Integer Linear Programming

Integer Linear Programming When you need discrete solutions

Discrete Variables • Linear programming allows us to solve large scale network problems. It gives answers in terms of continuous variables • However, there are many situations where we need solutions to network problems which are not allowed to fall in a continuous range

Design Problem • For example, if we are designing a network, we need to decide whether to place a link between two particular nodes • In this case a decision is either “yes” or “no”. We could represent this as a variable which had only two possible values, 0 or 1

Routing, Set-up Problems • A route could be represented by a logical variable which is 1 if a link is in the route, and 0 otherwise • The cost of using a link could include a “set-up” cost which would not be paid if the link was not used at all

Modular Equipment • Equipment, such as switches, have to be bought with quantised traffic capacity • For example, a switch could have capacity equal to 100, 200,…400 Gbps, but it might be impossible to purchase anything between these numbers

No Separate Method • No separate technique has been developed to solve this type of problem • The best solutions adapt the techniques of linear programming and look for discrete solutions in the vicinity of the continuous solution

LP with Discrete Variables • Only discrete values allowed

Branch and Bound • The most popular and successful method of obtaining discrete solutions today is called “branch and bound” • This method relies on our being able to solve a LP problem by the simplex method

Branch and Bound • The first step is to solve the problem as a LP problem, using the simplex method • This will, in general produce solutions with variables in a continuous range • We take one of these variables and constrain it to take the allowed value below its current one

Branch and Bound • This defines a new problem which can also be solved by the simplex method • However, it is also possible that the allowed value above the LP solution gives a better cost function • Therefore we should also try this constraint

Simple Example • Maximise z = x1+x2 with constraints • x1 + 3x2 ≤ 3 • 3x1 + x2 ≤ 6 • x1, x2 integers ≥ 0 • (This example comes from Eiselt and Sandblom)

Simple Example • The optimal solution to the LP problem is • x1 = 1.875 • x2 = 0.375 • z = 2.25 • We will choose x2 to constrain so that x2 ≤ 0

Simple Example • The new solution is • x1 = 2 • x2 = 0 • z = 2 • We still need to try solving with constraint x2 ≥ 0

Simple Example • The new solution is • x1 = 0 • x2 = 1 • z = 1 • Both of these solutions have given us integral values of x1 also • We choose the first solution because it has a higher z

Branching Tree • We do not need to explore any more values in the region • We can illustrate what we have just done with a “branching tree” x1 = 1.875 x2 = 0.375 z = 2.25 x2≥ 1 x2≤ 0 x1 = 0 x2 = 1 z = 1 x1 = 2 x2 = 0 z = 2

More Branching • In general we are not able to find suitable solutions after so few trials • The procedure may need to proceed through several branches • We now consider a more complicated example, also from Eiselt and Sandblom

Example 2 • Maximise z = 4x1 + 6x2 + 10x3 • such that • 2x1 + x2 + 2x3 ≤ 13 • 3x1 – 2x2 – 4x3 ≥ 7 • x1, x2, x3 are integers ≥ 0 • We will branch left (down) on the non-integer with the lowest subscript

x1 = 45/7, x2 = 34/7, x3 = 0, z1 = 402/7 n1 x1 ≤ 4 x1 ≥ 5 x1 = 4, x2 = 21/2, x3 = 0, z2 = 31 n2 x2 ≥ 3 x2 ≤ 2 x1 = 4, x2 = 2, x3 = 1/4, z3 = 301/2 n3 x3 ≤ 0 x3 ≥ 1 x1 = 4, x2 = 2, x3 = 0, z4 = 28 n4

Moving down the Tree • In this tree, node n1 is the LP solution, and represents an upper bound on the objective function • As we move through nodes n2, n3 and n4, we add more and more constraints on the variables • As we move down, z never increases

Bound • Node n4 is the first node we have found with all integer variables • Therefore, 28 is a lower bound on the solution • When we reach such a solution, we have reached a “bound” in the branch and bound technique

Other Branches • We have not yet explored all the possibilities near the LP solution • We now backtrack and explore some of the possibilities in the right hand branches • The next slide shows some of the results

x1 = 45/7, x2 = 34/7, x3 = 0, z1 = 402/7 n1 x1 ≤ 4 x1 ≥ 5 x1 = 4, x2 = 21/2, x3 = 0, z2 = 31 x1 = 5, x2 = 3, x3 = 0, z5 = 38 n2 n5 x2 ≥ 3 x2 ≤ 2 x1 = 4, x2 = 2, x3 = 1/4, z3 = 301/2 n6 No feasible solution n3 x3 ≤ 0 x3 ≥ 1 x1 = 4, x2 = 2, x3 = 0, z4 = 28 x1 = 4, x2 = 1/2, x3 = 1, z7 = 29 n4 n7

Further Branching • Node n5 has an all-integer solution, so we do not branch any further from this node • Node n6 does not have any feasible solutions: so no further branches • Node n7 does allow further branches

Fathoming • Notice that z5 = 38, and z7 = 29 • If we had evaluated node n5 before n7, we need not have proceeded any further, since it would be impossible for z7 to be more than z3, which is 301/2 • This is called “fathoming”

Binary Tree • The structure of solutions is called a “binary tree”. Every node in the tree represents a problem which must be solved by the simplex method • To reduce computing time, we try to keep the number of nodes in the tree as small as possible

Depth First Strategy • The “depth first” strategy takes the left hand branch until the tree can be “pruned” • Then it backtracks to the last node where it could take another branch (to the right) • Then it continues with left branches

Breadth First Strategy • The “breadth first” strategy evaluates all nodes at level 2 before going to examine all nodes at level 3 and so on • On the average, it has been found that “depth first” strategies are faster than “breadth first” strategies

Routing Problem • The capacity on each link in the network below is 10 Gbps • The three traffic flows, f1, f3 and f5 to node 4 from nodes 1, 3 and 5 have to choose one of the routes shown in the sketch 1 2 3 4 5 6

Routing Problem • The traffic must be distributed so that the following cost function is minimised • uk = utilisation in link k • dk = length of link k

Routing Problem • Let us introduce three variables, x1, x3 and x5 to indicate that the upper route is chosen when xj = 1, and the lower route is chosen when xj = 0 • Let us also name the nine links l1 to l9 as shown in the next slide • Assume traffic in both directions

Numbered Links 2 1 2 1 3 5 4 6 4 3 7 9 5 6 8

For link 1 For link 2 For link 3 For link 4 For link 5 For link 7 For links 8 and 9 x3f3 ≤ 10 x1f1 + x3f3 ≤ 10 x1f1 + x3f3 + x5f5 ≤ 10 (1-x1)f1 ≤ 10 x5f5 ≤ 10 (1-x3)f3 ≤ 10 (1-x1)f1 + (1-x3)f3 + (1-x5)f5 ≤ 10 Constraints

Cost Function • x1, x3, x5 ≥ 0 • Cost function is • = (x3f3d1 + (x1f1 + x3f3 )d2 +…+ ((1-x1)f1 + (1-x3)f3 + (1-x5)f5 )d9)/10 • This is a linear combination of x1, x3 and x5 • This states the problem in a form which can be solved using LP

Integer Linear Programming • The first solution will give fractional values for x1, x3 and x5 • Since we need these variables to be either 1 or zero, we select one arbitrarily (say x1) and solve again for x1 ≤ 0, and so we start the process of integer linear programming

Discrete Solution • Ultimately, we will get to the best possible solution with all three variables equal to either 1 or zero • This will then be the “best” way of distributing the routes in this network, for the set of flows f1, f3, f5

Realistic Problem • This was a problem with a reduced number of flows in a small network • A more realistic statement of the problem would have included flows from every node to every other node (6*5/2 = 15 flows) • We also allowed only two routes per flow

Increasing Complexity • As the network load changes, it will be more efficient to consider several routes for each flow • If we had four possible routes for one flow, we would then need a more complicated way to select the route to be used (see Lecture 6)

Single Route? • It is generally found that it is better to select a single route for each flow in networks with more than about 20 nodes • Below this number it is often best to split the traffic between at least two routes for a single flow

Small Network • This means that we could run the network better if we allowed x1, x3 and x5 to take fractional values • Then the first LP solution we had would be optimal for our small network, as long as 0 ≤ xi ≤ 1 for all i

OSPF • Note that OSPF allows multiple routes through a network

Integer Linear Programming