5.2 mixed integer linear programming
DESCRIPTION
5.2 Mixed Integer Linear Programming. 5.2.2 Implicit Enumeration. Assignment Problem. Assignment Problem. [Theorem ] Any basic feasible solution of the assignment problem has every xij equal to either zero or one. Implication: There are at most n variables that have the value 1. - PowerPoint PPT PresentationTRANSCRIPT
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5.2 Mixed Integer Linear Programming
5.2.2 Implicit Enumeration
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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
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Assignment LP Problem
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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.
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Forward steps: o-1-2-3Backtracking step: 3-2-4Pendant nodes: 3,4,6,7,10,11,13,14
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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.
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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.
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Standard Formmin. .
binary.
T
s t
c x
Ax bc 0x
Note that, if , then is the optimal solution b 0 x 0
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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
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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
.
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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
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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
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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
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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
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1 2
3 4 5
10
9feasible!
9 (incumbent solution)Node 2 is fathomed.
U
x xx x xz
z z
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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!
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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
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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
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.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.
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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.
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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
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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
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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
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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
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Solution
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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.
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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?
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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
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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