10.1 lecture_2d

8/9/2019 10.1 Lecture_2d

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Lecture 2d:

Integer Linear Programming

Jeff Chak-Fu WONG

Department of Mathematics

Chinese University of Hong Kong

MAT581SS

Mathematics for Logistics

Produced by Jeff Chak-Fu WONG 1

8/9/2019 10.1 Lecture_2d


TABLE OF CONTENTS

1. Land and Doig’s branch and bound method

BLE OF CONTENTS 2

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LAND AND DOIG’S BRANCH AND BOUND METHOD

AND AND DOIG’S BRANCH AND BOUND METHOD 3

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Branch and Bound is an efficient enumerative technique for

examining all possible integer feasible points.

This can be used to solve both, all as well as mixed, types of integer

programming problems.


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Let the given ILP be

max z =

n

j=1

cjxj

subject ton

j=1 aijxj = bi (i = 1, · · · , m)

xj ≥ 0 ( j = 1, · · · , n) (1)

and

xj integer for J 1 ⊂ J,

where J = {1, 2, · · · , n}.


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As the name suggests, each iteration consists of branching and

bounding.

1. The branching is done from the current node, a node being

identified as a LPP, where the initial node is the associated LLP of

the given problem.

2. Certain appropriate bounds are calculated so as to decide

if further branching is needed or not.

Though it is an enumerative technique, it is efficient in the sense thatthe exhaustive enumeration is seldom needed.


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Most of the times it is only partial enumeration as that node which

cannot further improve the current available solution is discarded

and therefore the corresponding region is no more required for

enumeration.


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The following are the details of the Branch and Bound method.

Step 1

Solve the associated LPP, call (LP )1, and let z1 be its optimal value.

• If the optimal solution of (LP )1 has integer components for

j ∈ J 1, i.e., it meets the integer requirements of the given ILP,stop; otherwise go to Step 2.

• Thus in Step 1, either we stop and get an optimal solution of the

given ILP or we have an upper bound z1 for the optimal

objective function value of the ILP.


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Step 2

• Select a variable xj which is constrained to be integer but is

currently having a fractional value β j in the optimal (LP )1

solution;

• Construct the following two LP’s:

(LP )2 (LP )3

max C

T X

max CT X

subject to subject to

AX = b AX = b

X ≥ 0 X ≥ 0

xj ≤ [β j ] xj ≥< β j >

Here

• [β j ] is the greatest integer less than or equal to β j ;

• < β j > is the smallest integer more than or equal to β j .


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Obviously

• [β j ] is the nearest integer in the left of β j ;

• < β j > is the nearest integer in the right of β j .

Example 1

If β j = 1.6, then [β j ] = 1 and < β j >= 2.

Sometimes [β j ] = 1 and < β j >= 2 are also called, respectively, the floor

and ceiling of β j


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• If we decide to identify problem (LP )k by the kth

node, thenthe above process is to branch from node 1 via the variable xj

to get two new nodes, node 2 and node 3.

• The new restrictions xj ≤ [β j ] and xj ≥< β j >, are mutually

exclusive so that (LP )2 and (LP )3 are dealt with separate LP’s.

• The optimal solution of the given ILP lies either in (LP )2 or (LP )3,

i.e., in one of the two branches as shown in Figure 1.

node 1

node 2 node 3

(LP)

(LP) (LP)

x <= [ ] x => < >

2 3

1

Optimal solution of

z optimal value of

(LP)

(LP)

1

11

j β β j j j

Figure 1:


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Step 3

• Obtain an optimal solution of (LP )2 and (LP )3.

• Clearly (LP )2 and (LP )3 should be solved by the dual simplex

method as both of these LP’s differ from (LP )1 with regard to an

additional constraint only.

• Suppose that optimal solutions of (LP )2 and (LP )3 still do not

meet the integer requirements of the given ILP.


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Step 4

• Select either (LP )2 or (LP )3 for further branching.

• Branch from these by adding a new constraint corresponding to a

variable which is constrained to be integer but is having a

fractional value.


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Step 5

• This process of branching and solving a sequence of LP’s is

continued until a solution is obtained for one of the LP’s which

meets the integer requirements, namely xj integer for j ∈ J 1.

• Let this LPP be (LP )k and zk denote its optimal value.

• Then zk becomes a lower bound for the optimal objective

function value of the given ILP.


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At this point, we can eliminate from consideration all those LP’s(nodes) which have the objective function value z less than this

lower bound, namely zk.

We say that all such nodes have been fathomed because it is not

possible to find a better solution (lower bound) from these LP’s than

what we have now.

• This is because, branching from a LPP means adding one more

constraint and hence further reducing its feasible region.


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Step 6

• Continue selecting a node (i.e. LP) and branching till all the

nodes have been fathomed.

• The fathomed node with the largest value of z will give the

optimal solution of the given ILP.


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Example 2 Use branch and bound method to solve the following ILP

max z = 7x1 + 9x2

subject to

−x1 + 3x2 ≤ 6

7x1

+ x2

≤ 35x1 ≤ 7

x2 ≤ 7

x1, x2 ≥ 0

x1 and x2 integer.


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Solution

• The first step is to solve the associated LPP, (LP )1.


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cj 7 9 0 0 0 0

x1 x2 ↓ x3 x4 x5 x6 θ

0 x3 ← 6 -1 3 1 0 0 0 6

3 = 2 1

3 × R1

0 x4 35 7 1 0 1 0 0 35/10 x5 7 1 0 0 0 1 0 7/1

0 x6 7 0 7 0 0 0 1 7/1

z(XB) = 0 -7 -9 0 0 0 0

cj 7 9 0 0 0 0

x1 x2 ↓ x3 x4 x5 x6 θ

9 x2 2 -1/3 1 1/3 0 0 0 6

3 = 2

0 x4 35 7 1 0 1 0 0 R2 − R1

0 x5 7 1 0 0 0 1 0

0 x6 7 0 7 0 0 0 1 R4 − 7 × R1

z(XB) = 0 -7 -9 0 0 0 0 R5 + 9 × R1


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cj 7 9 0 0 0 0

x1 ↓ x2 x3 x4 x5 x6 θ

9 x2 2 -1/3 1 1/3 0 0 0

0 x

4 ← 33 22/3 0 -1/3 1 0 0 33

22/3 = 4.5

3

22

×R2

0 x5 7 1 0 0 0 1 0 7/1

0 x6 5 1/3 0 1/3 0 0 1 5/(1/3) = 15

z(XB) = 18 -10 0 3 0 0 0

cj 7 9 0 0 0 0

x1 ↓ x2 x3 x4 x5 x6 θ

9 x2 2 -1/3 1 1/3 0 0 0 R1 + 1

3 × R2

7 x1 9/2 1 0 -1/22 3/22 0 0 33

22/3 = 4.50 x5 7 1 0 0 0 1 0 R3 − R2

0 x6 5 1/3 0 1/3 0 0 1 R4 − 1

3 × R2

z(XB) = 18 -10 0 3 0 0 0 R5 + 10 × R2


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cj 7 9 0 0 0 0

x1 x2 x3 x4 x5 x6 θ

9 x2 7/2 0 1 7/22 1/22 0 0

7 x1 9/2 1 0 -1/22 3/22 0 0

0 x5 5/2 0 0 1/22 -3/22 1 0

0 x6 7/2 0 0 -7/22 -1/22 0 1

z(XB) = 63 0 0 28/11 15/11 0 0

• The optimal solution of (LP )1 is (x1 = 9/2, x2 = 7/2) and the

optimal value is z1 = 63.

• As the solution of (LP )1 does not meet the integer

requirements, it is not optimal for the given ILP.

• However, z1 = 63 is an upper bound for the optimal objective

function value of the given ILP problem (LP )1 is depicted as

node 1 in Fig. 2.


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node 1

( LP ) 1

x = 9/2, x = 7/2

z = 63 ( upper bound )

1 2

1

Figure 2: (LP )1


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• The next step of the branch and bound method is to branch

from node 1 via a variable xj which is constrained to be integer

but is currently having a fractional value.

• In our example, x1 and x2 both are constrained to be integers

and both are having a fractional value so we may branch from

node 1 via x1 or x2.

• In practice, we choose that j for which the fractional part of

such xj is positive most, with the hope that this way we may get

a deeper cut.

• Here both x1 and x2 have equal fractional part so we can

branch either from x1 or x2.


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• – If we decide to branch from x1,– then for β 1 = 9/2, [β 1] = 4 and < β 1 >= 5

and therefore we get two new LP’s, (LP )2 and (LP )3, as

(LP )2 (LP )3

Same a (LP)1 but with Same as (LP)1 but with

one additional constraint one additonal constraint

x1

≤ 4. x1

≥ 5.

• These are identified as node 2 and node 3.

– Solving (LP )2 we get (x1 = 4, x2 = 10/3) and z2 = 58.

– Similarly solving (LP )3 we get (x1 = 5, x2 = 0) and z3 = 35.

• As the solution of (LP )3 meets the integer requirements, the

value z3 = 35 becomes a lower bound for the optimal objective

function value of the given ILP.


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node 1

node 2 node 3

( LP )

( LP ) ( LP )

1

2 3

x = 9/2, x = 7/2


1 2

1

x = 4, x = 10/3

z = 582

21

x <= 4 x => 51 1

Figure 3: (LP )2


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node 1

node 2 node 3

( LP )

( LP ) ( LP )

1

2 3

x = 9/2, x = 7/2


1 2

1

x = 4, x = 10/3

z = 58

x = 5, x = 0

z = 35 ( lower bound )2

21

3

1 2

x <= 4 x => 51 1

Figure 4: (LP )3


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(LP )2:

Consider

x1 ≤ 4 (2)

Putting

x1 = 122

x3 − 322

x4 + 92

into (2), and introducing the Gomorian slack, we have

1

22 x3 −

3

22 x4 + x7 = −

1

2 .


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cj 7 9 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7

9 x2 7/2 0 1 7/22 1/22 0 0 0

7 x1 9/2 1 0 -1/22 3/22 0 0 0

0 x5 5/2 0 0 1/22 -3/22 1 0 0

0 x

6 7/2 0 0 -7/22 -1/22 0 1 00 ← x7 -1/2 0 0 1/22 -3/22 0 0 1 − 22

3 ×R5

z(XB) = 63 0 0 28/11 15/11 0 0 0

↑

15/11−3/22


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cj 7 9 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7

9 x2 7/2 0 1 7/22 1/22 0 0 0 R1 − 122 ×R5

7 x1 9/2 1 0 -1/22 3/22 0 0 0 R2 − 322 ×R5

0 x

5 5/2 0 0 1/22 -3/22 1 0 0 R

3 +

3

22 ×R

5

0 x6 7/2 0 0 -7/22 -1/22 0 1 0 R4 + 122 ×R5

0 x4 11/3 0 0 -1/3 1 0 0 -22/3

z(XB) = 63 0 0 28/11 15/11 0 0 0 R6 − 1511 ×R5

↑

cj 7 9 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7

9 x2 10/3 0 1 1/3 0 0 0 1/3

7 x1 4 1 0 0 0 0 0 1

0 x5 3 0 0 0 0 1 0 -1

0 x6 11/3 0 0 -1/3 0 0 1 -1/3

0 x4 11/3 0 0 -1/3 1 0 0 -22/3

z(XB) = 58 0 0 3 0 0 0 10


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(LP )3:

Consider

x1 ≥ 5 (3)

Putting

x1 = 1

22 x3 − 3

22 x4 + 9

2

into (3), multiplying −1 on the both sides of (3), and introducing the

Gomorian slack, we have

− 122 x3 + 322 x4 + x7 = − 12 .


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cj 7 9 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7

9 x2 7/2 0 1 7/22 1/22 0 0 0

7 x1 9/2 1 0 -1/22 3/22 0 0 0

0 x5 5/2 0 0 1/22 -3/22 1 0 0

0 x6 7/2 0 0 -7/22 -1/22 0 1 00 ← x7 -1/2 0 0 -1/22 3/22 0 0 1 −22 ×R5

z(XB) = 63 0 0 28/11 15/11 0 0 0

↑

28/11−

1/22


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cj 7 9 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7

9 x2 7/2 0 1 7/22 1/22 0 0 0 R1 − 722 ×R5

7 x1 9/2 1 0 -1/22 3/22 0 0 0 R2 + 122 ×R5

0 x

5 5/2 0 0 1/22 -3/22 1 0 0 R

3 −

1

22 ×R

5

0 x6 7/2 0 0 -7/22 -1/22 0 1 0 R4 + 722 ×R5

0 x3 11 0 0 1 -3 0 0 -22

z(XB) = 63 0 0 28/11 15/11 0 0 0 R6 − 28

11 ×R5

↑

cj 7 9 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7

9 x2 0 0 1 0 1 0 0 7

7 x1 5 1 0 0 0 0 0 -1

0 x5 2 0 0 0 0 1 0 1

0 x6 7 0 0 0 -1 0 1 -7

0 x3 11 0 0 1 -3 0 0 -22

z(XB) = 35 0 0 0 9 0 0 56


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At this stage we check the objective function value of other LP’s.

In our case, there is only one node to be checked, i.e. node 2.

node 1

node 2 node 3

node 4 node 5

( LP )

( LP ) ( LP )

( LP ) ( LP )

1

2 3

4 5

x = 9/2, x = 7/2


1 2

1

x = 4, x = 10/3

z = 58

x = 5, x = 0


21

3

1 2

x <= 3 x => 4

x <= 4 x => 51 1

2 2

Figure 5: (LP )4 and (LP )5


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As the objective function value of (LP )2 is more than this lower

bound z3 = 35, this node is not fathomed and it is considered for

further branching.Had the value of (LP )2 been less than the lower bound z3 = 35, it

would have been a fathomed node and (x1 = 5, x2 = 0) would

have been taken as an optimal solution of the given ILP.

Branching from node 2, via x2 we get two LP’s, namely (LP )4 and

(LP )5.

• Problem (LP )5 is infeasible.

• An optimal solution of (LP )4 is (x1 = 4, x2 = 3) with the objective

function value z4 = 55.


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(LP )4:

Consider

x2 ≤ 3. (4)

Putting

x2 = − 13

x3 − 13

x7 + 103


−

1

3 x3 −

1

3 x7 + x8 = −

1

3 .


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cj 7 9 0 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7 x8

9 x2 10/3 0 1 1/3 0 0 0 1/3 0

7 x1 4 1 0 0 0 0 0 1 0

0 x5 3 0 0 0 0 1 0 -1 0

0 x6 11/3 0 0 -1/3 0 0 1 -1/3 00 x4 11/3 0 0 -1/3 1 0 0 -22/3 0

0 ← x8 -1/3 0 0 -1/3 0 0 0 -1/3 1 −3 × R6

z(XB) = 58 0 0 3 0 0 0 10 0

↑

3−1/3

= −9 10−1/3

= −30

cj 7 9 0 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7 x8

9 x2 10/3 0 1 1/3 0 0 0 1/3 0 R1 − 13 × R6

7 x1 4 1 0 0 0 0 0 1 0

0 x5 3 0 0 0 0 1 0 -1 0

0 x6 11/3 0 0 -1/3 0 0 1 -1/3 0 R4 + 13 × R6

0 x4 11/3 0 0 -1/3 1 0 0 -22/3 0 R5 + 13 × R6

0 x3 1 0 0 1 0 0 0 1 -3

z(XB) = 58 0 0 3 0 0 0 10 0 R7 − 3 × R6

↑


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cj 7 9 0 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7 x8

9 x2 3 0 1 0 0 0 0 0 1

7 x1 4 1 0 0 0 0 0 1 0

0 x5 3 0 0 0 0 1 0 -1 0

0 x6 4 0 0 0 0 0 1 0 -10 x4 4 0 0 0 1 0 0 -7 -1

0 x3 1 0 0 1 0 0 0 1 -3

z(XB) = 55 0 0 3 0 0 0 7 9

↑


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(LP )5:

Consider

x2 ≥ 4. (5)

Putting

x2 = − 13

x3 − 13

x7 + 103



13

x3 + 13

x7 + x8 = −53

.


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cj 7 9 0 0 0 0 0 0

x1 x2 x3 x4 x5 x6 x7 x8

9 x2 10/3 0 1 1/3 0 0 0 1/3 0

7 x1 4 1 0 0 0 0 0 1 0

0 x5 3 0 0 0 0 1 0 -1 0

0 x6 11/3 0 0 -1/3 0 0 1 -1/3 0

0 x4 11/3 0 0 -1/3 1 0 0 -22/3 0

0 ← x8 -5/3 0 0 1/3 0 0 0 1/3 1

z(XB) = 58 0 0 3 0 0 0 10 0

? ?


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Now all nodes have been fathomed and so we have to choose the

fathomed node with the maximum objective function value.

This gives (x∗1 = 4, x∗2 = 3) as an optimal solution of the given ILP and

the objective function value z∗ as z∗ = 55.


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See the above details in Fig 6.

node 1

node 2 node 3

node 4 node 5 infeasible

( LP )

( LP ) ( LP )

( LP ) ( LP )

1

2 3

4 5

x = 9/2, x = 7/2


1 2

1

x = 4, x = 10/3

z = 58

x = 5, x = 0

z = 35 ( lower bound )

2

21

3

1 2

z = 55

x = 4, x = 31 2

4

x <= 3 x => 4

x <= 4 x => 51 1

2 2

Figure 6:


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7 x + x = 35

x

x

x = 7

- x + 3 x = 6

x = 72

21

1 2

1

1

2

(9/2, 7/2)

Figure 7: (LP )1


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7 x + x = 35

x

x

x = 7

- x + 3 x = 6

x = 72

21

1 2

1

1

2

(9/2, 7/2)

x => 5x <= 4 11

(4, 10/3)

(5,0)



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7 x + x = 35

x

x

x = 7

- x + 3 x = 6

x = 72

21

1 2

1

1

2

(9/2, 7/2)

x => 5x <= 4 11

(4, 10/3)

(5,0)

x => 4

x <= 3

2

2(4, 3)



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Example 3 Use branch and bound method to solve the following integer

LPP

max z = 3x1 + 4x2

subject to

7x1 + 16x2 ≤ 523x1 − 2x2 ≤ 9

x1, x2 ≥ 0

x1 and x2 integer.


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Solution

cj : 3 4 0 0

CB XB x1 x2 ↓ x3 x4 θ

0 x3 ← 52 7 16 1 0 52

16 = 13

4

1

16 × R1

0 x4 9 3 -2 0 1

zj − cj z(XB) = 0 -3 -4 0 0

cj : 3 4 0 0

CB XB x1 x2 ↓ x3 x4 θ

4 x2 13/4 7/16 1 1/16 0 52

16 = 13

4

0 x4 9 3 -2 0 1 R2 + 2× R1

zj − cj z(XB) = 0 -3 -4 0 0 R3 + 4× R1


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cj : 3 4 0 0

CB XB x1 ↓ x2 x3 x4 θ

4 x2 13/4 7/16 1 1/16 0 13

/4

7/16 = 52/7

0 x4 ← 31/2 31/8 0 0 1 31/231/8 = 4 8

31 × R2

zj − cj z(XB) = 13 -5/4 0 1/4 0

cj : 3 4 0 0

CB XB x1 ↓ x2 x3 x4 θ

4 x2 13/4 7/16 1 1/16 0 R1 − 7

16 × R2

3 x1 4 1 0 1/31 8/31 31

/2

31/8 = 4

zj − cj z(XB) = 13 -5/4 0 1/4 0 R3 + 5

4 × R2


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cj : 3 4 0 0

CB XB x1 x2 x3 x4 θ

4 x2 3/2 0 1 3/62 -7/62

3 x1 4 1 0 1/31 8/31

zj − cj z(XB) = 18 0 0 9/31 10/31


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node 1

node 2 node 3

( LP )( LP )

x <= 1 x => 2

2 3

( LP ) 1

x = 4, x = 3/2

z = 18 ( upper bound )1

1 2

2 2



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(LP )2:

Consider

x2 ≤ 1 (6)

Putting

x2 = − 316

x3 + 762

x4 + 32

into (6) and introducing the Gomorian slack, we have

− 3

16

x3 + 7

62

x4 + x5 = −1

2

.


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cj : 3 4 0 0 0CB XB x1 x2 x3 x4 x5

4 x2 3/2 0 1 3/62 -7/62 0

3 x1 4 1 0 1/31 8/31 0

0 ← x5 -1/2 0 0 -3/62 7/62 1 −16

3 × R3

zj − cj z(XB) = 18 0 0 9/31 10/31 0

↑

9/31−3/62


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cj : 3 4 0 0 0

CB XB x1 x2 x3 x4 x5

4 x2 3/2 0 1 3/62 -7/62 0 R1 − 3

62 × R3

3 x1 4 1 0 1/31 8/31 0 R2 − 1

31 × R3

0 x3 31/3 0 0 1 -7/3 -62/3

zj − cj z(XB) = 18 0 0 9/31 10/31 0 R4 − 9

31 × R3

↑

cj : 3 4 0 0 0


4 x2 1 0 1 0 0 1

3 x1 11/3 1 0 0 1/3 2/30 x3 31/3 0 0 1 -7/3 -62/3

zj − cj z(XB) = 15 0 0 0 1 6


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(LP )3:

Consider

x2 ≥ 2 (7)

Putting

x2 = − 3

62 x3 + 7

62 x4 + 3

2



3

62 x3 −

7

62 x4 + x5 = −

1

2 .


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cj : 3 4 0 0 0CB XB x1 x2 x3 x4 x5

4 x2 3/2 0 1 3/62 -7/62 0

3 x1 4 1 0 1/31 8/31 0

0 ← x5 -1/2 0 0 3/62 -7/62 1 −62

7 × R3

zj − cj z(XB) = 18 0 0 9/31 10/31 0

↑

10/31−7/62


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cj : 3 4 0 0 0


4 x2 3/2 0 1 3/62 -7/62 0 R1 + 7

62 × R3

3 x1 4 1 0 1/31 8/31 0 R2 − 8

31 × R3

0 x4 31/7 0 0 -3/7 1 -62/7

zj − cj z(XB) = 18 0 0 9/31 10/31 0 R4 − 10

31 × R3

↑

cj : 3 4 0 0 0


4 x2 2 0 1 0 0 -1

3 x1 20/7 1 0 1/7 0 16/70 x4 31/7 0 0 -3/7 1 -62/7

zj − cj z(XB) = 116/7 0 0 3/7 0 20/7


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node 1

node 2 node 3

node 5node 4

( LP )

( LP )

( LP )

( LP )

x <= 1 x => 2

x <= 3 x => 4

2 3

4 5

( LP ) 1

x = 4, x = 3/2


1

z = 15

x = 11/3. x = 1 x = 20/7, x = 2

z = 116/72 3

1 12

2

2

2 2

1 1



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(LP )4:

Consider

x1 ≤ 3 (8)

Putting

x1 = −13

x4 − 23

x5 + 113


−1

3

x4 − 2

3

x5 + x6 = −2

3

.


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0

CB XB x1 x2 x3 x4 x5 x6

4 x2 1 0 1 0 0 1 0

3 x1 11/3 1 0 0 1/3 2/3 0

0 x3 31/3 0 0 1 -7/3 -62/3 0

0 ← x6 -2/3 0 0 0 -1/3 -2/3 1 −3 × R4

zj − cj z(XB) = 15 0 0 0 1 6 0

↑

1

−1/3

= −3 6

−2/3


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0


4 x2 1 0 1 0 0 1 0

3 x1 11/3 1 0 0 1/3 2/3 0 R2 − 1

3 × R4

0 x3 31/3 0 0 1 -7/3 -62/3 0 R3 + 7

3 × R4

0 x4 2 0 0 0 1 2 -3 −3 × R4

zj − cj z(XB) = 15 0 0 0 1 6 0 R5 − R4

↑

cj : 3 4 0 0 0 0


4 x2 1 0 1 0 0 1 0

3 x1 3 1 0 0 0 0 10 x3 15 0 0 1 0 -16 -7

0 x4 2 0 0 0 1 2 -3

zj − cj z(XB) = 13 0 0 0 0 4 3


8/9/2019 10.1 Lecture_2d


(LP )5:

Consider

x1 ≥ 4 (9)

Putting

x1 = −

1

3 x4 − 2

3 x5 + 11

3



1

3 x4 +

2

3 x5 + x6 = −

1

3 .


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0


4 x2 1 0 1 0 0 1 0

3 x1 11/3 1 0 0 1/3 2/3 0

0 x3 31/3 0 0 1 -7/3 -62/3 0

0 ← x6 -1/3 0 0 0 1/3 2/3 1

zj − cj z(XB) = 15 0 0 0 1 6 0

? ?


8/9/2019 10.1 Lecture_2d


node 1

node 2 node 3

node 6 node 7node 5node 4

( LP ) ( LP )( LP )

( LP )

( LP )

( LP )

x <= 1 x => 2

x <= 2 x => 3x <= 3 x => 4

2 3

4 5 6 7

( LP ) 1

x = 4, x = 3/2


1

z = 15

x = 11/3. x = 1 x = 20/7, x = 2

z = 116/7

x = 3, x = 1

2 3

1 1

1

2

2

2

2

2 2

1 1 1 1


infeasible



8/9/2019 10.1 Lecture_2d


(LP )6:

Consider

x1 ≤ 2 (10)

Putting

x1 = − 13

x3 − 177

x5 + 207


−1

3

x3 − 17

7

x5 + x6 = −6

7

.


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0


4 x2 2 0 1 0 0 -1 0

3 x1 20/7 1 0 1/7 0 16/7 0

0 x4 31/7 0 0 -3/7 1 -62/3 0

0 ← x6 -6/7 0 0 -1/7 0 -16/7 1 − 716 × R4

zj − cj z(XB) = 116/7 0 0 3/7 0 20/7 0

↑

3/7−1

/7

= −3 20/7−16

/7


8/9/2019 10.1 Lecture_2d


cj

: 3 4 0 0 0 0


4 x2 2 0 1 0 0 -1 0 R1 + ×R4

3 x1 20/7 1 0 1/7 0 16/7 0 R2 − 16

7 × R4

0 x4 31/7 0 0 -3/7 1 -62/3 0 R3 + 62

7 × R4

0 x5 3/8 0 0 1/16 0 1 -7/16

zj − cj z(XB) = 116/7 0 0 3/7 0 20/7 0 R5 − 20

7 × R4

↑

cj : 3 4 0 0 0 0


4 x2 19/8 0 1 1/16 0 0 -7/16

3 x1 2 1 0 0 0 0 10 x4 31/4 0 0 1/8 1 0 -31/8

0 x5 3/8 0 0 1/16 0 1 -7/16

zj − cj z(XB) = 31/2 0 0 1/84 0 0 5/4


8/9/2019 10.1 Lecture_2d


(LP )7:

Consider

x1 ≥ 3 (11)

Putting

x1 = −

1

3 x3 −

17

7 x5 +

20

7



1

3 x3

+

17

7 x5

+ x6

=−

1

7 .


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0


4 x2 2 0 1 0 0 -1 0

3 x1 20/7 1 0 1/7 0 16/7 0

0 x4 31/7 0 0 -3/7 1 -62/3 0

0 ← x6 -1/7 0 0 1/3 0 16/7 1

zj − cj z(XB) = 116/7 0 0 3/7 0 20/7 0

? ?


8/9/2019 10.1 Lecture_2d


node 1

node 2 node 3

node 9node 8


( LP )

( LP ) ( LP )

( LP )( LP )

( LP )

( LP )

( LP )

x <= 1 x => 2

x <= 2 x => 3

x <= 2 x => 3

x <= 3 x => 4

2 3

4 5 6 7

8

( LP ) 1

9

x = 4, x = 3/2


1

z = 15

x = 11/3. x = 1 x = 20/7, x = 2

z = 116/7

z = 31/2

x = 2, x = 19/8

infeasiblex = 3, x = 1

6

2 3

1 1

1

1

2

2

2

2

2

2 2

1 1 1 1

2 2


infeasible



8/9/2019 10.1 Lecture_2d


(LP )8:

Consider

x2 ≤ 2 (12)

Putting

x2 = − 116

x3 + 716

x6 + 198


− 1

16

x3 + 7

16

x6 + x7 = −3

8

.


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0 0

CB XB x1 x2 x3 x4 x5 x6 x7

4 x2 19/8 0 1 1/16 0 0 -7/16 0

3 x1 2 1 0 0 0 0 1 0

0 x4 31/4 0 0 1/8 1 0 -31/8 0

0 x5

3/8 0 0 1/16 0 1 -7/16 00 ← x7 -3/8 0 0 -1/16 0 0 7/16 1 −16 × R5

zj − cj z(XB) = 31/2 0 0 1/4 0 0 5/4 0

↑

1/4−1/16


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0 0


4 x2 19/8 0 1 1/16 0 0 -7/16 0 R1 − 1

16 × R5

3 x1 2 1 0 0 0 0 1 0

0 x4 31/4 0 0 1/8 1 0 -31/8 0 R3 − 1

8 × R5

0 x5 3/8 0 0 1/16 0 1 -7/16 0 R4 − 116 × R5

0 ← x7 6 0 0 1 0 0 -7 -16

zj − cj z(XB) = 31/2 0 0 1/4 0 0 5/4 0 R6 − 1

4 × R5

↑


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0 0


4 x2 2 0 1 0 0 0 0 1

3 x1 2 1 0 0 0 0 1 0

0 x4 7 0 0 0 1 0 -3 20 x5 0 0 0 0 0 1 0 1

0 x3 6 0 0 1 0 0 -7 -16

zj − cj z(XB) = 14 0 0 0 0 0 3 4


8/9/2019 10.1 Lecture_2d


(LP )9:

Consider

x2 ≥ 3 (13)

Putting

x2 = −

1

16 x3 +

7

16 x6 +

19

8



1

16x3 −

7

16x6 + x7 = −

5

8.


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0 0


4 x2 19/8 0 1 1/16 0 0 -7/16 0

3 x1 2 1 0 0 0 0 1 0

0 x4 31/4 0 0 1/8 1 0 -31/8 0

0 x

5 3/8 0 0 1/16 0 1 -7/16 0

0 ← x7 -5/8 0 0 1/16 0 0 -7/16 1 − 16

7 × R5

zj − cj z(XB) = 31/2 0 0 1/4 0 0 5/4 0

↑

5/4−7/16


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0 0


4 x2 19/8 0 1 1/16 0 0 -7/16 0 R1 + 7

16 × R5

3 x1 2 1 0 0 0 0 1 0 R1 − R5

0 x4 31/4 0 0 1/8 1 0 -31/8 0 R3 + 31

8 × R5

0 x5 3/8 0 0 1/16 0 1 -7/16 0 R4 + 716 × R5

0 ← x7 10/7 0 0 -1/7 0 0 1 -16/7

zj − cj z(XB) = 31/2 0 0 1/4 0 0 5/4 0 R6 − 5

4 × R5

↑


8/9/2019 10.1 Lecture_2d


cj : 3 4 0 0 0 0 0


4 x2 3 0 1 0 0 0 0 -1

3 x1 4/7 1 0 1/7 0 0 0 16/7

0 x4

93/7 0 0 -3/7 1 0 0 -62/70 x5 1 0 0 0 0 1 0 -1

0 x6 10/7 0 0 -1/7 0 0 1 -16/7

zj − cj z(XB) = 96/7 0 0 3/7 0 0 0 20/7


In Fig. 14, all the nodes have been fathomed.

8/9/2019 10.1 Lecture_2d


node 1

node 2 node 3

node 9node 8


( LP )

( LP ) ( LP )

( LP )( LP )

( LP )

( LP )

( LP )

x <= 1 x => 2

x <= 2 x => 3

x <= 2 x => 3

x <= 3 x => 4

2 3

4 5 6 7

8

( LP ) 1

9

x = 4, x = 3/2


1

z = 15

x = 11/3. x = 1 x = 20/7, x = 2

z = 116/7

z = 31/2

x = 2, x = 19/8

infeasiblex = 3, x = 1

z = 14

x = 2, x = 2 x = 4/7, x = 3

z = 96/78 9

6

2 3

1 1

1

1

2

2

2

2

2 2

2

2 2

1 1 1 1

2 2

1 1


infeasible

Figure 14:


8/9/2019 10.1 Lecture_2d


Therefore an optimal solution of the given ILP is (x∗1

= 2, x∗2

= 2) and the

optimal value is z∗ = 14.


8/9/2019 10.1 Lecture_2d


For example,

• if we plot the feasible region of (LP )1, (LP )2 and (LP )3

• then we can visualize that the optimal solution of the ILP is either in

(LP )1 or (LP )2, i.e. either in the branch x2 ≤ 1 or in the branch x2 ≥ 2.

Fig 16 illustrates this point.


x2

8/9/2019 10.1 Lecture_2d


3 x - 2 x = 921

7 x + 16 x = 52

x

1 2

1

(4, 3/2)

Figure 15: (LP )1


x2

8/9/2019 10.1 Lecture_2d


3 x - 2 x = 921

7 x + 16 x = 52

x

1 2

1

(4, 3/2)

x => 2

x <= 1

2

2

(11/3. 1)

(20/7. 2)



x2

8/9/2019 10.1 Lecture_2d


3 x - 2 x = 921

7 x + 16 x = 52

x

1 2

1

(4, 3/2)

x => 2

x <= 1

2

2

(11/3. 1)

(20/7. 2)

x <= 3 x => 411

(3, 1)



x2

8/9/2019 10.1 Lecture_2d


3 x - 2 x = 921

7 x + 16 x = 52

x

1 2

1

(4, 3/2)

x => 2

x <= 1

2

2

(11/3. 1)

(20/7. 2)

x <= 21

x => 31

(2, 19/8)



x2

8/9/2019 10.1 Lecture_2d


3 x - 2 x = 921

7 x + 16 x = 52

x

1 2

1

(4, 3/2)

(11/3. 1)

(20/7. 2)

(2, 19/8)

x => 3

x <= 2

2

2

(4/7, 3)

(2, 2)



10.1 lecture_2d

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