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Integer Programming

Fabio Furini

December 8, 2014

Integer Programming 1


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Outline

1 General Definitions

2 Branch and bound3 Cutting planes



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Optimization Problems

Optimization Problems



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Integer Linear Programming – ILP

NotationAn integer linear program is a linear program with the additionalconstraint that the variables are required to be integer.

min cTx (1)Ax ≥ b (2)

x ≥ 0 (3)

x ∈ N (4)

where A ∈ Rm X n , b ∈ Rm, c ∈ Rn.Write an example of the matrix and the vectors...



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NotationEquivalent way an writing an ILP:

min

n

i =1

c i x i (5)n

i =1

Aij x i ≥ b j j = 1, . . . , m (6)

x i ≥ 0 i = 1, . . . , n (7)x i ∈ N i = 1, . . . , n (8)



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Mixed – ILP

An ILP when some variables are restricted to be integer and some are not,the problem is called: mixed integer linear program.

min cTx (9)

Ax ≥ b (10)x ≥ 0 (11)

x j ∈ N j = 1, . . . , p (12)

(Mixed) Binary ILPAn ILP where the integer variables are restricted to be 0 or 1.

x ∈ {0, 1}



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Modeling Potential

Modeling with integer variables is useful in a wide variety of applications/problems:

model logical requirements (boolean conditions)

model managerial decisions

select among different resources

allocate scarce resources

manage portfolios and projects

model fixed costsmany others . . .



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Solvers

Commercial solvers:

CPLEX

GUROBI

XPRESSEXCEL

Open solvers:

SCIP

CBC

GLPKLPSOLVE

Libre Office

Problems with more than a few thousand variables are often not possibleto solve unless they show a specific exploitable structure.



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Exact Algorithms

Branch and Bound

Cutting planes

Branch and CutBranch and Price

Dynamic Programming

A combination of the above methods



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Heuristic Algorithms

Genetic algorithm

Tabu Search

Simulated Annealing

Local Search

Ant Colony Optimization

Metaheuristic

A combination of the above methods



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Example 1

Suppose we wish to invest 19,000 e and we have four investmentopportunities:

1 investment of 6,700 e and has a net present value of 8,000 e

2 investment of 10,000 e and has a net present value of 11,000 e3 investment of 5,500 e and has a net present value of 6,000 e

4 investment of 3,400 e and has a net present value of 4,000 e

Into which investments should we place our money so as to maximize our

total present value? Such problems are called capital budgeting problems.



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Example 1

Decision variable:

x j =

1 then we will make investment j0 otherwise

max8, 000x 1 + 11, 000x 2 + 6, 000x 3 + 4, 000x 4 (13)

6, 700x 1 + 10, 000x 2 + 5.500x 3 + 3, 400x 4 ≤ 19, 000 (14)

x j ∈ {0, 1} i = 1, 2, 3, 4 (15)



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Example 1

The optimal linear programming solution is:

the optimal linear programming sol is: x 1=1, x 2=0.89, x 3= 0, x 4 = 1

the objective function value is 21,790 e

this solution is not integral

rounding x 2 down to 0 gives a feasible sol with a value of 12,000 e

the optimal solution is: x 1=0, x 2=1, x 3= 1, x 4 = 1

the objective function value is 21,000 e



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Example 1

There are a number of additional constraints we might want to add:

We can only make two investments

If investment 2 is made, then investment 4 must also be made

If investment 1 is made, then investment 3 cannot be made andvice-versa

x 1 + x 2 + x 3 + x 4 ≤ 2

x 2 − x 4 ≤ 0

x 1 + x 3 ≤ 1



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Example 2

As the project team leader, you must determine the best selection of 5 outof 10 possible projects. Project profits p 1, p 2, . . . , p 10.

x j =

1 then we execute project j0 otherwise

max10

j =1

p j x j (16)

10

j =1

x j = 5 (17)

x j ∈ {0, 1} i = 1, . . . , 10 (18)



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Example 2 – additional operational constraints

If project 2 is executed, then project 3 must also be executed

x 2 ≤ x 3 (19)If project 2 is executed, then project 4 cannot be executed

x 4 ≤ 1 − x 2 (20)

Executing projects 1 and 6 will allow you to execute project 7

x 7 ≤ x 1, x 7 ≤ x 6 (21)

Executing projects 1 or 6 will allow you to execute project 8

x 8 ≤ x 1 + x 6 (22)

Executing projects 2 and 3 will prevent you executing project 9x 9 ≤ 2 − (x 2 + x 3) (23)

Executing projects 2 or 3 will prevent you executing project 10

x 10 ≤ 1 − x 2, x 10 ≤ 1 − x 3 (24)


B h d b d


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Branch and bound

Branch and bound


B h d b d


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Branch and bound

Branch and bound has been the most effective technique for solvingMILPs in the following forty years or so. It was proposed in 1960 byLand and Dong.

It is based on dividing the problem into a number of smaller problems

(branching) and evaluating their quality based on solving theunderlying linear programs (bounding).

However, in the last ten years, cutting planes have made a resurgenceand are now efficiently combined with branch and bound into an

overall procedure called branch and cut. This term was coined byPadberg and Rinaldi 1987.


B h d b d


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Branch and bound

All these approaches involve solving a series of linear programs:

(MILP) min cTx (25)

Ax ≥ b (26)

x ≥ 0 (27)

x j ∈N

j = 1, . . . , p (28)Linear Programming Relaxation:

(LP) min cTx (29)

Ax ≥ b (30)

x ≥ 0 (31)

The best known (and most successful) methods for solving LPs are the

interior-point and simplex methods .Integer Programming 19

B h d b d


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Branch and bound

Since LP is less constrained than MILP, the following are immediate:

The optimal objective value for LP is less than or equal to the optimalobjective for MILP

If LP is infeasible, then so is MILP

If the optimal solution x of LP satisfies x j integer for j = 1, ..., p ,then x is also optimal for MILP

it gives a bound on the optimal value of MILP


B h d b d l 1


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Branch and bound example 1

We first explain branch and bound by solving the following pure integerlinear program:

max x 1 + x 2 (32)

−x 1 + x 2 ≤ 2 (33)8x 1 + 2x 2 ≤ 19 (34)

x 1, x 2 ≥ 0 (35)

x 1, x 2 ∈ N (36)

The first step is to solve the linear programming relaxation:x 1 = 1.5, x 2 = 3.5 with objective value 5.




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How can we cut this sol while preserving the feasible integral solutions?




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One way is to branch, creating two linear programs:

x 1 ≤ 1

x 1 ≥ 2Clearly, any solution to the integer program must be feasible to one orthe other of these two problems.

We will solve both of these linear programs.




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max x 1 + x 2 (37)

−x 1 + x 2 ≤ 2 (38)

8x 1 + 2x 2 ≤ 19 (39)x 1 ≤ 1 (40)

x 1, x 2 ≥ 0 (41)

The solution is

x 1 = 1, x 2 = 3 withobjective value 4.

max x 1 + x 2 (42)

−x 1 + x 2 ≤ 2 (43)

8x 1 + 2x 2 ≤ 19 (44)x 1 ≥ 2 (45)

x 1, x 2 ≥ 0 (46)

The solution is

x 1 = 2, x 2 = 1.5 withobjective value 3.5.




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These problems can be arranged in a branch-and-bound tree. Each node

of the tree corresponds to one of the problems that were solved:

Figure: B&B tree of example 1




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We can stop the enumeration at a node of the branch-and-bound tree forthree different reasons (when they occur, the node is said to be pruned).

Pruning by integrality occurs when the corresponding linear programhas an optimum solution that is integral.

Pruning by bounds occurs when the objective value of the linearprogram at that node is worse than the value of the best feasiblesolution found so far.

Pruning by infeasibility occurs when the linear program at that node

is infeasible.




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max3x 1 + x 2 (47)

−x 1 + x 2 ≤ 2 (48)

8x 1 + 2x 2 ≤ 19 (49)

x 1, x 2 ≥ 0 (50)x 1, x 2 ∈ N (51)

Exercise: Execute the branch and bound algorithm

Point of intersection among two lines: ax 1 + bx 2 + c = 0

rx 1 + sx 2 + t = 0

(α, β ) = ((cs - bt)/(br - as), (at - cr)/(br - as))




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Figure: B&B tree of example 2


Branch and bound – Algorithm


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Branch and bound Algorithm

The branch-and-bound algorithm keeps a list of linear programmingproblems obtained by relaxing the integrality requirements on thevariables and imposing constraints such as:

x j ≤ u j x j ≥ l j

Each such linear program corresponds to a node of thebranch-and-bound tree.

For a node N i , let z i denote the value of the corresponding linearprogram. Let L denote the list of nodes that must still be solved. Letz U denote an upper bound on the optimum value (initially, the boundz U can be derived from a heuristic solution of MILP, or it can be setto +∞).


Branch and bound – Algorithm


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Branch and bound Algorithm

0 Initialize L = MILP, z U = +∞, x ∗ = ∅.

1 Terminate? If L = ∅, the solution x ∗ is optimal.

2 Select node Choose and delete a problem N i from L.

3 Bound Solve N i . If it is infeasible, go to Step 1. Else, let x i be itssolution and z i its objective value.

4 PruneIf z i ≥ z U , go to Step 1.If x i is not feasible for MILP, go to Step 5.If x i is feasible for MILP, let z U = z i , x

∗ = x i . Delete from L allproblems with z j ≥ z U . Go to Step 1.

5 Branch From N i , construct linear programs N 1i , ..., N

k i with smaller

feasible regions whose union contains all the feasible solutions of (MILP) in N i . Add N

1i , ..., N

k i to L and go to Step 1.


Branch-and-bound – Additional Elements


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Branch and bound Additional Elements

Formulation (so that the gap z I - z LP is small)

Branching

Node selection

Heuristics (to find a good upper bound z U )


Binary Branching (branching on a variable)


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y g ( g )

Problem N i is a linear program. A way of dividing its feasible region is toimpose bounds on a variable. Let x i j be one of the fractional values for

j = 1, . . . , p , in the optimal solution x i of N i .From problem N i , we can construct two linear programs (satisfy the

requirements of Step 5):N −ij = N i with the additional bound x j ≤ x

i j

N +ij = N i with the additional bound x j ≥ x i

j

The advantage of branching on a variable is that the number of

constraints in the linear programs does not increase, since linearprogramming solvers treat bounds on variables implicitly.


Strong Branching


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g g

Compute for all the fractional variables:

D +ij = increase of the objective values from N i to N +ij

D −ij = increase of the objective values from N i to N −

ij

branch on the variable j = 1, . . . , p such that min

(D −

ij , D

+

ij ) is the largest(thus tightening the lower bound for pruning). Others have proposed tochoose j such that (D +ij + D

−

ij ) is the largest. Combining these two criteriais even better, with more weight on the first.But the computing time of doing it at each node N i , for every fractional

variable x i j , may be too high. Significantly more time should be spent onthese evaluations towards the top of the tree.


Pseudocost Branching


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g

Let f i j = x i

j − x i

j be the fractional part of x i

j , for j = 1, . . . , p . For an

index j such that f i j > 0, define:

down pseudocost P − j = D −

ij

f i j

up pseudocost P + j = D +

ij

1−f i j

It can be observed that the pseudocosts tend to remain fairly constantthroughout the branch-and-bound tree. Therefore the pseudocosts need

not be computed at each node of the tree. They are estimated instead.


Pseudocost Branching


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g

A good way of initializing the pseudocosts is through strongbranching at the root node or other nodes of the tree when newvariables become fractional for the first time

To update the pseudocost P j , we average the observations over allthe nodes of the tree where x j was branched on.

The estimated pseudocosts P − j and P + j are used to computeestimates of D −ij and D

+ij at node N i

D −ij = P −

j f i

j

D +ij = P +

j (1 − f i

j )

Among these candidates, the branching variable x j is chosen to be theone with largest min(D −ij , D

+ij ) (or other criteria such as those

mentioned earlier).


Node selection


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How does one choose among the different problems N i available in Step 2of the algorithm?

Two goals need to be considered:finding good feasible solutions (thus decreasing the upper bound z U )

proving optimality of the current best feasible solution (by increasingthe lower bound as quickly as possible)


Node selection


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Best estimate criterion

For the first goal, we estimate the value of the best feasible solutionin each node N i :

E i = z i +

p j =1

min(P − j f i

j , P +

j (1 − f i

j ))

This corresponds to rounding the noninteger solution x i to a nearbyinteger solution and using the pseudocosts to estimate thedegradation in objective value.

Depth-first search strategy

It is an effective strategy to quickly look for a feasible solution.Breadth-first search strategyIt is an effective strategy to increase the global lower bound of theproblem.



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Local Branching


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This heuristic is particularly suited for MILPs that are too large to solve to

optimality, but where the linear programming relaxation can be solved inreasonable time. For simplicity, assume that all the integer variables are0,1 valued.

Let x̃ be a feasible solution of (MILP) (found by a diving heuristic, for

example).The idea is to define a neighborhood of x̃ as follows:p

j =1 |x j − x̃ j | ≤ k p j ∈I :x̃ j =0

x j +p

j ∈I :x̃ j =1(1 − x j ) ≤ k (linearised)

where k is an integer chosen by the user (for example k = 20)

Instead of getting lost in a huge enumeration tree, the search isrestricted to the neighborhood of x̃ by this constraint


Practical considerations


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The most successful node selection strategy may differ depending onthe application.

For this reason, most MILP solvers have several node selectionstrategies available as options.

The default strategy is usually a combination of the best estimatecriterion (or a variation) and depth-first search.


Cutting planes


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Cutting planes


Gomory Cuts


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Suppose we are given an instance of ILP and begin by solving theLP-relaxation, producing an optimal basic feasible solution x B associatedwith basis B .

If the solution is integer then stop.Otherwise add a cutting plane such that x B :

is not feasibleall feasible integer points remain feasible

Then, repeat above processes.


Gomory Cuts


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Consider any equation in the optimized final tableau, say the i-th equation:

x i

B +

j /∈B

y ij

x j

= y i 0

(1G )

for some i , 0 ≤ i ≤ m. Due to nonnegativity of x

j /∈B

y ij x j ≤ j /∈B

y ij x j

Therefore we have:

x i B + j /∈B

y ij x j ≤ y i 0 (2G )

Subtracting (2G) from (1G), we get the constraint: j /∈B

(y ij − y ij )x j ≥ y i 0 − y i 0

called the Gomory cut corresponding to the source or generating row i .


Gomory Cuts – Example


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max x 1 + x 2 (52)

−x 1 + x 2 ≤ 2 (53)

8x 1 + 2x 2 ≤ 19 (54)

x 1, x 2 ≥ 0 (55)

x 1, x 2 ∈ {0, 1} (56)

We first add slack variables x 3 and x 4 to turn the inequality constraintsinto equalities. The problem becomes:

max z − x 1 − x 2 = 0 (57)

−x 1 + x 2 + x 3 = 2 (58)

8x 1 + 2x 2 + x 4 = 19 (59)

x 1, x 2, x 3, x 4 ≥ 0 (60)

x 1, x

2, x

3, x

4 ∈ {0, 1} (61)




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Solving the linear programming relaxation by the simplex method, we getthe optimal tableau:

z + 0.6x 3 + 0.2x 4 = 5 (62)

x 2 + 0.8x 3 + 0.1x 4 = 3.5 (63)

x 1 − 0.2x 3 + 0.1x 4 = 1.5 (64)

x 1, x 2, x 3, x 4 ≥ 0 (65)

The corresponding basic solution is x 3 = x 4 = 0, x 1 = 1.5, x 2 = 3.5 andz = 5.




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This solution is not integer. Let us generate the Gomory mixed integer cutcorresponding to the equation:

x 2 + 0.8x 3 + 0.1x 4 = 3.5 (66)

we get the GMI cut:

8x 3 + x 4 ≥ 5. (67)

We can express the above Gomory cut in the space (x 1 , x 2 ). This yields:

x 2 ≤ 3. (68)



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