5.2 Mixed Integer Linear Programming

5.2.2 Implicit Enumeration

Assignment Problem – LP Model

1 1

1

1

Optimally assign workers to jobs

min

. .

1 1,2, ,

1 1,2, ,

0 , 1,2, ,

n n

ij ijj i

n

ijj

n

iji

ij

n n

c x

s t

x i n

x j n

x i j n

Assignment LP Problem

Basic ConceptThe basic idea of implicit enumeration is to

explicitly enumerate only a small subset of all possible solutions while concluding that it is not necessary to explicitly investigate the remaining solutions. This is because they are either

1. infeasible or 2.will result in an objective value that is inferior

to the best integer solution already found.

Forward steps: o-1-2-3Backtracking step: 3-2-4Pendant nodes: 3,4,6,7,10,11,13,14

Search Rules• A forward step is defined as the branching process of

fixing a free variable to be 1.• The backtracking step is defined as the process to

trace back to the origin until encounter the 1st node with only one descending branch. At this node, the second descending branch is searched by changing the corresponding variable from 1 to 0.

• The search process is continued until all pendant nodes are fathomed and each non-pendant node has exactly 2 descending branches.

Additional Terminologies• Completion: Given a node and a partial

solution, a completion of the partial solution is a solution in which values are specified for all the remaining free variables.

• Fathom: A partial solution is fathomed by either (1) demonstrating that there are no improving feasible completions or (2) finding the best feasible solution.

Standard Formmin. .

binary.

T

s t

c x

Ax bc 0x

Note that, if , then is the optimal solution b 0 x 0

Zero Completion TestThe objective function can be minimized ifall free variables are set to zero.

If (smallest upper bound) fathomedIf otherwise (1) fathomed (2)

T

U

U

U

z

z z z

z z zz z z

c x

b 0

additional branching

Infeasibility TestIf no feasible completions (may or may not be

zero completion), then the node should be fathomed. 1 2 3 4 5 6

1 3

2 4 5 6

2 4 5 6

max

: 4 2 3 3 2Current partial solution: 1 and 1

2 3 5Introduce slack variable

5 2 35 2(1) (0) 1 3(0) 2 0

All completions are infeasible. fathomed

x x x x x xx x

x x x x

s x x x xs

Example

.

Example1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

min 4 5 6 2 3. .4 2 3 2 1

5 2 2 2 5 binary

z x x x x xs t

x x x x xx x x x x

x

Subproblem P0All variables are free.

1 1 2 3 4 5

1,max

2 1 2

additional bran

Zero completion test:

Infeasibi

Zero completion is infeasible to both constraints.

0

1 4 2 3 21 4(1) 2(1) 3(0) 2(1) (

lity

0) 7 0

chin

te t

5 5 2

s

g

:

U

U

z

z z

s x x x x xs

s x x

3 4 5

2,max

2 25 (1) 5(1) 2(1) 2(0) 2(1)

inconclus

5 0

ive

x x xs

1 1x

0

12 3 4 5

2 3 4 5

2 3 4 5

min 4 5 6 2 3. .2 3 2 35 2 2 2 4 binary

z x x x xs t

x x x xx x x x

x

1 2 3 4 5

1,max

2 2 3 4 5

2,m

additional bran

Zero Completion Test:

Infeasibili

The zero completion is infeasible to the 2nd constraint.4

3 2 3 23 2(1) 3(0)

ty Test:

2(1) (0) 7 0

ch g

4

i

5 2 2

n

2

Uz z

s x x x xs

s x x x xs

ax 4 5(1) 2(1) 2(0) 2(1) 5

inconclusiv

0

e

1 2

3 4 5

10

9feasible!

9 (incumbent solution)Node 2 is fathomed.

U

x xx x xz

z z

3 1 2

3 4 5

3 4 5

3 4 5

1, 0

min 4 6 2 3. .

3 2 32 2 2 4 binary

P x x

z x x xs tx x x

x x x

Subproblem

x

Backtracking!

1 3 4 5

1,max

2 3 4 5

2,max

Additional B

Zero Completion Test:

Infeasi

The zero completion is infeasible.The trivial lower bound 4 9.

3 3

ranchibility Te

23 3(0) 2(1) (0) 5 0

4 2 2 24 2(1

g

)

st:n

Uz z

s x x xs

s x x xs

inconclusive

2(0) 2(1) 0

4 4 1 2 3

4 5

4 5

4 5

1, 0, 1

min 10 2 3. .2 0

2 2 2 binary

P x x x

z x xs t

x xx x

Subproblem

x

.The trivial lower bound is 10 9.

since additional branching can never produce an integer

Zero completion is infeasible

Node 4 can be fathomed solution as good as the

incumbent solution found in sub

Uz z

problem 2.

1 4 5

1,max

2 4 5

2,max

Zero Completion Test:

Infeasibil

The zero completion is infeasible.The trivial lower bound is 4 9.

3 23 2(1) (0) 5 0

4 2 24 2(0) 2(1) 2

Inf

ity Test

easible Node 5

0

is

:

Uz z

s x xs

s x xs

fathomed.

1 2 3 4 5

1,max

2 2 3 4

additional branching

Zero Completion Test:

Infeasibility Test:

The zero completion is infeasible.The trivial lower bound is 0 9.

1 2 3 21 2(1) 3(0) 2(1) (0) 3 0

5 5 2 2

Uz z

s x x x xs

s x x x

5

2,max

25 5(1) 2(1) 2(0) 2(1) 4

Inconclusi e

0

v

xs

1 2

3 4 5

Zero Completion Test:

fea

0, 10

5 9Since this test is , the new upper bound is

incumbent solutionNode 7 fathomed.

5sible

U

U

x xx x xz z

z

8 1 2

3 4 5

3 4 5

3 4 5

0, 0

min 6 2 3. .

3 2 12 2 2 5 binary

P x x

z x x xs tx x x

x x x

Subproblem

x

1 3 4 5

1,max

2 3 4 5

2,max

Zero Completion Test:

Infeasibility Test:

The zero completion is infeasible.The trivial lower bound is 0 5.

1 3 21 3(0) 2(1) (0) 1 0

5 2 2 25 2(1) 2(0) 2(1) 1 0

Uz z

s x x xs

s x x xs

Infeasible and fathomed

Solution

In a plant we have 2 production units designated as number 1 and 2, making product 1 and 2, respectively, from the 3 feed stocks (A, B and C). Unit 1 has a maximum capacity of 8000 lb/day, and unit 2 of 10000 lb/day.

To make 1 lb of product 1 requires 0.4 lb of A and 0.6 lb of B; to make 1 lb of product 2 requires 0.3 lb of B and 0.7 lb of C. A maximum 6000 lb/day of B is available, but there are no limits on the available amounts of A and C.

Blending Products including Batch Sizes

Assume the net revenue after expenses from the manufacture of product 1 is $0.16/lb, and of product 2 is $0.20/lb.

How much of products 1 and 2 should be produced per day, assuming that each must be made in batches of 2000lb?

1 2

1 2

1 2

1 2 1 2

Let and be the amounts of products 1 and 2 in two thousands of pounds per day.

max 0.16 2000 0.2 2000

. . 1, 2

0.6 2000 0.3 2000 60000 4, 0 5 , inte

i i

x x

f x x

s tx y i

x xy y y y

gers

1 2

1 2

1 2

1 2 1 2

Let and be the amounts of products 1 and 2 in two thousands of pounds per day.

max 0.16 0.2. .

1, 20.6 0.3 30 4, 0 5 , integers

i i

x x

f x xs tx y i

x xy y y y