1
1.1
Solutions for Problems I
Solution to Problem 1
For this problem the supply vector is (13, 11, 11, 15) and the demand vector
is (21, 12, 17). The components of both the supply vector and the demand
vector add up to 50.
The costs are as specified in the following cost matrix:—


8 4 6
 3 7 2 


 13 8 6  .
5 7 2
We fill in the row sums (or supplies), the column sums (or demands) and
the costs ci,j for the given problem. The resultant tableau looks as follows:—
ci,j & xi,j
1
1
8
2
3
2
4
?
?
7
?
3
13
8
dj
7
?
21
?
13
?
11
?
11
6
?
5
si
2
?
?
4
3
6
2
?
12
? 15
17 50
We apply the minimum cost method to find an initial basic solution.
The cells with lowest cost are the cells (2, 3) and (4, 3). We could assign to
either of these cells the maximum value possible which would be the minimum
of the corresponding row and column sums required. Thus we could assign to
(2, 3) the minimum of 11 and 17, or we could assign to (4, 3) the minimum of
15 and 17. Given our objective of minimizing cost it makes sense to choose so
as to obtain the largest value of xi,j for which ci,j = 2. Accordingly we assign
x4,3 = 15. This forces x4,1 = 0 and x4,2 = 0, completing the 4th row. The
pair (4, 3) is the first element of the current basis B that we have identified.
The next undetermined cell of lowest cost is (2, 3) which has cost equal
to 2. The value x2,3 that we assign to this cell must satisfy the inequalities
x2,3 ≤ s2 = 11 and x2,3 ≤ d3 − x4,3 = 17 − 5 = 2. Accordingly we assign
x2,3 = 2. This forces x1,3 = 0 and x3,3 = 0, completing the third column.
The pair (2, 3) is added to the current basis.
1
The next undetermined cell of lowest cost is (2, 1), which has cost equal
to 3. The value x2,1 that we assign to this cell must satisfy the inequalities
x2,1 ≤ s2 − x2,3 = 11 − 2 = 9 and x2,1 ≤ d1 − x4,1 = 21 − 0 = 21. Accordingly
we assign x2,1 = 9. Thus forces x2,2 = 0, completing the second row. The
pair (2, 1) is added to the current basis.
The next undetermined cell of lowest cost is (1, 2), which has cost equal
to 4. The value x1,2 that we assign to this cell must satisfy the inequalities
x1,2 ≤ s1 − x1,3 = 13 − 0 = 13 and x1,2 ≤ d2 − x2,2 − x4,2 = 12 − 0 − 0.
Accordingly we assign x1,2 = 12. Thus forces x3,2 = 0, completing the third
column. The pair (1, 2) is added to the current basis.
We next assign x1,1 = 1, adding (1, 1) to the current basis, and finally
assign x3,1 = 11, adding (3, 1) to the current basis.
The values of the elements xi,j of the initial basic feasible solution are
tabulated (with basis elements marked by the • symbol) as follows:—
ci,j & xi,j
1
2
3
4
dj
1
8
2
4
3
6
si
•
•
1
12
0 13
3 • 7
2 •
9
0
2 11
13 • 8
6
11
0
0 11
5
7
2 •
0
0
15 15
21
12
17 50
The initial basis is B where
B = {(1, 1), (1, 2), (2, 1), (2, 3), (3, 1), (4, 3)}.
The basic feasible solution is represented by the 6 × 5 matrix X, where


1 12 0
 9 0 2 

X=
 11 0 0  .
0 0 15
The cost of this initial feasible basic solution is
8 × 1 + 4 × 12 + 3 × 9 + 2 × 2 + 13 × 11
+ 2 × 15
= 8 + 48 + 27 + 4 + 143 + 30
= 260.
2
We next determine whether the initial basic feasible solution found by
the Minimum Cost Method is an optimal solution, and, if not, how to adjust
the basis go obtain a solution of lower cost.
We determine u1 , u2 , u3 , u4 and v1 , v2 , v3 such that ci,j = vj − ui for all
(i, j) ∈ B, where B is the initial basis.
We seek a solution with u1 = 0. We then determine qi,j so that ci,j =
vj − ui + qi,j for all i and j.
We therefore complete the following tableau:—
ci,j & qi,j
1
1
8
2
• 4 •
0
0
3 • 7
0
?
13 • 8
0
?
5
7
0
0
?
?
2
3
4
vj
3
6
?
2 •
0
6
?
2 •
?
?
ui
0
?
?
?
We are looking for a solution with u1 = 0. Now u1 = 0, (1, 1) ∈ B and
(1, 2) ∈ B force v1 = 8 and v2 = 4.
Then v1 = 8, (2, 1) ∈ B and (3, 1) ∈ B force u2 = 5 and u3 = −5.
Then u2 = 5 and (2, 3) ∈ B force v3 = 7.
Then v3 = 7 and (4, 3) ∈ B force u4 = 5.
Computing the numbers qi,j such that ci,j + ui = vj + qi,j , we find that
q1,3 = 9, q2,2 = 8, q3,2 = −1, q3,3 = −6, q4,1 = 2 and q4,2 = 8.
The completed tableau is as follows:—
ci,j & qi,j
1
2
3
4
vj
1
8
2
3
ui
• 4 • 6
0
0
0
−1
3 • 7
2 •
5
0
8
0
13 • 8
6
−5
0
−1
−6
5
7
2 •
5
2
8
0
8
4
7
The initial basic feasible solution is not optimal because some of the
quantities qi,j are negative. Indeed q1,3 = −1 q3,2 = −1 and q3,3 = −6. The
most negative of these is q3,3 . We therefore seek to bring (3, 3) into the basis.
3
The procedure for achieving this requires us to determine a 3×3 matrix Y
satisfying the following conditions:—
• y3,3 = 1;
• yi,j = 0 when (i, j) 6∈ B ∪ {(3, 3)};
• all rows and columns of the matrix Y sum to zero.
Accordingly we fill in the following tableau with those coefficients yi,j of
the matrix Y that correspond to cells in the current basis (marked with the
• symbol), so that all rows sum to zero and all columns sum to zero:—
yi,j
1
2
3
4
1
2
3
? • ? •
? •
? •
? •
1 ◦
? •
0
0
0
0
0
0
0
0
The constraints describe above result in the following completed tableau:—
yi,j
1
2
3
4
1
2
3
0 • 0 •
1 •
−1 •
−1 •
1 ◦
0 •
0
0
0
0
0
0
0
0
We now determine those values of λ for which X+λY is a feasible solution,
where


1
12
0
 9+λ 0 2−λ 
.
X + λY = 
 11 − λ 0
λ 
0
0
15
In order to drive down the cost as far as possible, we should make λ as
large as possible, subject to the requirement that all the coefficients of the
4
above matrix should be non-negative numbers. Accordingly we take λ = 2.
Our new basic feasible solution X is then as follows:—


1 12 0
 11 0 0 

X=
 9 0 2 .
0 0 15
We regard X of as the current feasible basic solution.
The cost of the current feasible basic solution X is
8 × 1 + 4 × 12 + 3 × 11 + 13 × 9 + 6 × 2
+ 2 × 15
= 8 + 48 + 33 + 117 + 12 + 30
= 248.
The cost has gone down by 12, as one would expect (the reduction in the
cost being −λq3,3 where λ = 2 and q3,3 = −6).
The current basic feasible solution X is associated with the basis B where
B = {(1, 1), (1, 2), (2, 1), (3, 1), (3, 3), (4, 3)}.
We now compute, for the current feasible basic solution We determine, for
the current basis B values u1 , u2 , u3 , u4 and v1 , v2 , v3 such that ci,j = vj − ui
for all (i, j) ∈ B.
We seek a solution with u1 = 0. We then determine qi,j so that ci,j =
vj − ui + qi,j for all i and j.
We therefore complete the following tableau:—
ci,j & qi,j
1
2
3
4
vj
1
8
2
• 4 •
0
0
3 • 7
0
?
13 • 8
0
?
5
7
?
?
?
?
3
6
ui
0
?
2
?
6 •
0
2 •
0
?
?
?
?
Now u1 = 0, (1, 1) ∈ B and (1, 2) ∈ B force v1 = 8 and v2 = 4.
Then v1 = 8, (2, 1) ∈ B and (3, 1) ∈ B force u2 = 5 and u3 = −5.
5
Then u3 = −5 and (3, 3) ∈ B force v3 = 1.
Then v3 = 1 and (4, 3) ∈ B force u4 = −1.
Computing the numbers qi,j such that ci,j + ui = vj + qi,j , we find that
q1,3 = −1, q2,2 = 8, q2,3 = 6, q3,2 = −1, q4,1 = −4 and q4,2 = −2.
The completed tableau is as follows:—
ci,j & qi,j
1
1
8
2
3
3
13
4
5
vj
8
2
4
•
0
•
0
•
0
•
0
7
3
6
ui
0
−1
2
8
8
6
−1
7
−4
2
−2
4
5
6
•
0
•
0
−5
−1
1
The current basic feasible solution is not optimal because some of the
quantities qi,j are negative. Indeed q1,3 = −1 q3,2 = −1 and q4,1 = −4 and
q4,2 = −2. The most negative of these is q4,1 . We therefore seek to bring
(4, 1) into the basis.
The procedure for achieving this requires us to determine a 3×3 matrix Y
satisfying the following conditions:—
• y4,1 = 1;
• yi,j = 0 when (i, j) 6∈ B ∪ {(4, 1)};
• all rows and columns of the matrix Y sum to zero.
Accordingly we fill in the following tableau with those coefficients yi,j of
the matrix Y that correspond to cells in the current basis (marked with the
• symbol), so that all rows sum to zero and all columns sum to zero:—
yi,j
1
2
3
4
1
?
?
?
1
0
2
• ? •
•
•
◦
0
3
0
0
? • 0
? • 0
0
0
The constraints describe above result in the following completed tableau:—
6
yi,j
1
2
3
4
1
2
3
0 • 0 •
0 •
−1 •
1 •
1
−1 •
0
0
0
0
0
0
0
0
We now determine those values of λ for which X+λY is a feasible solution,
where


1
12
0
 11

0
0

X + λY = 
 9 − λ 0 2 + λ .
0 + λ 0 15 − λ
In order to drive down the cost as far as possible, we should make λ as
large as possible, subject to the requirement that all the coefficients of the
above matrix should be non-negative numbers. Accordingly we take λ = 9.
Our new basic feasible solution X is then as follows:—


1 12 0
 11 0 0 

X=
 0 0 11  .
9 0 6
We regard X of as the current feasible basic solution.
The cost of the current feasible basic solution X is
8 × 1 + 4 × 12 + 3 × 11 + +6 × 11 + 5 ∗ 9
+2×6
= 8 + 48 + 33 + 66 + 45 + 12
= 212.
The cost has gone down by 36, as one would expect (the reduction in the
cost being −λq3,3 where λ = 9 and q3,3 = −4).
The current basic feasible solution X is associated with the basis B where
B = {(1, 1), (1, 2), (2, 1), (3, 3), (4, 1), (4, 3)}.
We now compute, for the current feasible basic solution We determine, for
the current basis B values u1 , u2 , u3 , u4 and v1 , v2 , v3 such that ci,j = vj − ui
for all (i, j) ∈ B.
7
We seek a solution with u1 = 0. We then determine qi,j so that ci,j =
vj − ui + qi,j for all i and j.
We therefore complete the following tableau:—
ci,j & qi,j
1
2
3
4
vj
1
8
2
• 4 •
0
0
3 • 7
0
?
13
8
?
?
5 • 7
0
?
?
?
3
6
ui
0
?
2
?
?
6 •
0
2 •
0
?
?
?
Now u1 = 0, (1, 1) ∈ B and (1, 2) ∈ B force v1 = 8 and v2 = 4.
Then v1 = 8, (2, 1) ∈ B and (4, 1) ∈ B force u2 = 5 and u4 = 3.
Then u4 = 3 and (4, 3) ∈ B force v3 = 5.
Then v3 = 5 and (3, 3) ∈ B force u3 = −1.
Computing the numbers qi,j such that ci,j + ui = vj + qi,j , we find that
q1,3 = 1, q2,2 = 8, q2,3 = 2, q3,1 = 4, q3,2 = 3 and q4,2 = 6.
The completed tableau is as follows:—
ci,j & qi,j
1
2
3
4
vj
1
8
2
• 4 •
0
0
3 • 7
0
8
13
8
4
3
5 • 7
0
6
8
4
3
6
ui
0
1
2
5
2
6 • −1
0
2 • 3
0
5
All numbers qi,j are non-negative for the current feasible basic solution.
This solution is therefore optimal. We conclude that X is an optimal basic
solution, where


1 12 0
 11 0 0 

X=
 0 0 11  .
9 0 6
8
1.2
Direct Solution to Problem 2
The problem is to minimize cT x subject to constraints Ax = b, and x ≥ 0,
where
3 4 7 3 5 4
10
A=
, b=
2 5 3 4 2 6
9
and
cT =
8 4 2 4 3 3
xT =
,
x1 x2 x3 x4 x5 x6
.
We have an initial solution x = (2, 1, 0, 0, 0, 0) with initial basis B = {1, 2}
and initial cost 20. We find p ∈ R2 to satisfy the matrix equation
8 4 = c1 c2 = pT MB ,
where
MB =
3 4
2 5
MB−1
,
1
=
7
5 −4
−2 3
,
and thus
T
p =
c1 c2
MB−1
1
=
7
8 4
5 −4
−2 3
32
7
=
− 20
7
.
Then
T
T
c −p A =
8 4 2 4 3 3
−
8 4 2 4 3 3 −
12
− 99
0 0 − 150
7
7
7
=
=
32
7
− 20
7
8 4
13
164
7
3 4 7 3 5 4
2 5 3 4 2 6
16
120
8
7
7
7
7
Let x be a feasible solution, where Let x = (x1 , x2 , x3 , x4 , x5 , x6 ), Then Ax =
B and xj ≥ 0 for j = 1, 2, 3, . . . , 6. Then
cT x = pT Ax + qT x = pT b + qT x = 28 + qT x
150
12
99
13
= 20 −
x3 + x4 − x5 + x6 .
7
7
7
7
where qT = cT − pT A We look for a basis that includes 3. Now
t1,3
(3)
(1)
(2)
a = t1,3 a + t2,3 a = MB
.
t2,3
Therefore
t1,3
t2,3
=
MB−1 a(3)
1
=
7
5 −4
−2 3
9
7
3
=
23
7
− 57
.
It follows that
2−
23
λ
7
1 + 75 λ λ 0 0 0
is a feasible solution of the problem whenever all components are non-negative.
λ = 0.
We obtain another basic solution on determining λ such that 2 − 23
7
14
We find that λ = 23
and the new basic feasible solution is
14
33
0 0 0
0 23
23
We now let this row vector represent the current basic solution. The current
cost is then 180
and the current basis B is given by B = {2, 3}. Now let MB
23
now consist of the 2rd and 3th columns of the matrix A. We find that
1
3 −7
4 7
−1
.
MB =
, MB = −
5 3
23 −5 4
We then let
T
p =
c2 c3
MB−1
1
=−
23
4 2
3 −7
−5 4
2
− 23
=
20
23
.
Then
T
T
c −p A =
8 4 2 4 3 3
−
2
− 23
20
23
8 4 2 4 3 3 − 34
4 2
23
150
18
39
43
0 0 23 23 − 23
23
=
=
3 4 7 3 5 4
2 5 3 4 2 6
30
112
74
23
23
23
Because the 6th component of the vector q is negative, the current basic
feasible solution is not optimal, and we will get a better solution by bringing
6 into the basis. We therefore look for a basis that includes 6. Now
t2,6
(6)
(2)
(3)
a = t2,6 a + t3,6 a = MB
.
t3,6
Therefore
t2,6
t3,6
=
MB−1 a(6)
1
=−
23
3 −7
−5 4
4
6
=
30
23
4
− 23
.
It follows that
0
33
23
−
30
λ
23
14
23
+
4
λ
23
0 0 λ
is a feasible solution of the problem whenever all components are non-negative.
33
We obtain another basic solution on determining λ such that 23
− 30
λ = 0.
23
11
33
We find that λ = 30 = 10 and the new basic feasible solution is
8
0 0 10
0 0 11
10
10
We now let this row vector represent the current basic solution. The current
and the current basis B is given by B = {3, 6}. Now let MB
cost is then 49
10
now consist of the 3rd and 6th columns of the matrix A. We find that
1
7 4
6 −4
−1
MB =
, MB =
.
3 6
30 −3 7
We then let
T
p =
c3 c6
MB−1
1
=
30
2 3
6 −4
−3 7
=
3
30
13
30
.
Then
T
T
c −p A =
=
=
3
30
13
30
8 4 2 4 3 3 − 35
30
205
43
59
49
0
0
30
30
30
30
77
30
8 4 2 4 3 3
−
3 4 7 3 5 4
2 5 3 4 2 6
41
2 61
3
30
30
Because all components of this last row vector are non-negative, the argument presented at the end of the first iteration now demonstrates that the
current basic feasible solution is optimal.
1.3
Solution to Problem 2 using the Simplex Tableau
Algorithm
The problem is to minimize cT x subject to constraints Ax = b, and x ≥ 0,
where
3 4 7 3 5 4
10
A=
, b=
2 5 3 4 2 6
9
and
cT =
8 4 2 4 3 3
,
xT =
x1 x2 x3 x4 x5 x6
.
We have an initial solution x = (2, 1, 0, 0, 0, 0) with initial basis B = {1, 2}
and initial cost 20. This basis determines a 2 × 2 matrix MB consisting of
columns of A corresponding the basis elements. Because B = {1, 2}, the
matrix MB consists of the first two columns of A. Computing the inverse we
find that
1
3 4
5 −4
−1
MB =
, MB =
.
2 5
7 −2 3
The costs associated with the elements of the basis B constitute a row vector
cB , where
cTB = 8 4 .
11
In order to set up the tableau for the initial basic feasible solution we need
to compute the matrix MB−1 A, the column vector MB−1 b, the initial cost C
where C = cTB MB−1 b and the row vector −qT , where −qT = cTB MB−1 A − cT .
We find that
1
5 −4
3 4 7 3 5 4
−1
MB A =
2 5 3 4 2 6
7 −2 3
!
1
17
4
1 0 23
−
−
7
7
7
17
=
,
0 1 − 57 76 − 47 10
7
1
5 −4
10
2
−1
=
,
MB b =
9
1
7 −2 3
2
−1
T
8 4
C = cB MB b =
= 20
1
!
23
1
17
4
1
0
−
−
7
7
7
17
8 4
−qT = cTB MB−1 A − cT =
0 1 − 75 67 − 74 10
7
− 8 4 2 4 3 3
16
120
8
8 4 164
=
− 8 4 2 4 3 3
7
7
7
17
99
0 0 150
− 12
− 13
=
7
7
7
17
The tableau corresponding to the initial basic solution is thus as follows:—
(1)
a
a(2)
a(1)
1
0
a(2)
0
1
0
0
a(3)
23
7
− 57
150
7
a(4)
− 71
6
7
− 12
7
a(5)
17
7
− 47
99
7
a(6)
− 47
10
7
− 13
7
b
2
1
20
The initial basic feasible solution is not optimal because the components
in the criterion row in the first six columns are not all non-positive. We could
get a solution with lower cost by bringing a(3) or a(5) in to the basis.
Suppose that we were to bring a(3) into the basis. Then the third column
would be our pivot column. We must then select the pivot row so that the
pivot element is strictly positive and such that the ratio of the coefficient in
the b column to the corresponding element in the pivot column is minimized
subject to the corresponding coefficient in the pivot column being positive.
In this instance there would only be one positive coefficient in the pivot row,
and we would choose the a(1) row as the pivot row, and the cost would be
reduced by 2 × 150
÷ 23
which has value 300
.
7
7
23
(5)
Suppose we were to bring a into the basis. Then the fifth column would
be our pivot column, and we would need to select the a(1) row as the pivot
12
row in order to ensure that the pivot element is positive. The cost reduction
9
9
which has value 17
. Now 300
> 17
. Therefore we get the
would be 1 × 97 ÷ 17
7
23
(3)
greatest cost reduction by bringing a into the basis in place of a(1) .
Thus the pivot column is the a(3) column, the pivot row is the a(1) row.
We transform the tableau by dividing elements of the pivot row by the
pivot element and subtracting appropriate multiples of the pivot row from
the other rows so as to clear the elements in the pivot column other than the
pivot element. We obtain the following tableau:—
a(1)
(3)
a
a(2)
7
23
5
23
− 150
23
a(2)
0
1
a(3)
1
0
0
0
a(4)
1
− 23
19
23
− 18
23
a(5)
17
23
1
− 23
39
− 23
a(6)
4
− 23
30
23
43
23
b
14
23
33
23
160
23
The coefficient in the criterion row in the a(6) column is positive. It follows
that the basic feasible solution represented by this tableau is not optimal, and
that we would lower the cost by bringing a(6) into the basis. This column will
be our pivot column for the next iteration. The pivot element in the pivot
column and pivot row must be positive, and therefore the pivot row is the
. As usual we transform the tableau
a(2) row, and the new pivot element is 30
23
by dividing elements of the pivot row by the pivot element and subtracting
appropriate multiples of the pivot row from the other rows so as to clear the
coefficients occuring in the pivot column.
We obtain the following tableau:—
a(3)
a(6)
a(1)
a(2)
10
30
5
30
− 205
30
4
30
23
30
− 43
30
a(3)
1
0
0
a(4)
a(5)
2
30
19
30
59
− 30
22
30
1
− 30
− 92
30
a(6)
0
1
0
b
24
30
33
30
147
30
The quantities in the criterion row in first six columns are all nonpositive and therefore this tableau represents an optimal solution.
The optimal solution x satisfies
33
8
0 0 30
0 0 11
xT = 0 0 24
= 0 0 10
.
30
10
1.4
Solution to Problem 2 using the Revised Simplex
Algorithm
The problem is to minimize cT x subject to constraints Ax = b, and x ≥ 0,
where
10
3 4 7 3 5 4
A=
, b=
2 5 3 4 2 6
9
13
and
cT =
8 4 2 4 3 3
,
xT =
x1 x2 x3 x4 x5 x6
.
We have an initial solution x = (2, 1, 0, 0, 0, 0) with initial basis B = {1, 2}
and initial cost 20. This basis determines a 2 × 2 matrix MB consisting of
columns of A corresponding the basis elements. Because B = {1, 2}, the
matrix MB consists of the first two columns of A. Computing the inverse we
find that
1
5 −4
3 4
−1
.
MB =
, MB =
2 5
7 −2 3
The costs associated with the elements of the basis B constitute a row vector
cB , where
cTB = 8 4 .
We calculate the components of the 2-dimensional vector p that satisfies
pT = cTB MB−1 and also the cost C of the initial basic solution where C = bT b.
Now
−1 1
5 −4
T
8 4
− 20
p = c1 c2 MB =
= 32
7
7
−2 3
7
× 10 − 20
×9 =
and C = 32
7
7
solution is as follows:
a(1)
a(2)
b e(1)
5
10
7
9 − 27
20
32
7
140
7
= 20. The tableau for the initial basic
e(2)
− 47
3
7
− 20
7
We next compute components of the row vector −qT , where −qT = cTB MB−1 A−
cT , looking for one that is positive. Denoting the jth component of this row
vector by qj , we are guaranteed that q1 = 0 and q2 = 0, since these correspond to elements of the initial basis. We find −q3 = 150
. This number is
7
positive, so we bring the third column a(3) of the matrix A into the basis. The
entries in the pivot column that we add to the tableau are the components
of the vector MB−1 a(3) , where
5
23 − 47
7
−1 (3)
7
7
MB a =
=
.
2
3
3
−7 7
− 57
Accordingly we bring the pivot column determined by a(3) into the tableau
to obtain the following tableau:—
14
a(3)
a(1)
a(2)
23
7
− 57
150
7
b
2
1
e(1)
20
32
7
5
7
− 27
e(2)
− 74
3
7
− 20
7
Because the entry in the pivot column in the a(1) row is the only positive
element in the basis rows of the pivot column, this must be our pivot element,
and therefore the pivot row is the first row of the tableau. We therefore
transform the tableau by dividing elements of the pivot row by the pivot
element, and subtracting appropriate multiplies of the pivot row from the
other rows to clear the coefficients in those rows of the pivot column. We
obtain the following tableau:—
a(3)
a(2)
a(3)
1
0
0
b
e(1)
14
23
33
23
160
23
5
23
3
− 23
2
− 23
e(2)
4
− 23
7
23
20
23
We next compute components of the row vector −qT , where −qT = cTB MB−1 A−
cT , looking for one that is positive. We are guaranteed that q2 = 0 and q3 = 0,
because the current basis consists of the vectors 2 and 3. Calculating other
, −q4 = − 18
, −q5 = − 39
, −q6 = 43
. Now −q6 is
values we find −q1 = − 150
23
23
23
23
(6)
positive. Accordingly we should bring a into the tableau. Computing the
new pivot column, we obtain the following tableau:—
a(3)
a(2)
a(6)
4
− 23
30
23
43
23
b
e(1)
14
23
33
23
160
23
5
23
3
− 23
2
− 23
e(2)
4
− 23
7
23
20
23
The new pivot row must be the a(2) row, since this is the only row with
a positive coefficient in the pivot column. Accordingly we transform the
23
, adding elements of
tableau by multiplying elements of the pivot row by 30
4
(3)
the pivot row multiplied by 30 to the a row, and subtracting elements of
the pivot row multiplied by 43
to obtain the following tableau:—
30
a(3)
a(2)
a(6)
0
1
0
b
e(1)
24
30
33
30
147
30
6
30
3
− 30
3
30
e(2)
4
− 30
7
30
13
30
15
We next compute components of the row vector −qT , where −qT = cTB MB−1 A−
cT , looking for one that is positive. We are guaranteed that q3 = 0 and
, −q2 = − 43
q6 = 0, because the current basis is {3, 6}. We find −q1 = − 205
30
30
59
49
−q4 = − 30 , −q5 = − 30 . These values are all non-positive (and thus the
components of the vector q are all non-negative.
It follows that we have now found an optimal solution, which is
11
8
0, 0, 10
.
, 0, 0, 10
16