2 3 0
−5 6 0
= [ 1 5 0]
−4 4 0
So,
0
𝑺 ∗ 𝑨 + 𝑺 ∗ 𝜟𝑨 = [0
1
1 5
−5 6
]
+
[
0 2
1 5
0 4
−4 4
0
−5 7
]
=
[
0
1 5
0
−3 4
5
2]
4
Revised final tableau:
17
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
-2
3
2
1
1
0
18
x2
(1)
0
-5
7
5
1
3
0
24
x6
(2)
0
1
5
2
0
1
1
7
x1
(3)
0
-3
4
4
1
2
0
21
Since 𝑥1 is a basic variable, we have to do elementary row operations to turn the coefficient of 𝑥1
in Row-3 of the optimal tableau into “1’ and in other rows into “0”.
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
1/3
-2/3
1/3
-1/3
0
4
x2
(1)
0
0
1/3
-5/3
-2/3
-1/3
0
-11
x6
(2)
0
0
19/3
10/3
1/3
5/3
1
14
x1
(3)
0
1
-4/3
-4/3
-1/3
-2/3
0
-7
Since 𝑥2 is a basic variable, we have to do elementary row operations to turn the coefficient of 𝑥2
in Row-1 of the optimal tableau into “1’ and in other rows into “0”.
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
1
1
0
0
15
x2
(1)
0
0
1
-5
-2
-1
0
-33
x6
(2)
0
0
0
35
13
8
1
223
x1
(3)
0
1
0
-8
-3
-2
0
-51
18
The current basic solution is super-optimal, but infeasible. Using dual Simplex, we enter 𝑥3 into
the basis to replace 𝑥1 . We get the following tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
1
1
0
0
15
x2
(1)
0
-1/2
1
-1
-1/2
0
0
-15/2
x6
(2)
0
4
0
3
1
0
1
19
x5
(3)
0
-1/2
0
4
3/2
1
0
51/2
The current basic solution is super-optimal, but infeasible. Using dual Simplex, we enter 𝑥1 into
the basis to replace 𝑥2 . We get the following tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
0
1
1
0
0
15
x1
(1)
0
1
-2
2
1
0
0
15
x6
(2)
0
0
8
-5
-3
0
1
-41
x5
(3)
0
0
-1
5
2
1
0
33
The current basic solution is super-optimal, but infeasible. Using dual Simplex, we enter 𝑥4 into
the basis to replace 𝑥6 . We get the following tableau:
19
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
8/5
0
2/5
0
1/5
34/5
x1
(1)
0
1
6/5
0
-1/5
0
2/5
-7/5
x3
(2)
0
0
-8/5
1
3/5
0
-1/5
41/5
x5
(3)
0
0
7
0
-1
1
1
-8
The current basic solution is super-optimal, but infeasible. Using dual Simplex, we enter 𝑥4 into
the basis to replace 𝑥5 . We get the following tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
Right Side
Z
(0)
1
0
22/5
0
0
2/5
3/5
18/5
x1
(1)
0
1
-1/5
0
0
-1/5
1/5
1/5
x3
(2)
0
0
13/5
1
0
3/5
2/5
17/5
x4
(3)
0
0
-7
0
1
-1
-1
8
The current basic solution is feasible and optimal.
𝑥1 =
1
,
5
𝑥2 = 0,
𝑥3 =
17
,
5
𝑥4 = 8,
𝑥5 = 0,
𝑥6 = 0, 𝑍 =
18
.
5
20
(f) Introduce a new constraint into the original LP:
2𝑥1 + 𝑥2 + 3𝑥3 ≤ 60.
Solution
(f)
The current basic solution does not satisfy the new constraint. We introduce a slack variable 𝑥7 to
standardize the LP with this new constraint and revise the final Simplex tableau.
New tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
x7
Right Side
Z
(0)
1
0
0
2
1
1
0
0
18
x2
(1)
0
0
1
5
1
3
0
0
24
x6
(2)
0
0
0
2
0
1
1
0
7
x1
(3)
0
1
0
4
1
2
0
0
21
x7
(4)
0
2
1
3
0
0
0
1
60
We do elementary row operations to obtain the tableau in a proper form
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
x7
Right Side
Z
(0)
1
0
0
2
1
1
0
0
18
x2
(1)
0
0
1
5
1
3
0
0
24
x6
(2)
0
0
0
2
0
1
1
0
7
x1
(3)
0
1
0
4
1
2
0
0
21
x7
(4)
0
0
0
-10
-3
-7
0
1
-6
21
The current basic solution is super-optimal, but infeasible. Using dual Simplex, we enter 𝑥5 into
the basis to replace 𝑥7 . We get the following tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
x7
Right Side
Z
(0)
1
0
0
4/7
4/7
0
0
1/7
120/7
x2
(1)
0
0
1
5/7
-2/7
0
0
3/7
150/7
x6
(2)
0
0
0
4/7
-3/7
0
1
1/7
43/7
x1
(3)
0
1
0
8/7
1/7
0
0
2/7
135/7
x5
(4)
0
0
0
10/7
3/7
0
0
-1/7
6/7
𝑥6 =
43
120
, 𝑍=
.
7
7
The current basic solution is feasible and optimal.
𝑥1 =
135
,
7
𝑥2 =
150
,
7
𝑥3 = 0,
𝑥4 = 0,
𝑥5 =
6
,
7
22
(g) Introduce a new activity into the original LP. Let’s define this activity by 𝑥7 :
𝑓𝑟𝑜𝑚
𝑐7
0
𝑎17
0
[𝑎 ] = [ ]
27
0
𝑎37
0
𝑐7
4
𝑎17
2
[𝑎 ] = [ ]
1
27
𝑎37
3
𝑡𝑜
where 𝑎17 , 𝑎27 , 𝑎37 are the coefficients of 𝑥7 in constraints 1,2,3 and 𝑐7 is the coefficient of 𝑥7 in
the objective function.
Solution
(g)
Introduce a new activity into the original LP. Let’s define this activity by 𝑥7 :
𝑓𝑟𝑜𝑚
𝑐7
0
𝑎17
0
[𝑎 ] = [ ]
27
0
𝑎37
0
𝑐7
4
𝑎17
2
[𝑎 ] = [ ]
1
27
𝑎37
3
𝑡𝑜
where 𝑎17 , 𝑎27 , 𝑎37 are the coefficients of 𝑥7 in constraints 1,2,3 and 𝑐7 is the coefficient of 𝑥7 in
the objective function.
1
𝑺∆𝑨 = [0
1
3 0 𝟐
𝟓
1 1] [𝟏] = [𝟒]
2 0 𝟑
𝟒
𝟐
𝒚∆𝑨 − ∆𝒄 = [𝟏 𝟏 𝟎] [𝟏] − 𝟒 = −𝟏
𝟑
New tableau:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
x7
Right Side
Z
(0)
1
0
0
2
1
1
0
-1
18
x2
(1)
0
0
1
5
1
3
0
5
24
x6
(2)
0
0
0
2
0
1
1
4
7
x1
(3)
0
1
0
4
1
2
0
4
21
23
The current basic solution is feasible but sub-optimal. Tableau after re-optimization:
Basic
Variable
Coefficient of:
Eq
Z
x1
x2
x3
x4
x5
x6
x7
Right Side
Z
(0)
1
0
0
5/2
1
5/4
1/4
0
79/4
x2
(1)
0
0
1
5/2
1
7/4
-5/4
0
61/4
x7
(2)
0
0
0
1/2
0
1/4
1/4
1
7/4
x1
(3)
0
1
0
2
1
1
-1
0
14
The current basic solution is feasible and optimal.
𝑥1 = 14,
𝑥2 =
61
,
4
𝑥3 = 0, 𝑥4 = 0, 𝑥5 = 0,
𝑥6 = 0,
𝑥7 =
7
79
, 𝑍= .
4
4
The new modified LP is:
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 2𝑥1 − 𝑥2 + 𝑥3 + 4𝑥7
subject to:
3𝑥1 − 2𝑥2 + 2𝑥3 + 2𝑥7 ≤ 15
−𝑥1 + 𝑥2 + 𝑥3 + 𝑥7 ≤ 3
𝑥1 − 𝑥2 + 𝑥3 + 3𝑥7 ≤ 4
𝑥1 , 𝑥2 , 𝑥3 , 𝑥7 ≥ 0.
24