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PROGRAMMING Outline 1. INTEGER Introduction 2. Classes of integer optimization models of increasing complexity: a) b) c) d) Binary (0-1) variables Integer variables Linear systems with integer/real variables Non-linear systems with integer/real variables 5/15/2008 5/15/2008 Outline (cont.) 3. A Simple Capital Investment Modeling Problem 4. Solution Methods a) b) c) d) e) 5/15/2008 Outline (cont.) 5. Modeling with integer variables 6. IP solvers and GAMS Enumeration Implicit enumeration Branch and bounds LP Relaxations Cutting planes 5/15/2008 Relation to Material in Text INTRODUCTION Cornuejols and Ttnc discuss integer programming in Chapter 11. In Chapter 12 they discuss three financial application: combinatorial auctions, the lockbox problem, and index funds. They also discuss a variant of portfolio optimization. See also Section 1.1.4. 5/15/2008 5/15/2008 1 Some Useful Links J.E. Beaseleys Notes on IP gives a good introduction to IP. Michael Tricks Presentation: Formulations and Reformulations in Integer Programming indicates some important modeling issues in IP. In particular, when should the user use modeling tricks and when should he/she let the solver find them? Links to these are in the Links folder of our class web site. 5/15/2008 Some Useful References Alexaner Schrijver, Theory of Linear and Integer Programming, Wiley-Interface, 1986 (Chapters 19-21 on Total Unimodularity) 5/15/2008 Good News! Even in the simplest cases of linear programs, say D = {x|0x1}, there is an (uncountably) infinite number of possible solutions. On the other hand, for problems where the decision variables are integers, and D is bounded (which is often the case) there is a finite number of possible decisions. That is the good news! 5/15/2008 Bad News! Unfortunately, the finite number can be extremely large. Suppose we have n binary variables, (variables that can only take on two values say 0 and 1) x1,x2,,xn . Clearly, D for such problems is bounded. 5/15/2008 More Bad News! If n =1, we can have, at most, two decisions, x1=0, or x1=1. For n=2 we can make at most four decisions, {x1=0, x2=0}, {x1=0, x2=1}, {x1=1, x2=0}, or {x1=1, x2=1}. For n = 10 about 1000 decisions; for n = 20, about 1,000,000; for n = 30 about 1,000,000,000 decisions. Some say for n=266, there would be more possible decisions than atoms in the universe! Recall that linear programs are routinely solved with hundreds of thousands of variables! 5/15/2008 Classes of integer optimization models of increasing complexity: a) b) c) d) Binary (0-1) variables Integer variables Linear systems with integer/real variables Non-linear systems with integer/real variables 5/15/2008 2 A Simple Capital Budgeting Problem (Text pp. 193-194) A SIMPLE CAPITAL BUDGETING PROBLEM We wish to invest $19,000 We have 4 investment opportunities: Investment 1: Price $6,700, NPV $8,000 Investment 2: Price $10,000, NPV $11,000 Investment 3: Price $5,500, NPV $6,000 Investment 4: Price $3,400, NPV $4,000 We wish to maximize our total present value. Each investment is take it or leave it. We cannot make partial investments in any of the 4 investments. 5/15/2008 5/15/2008 A Simple Capital Budgeting Problem (cont.) This suggests the following binary (0-1) optimization problem: Max 8x1 + 11x2+ 6x3+ 4x4 s.t. 6.7x1 + 10x2+5.5x3+3.4x419 xj {0,1} A Simple Capital Budgeting Problem (cont.) Since we have only 4 binary variables, there are at most 24 = 16 possible choices. In this case we may simply examine each possibility and choose the best one. This solution method is called enumeration. We will organize this in a tree structure illustrated in the next slide. This structure can be generalized to be used in more sophisticated solution methods. 5/15/2008 5/15/2008 A Simple Capital Budgeting Problem (cont.) The notation (1,0,?,?) means, for example, at this point we are committed to taking investment 1, and to not taking investment 2, while we are, for now, undecided about investments 3, and 4. The notation <8, 6.7> indicates that for the decisions we are committed to, the return would be $8,000, and the price, $6,700. 5/15/2008 5/15/2008 3 A Simple Capital Budgeting Problem (cont.) Notice, at node (1,1,1,?), we have <25, 22.2> indicating a NPV of $25,000, unfortunately, it costs $22,200, which is more than our budget. We cannot make this any better by a choice for asset 4. So the node and its two descendants are not relevant to finding the optimal solution. Similarly, the node (1,1,0,1) also costs too much ($21,000) and is not, therefore, relevant. 5/15/2008 A Simple Capital Budgeting Problem (cont.) So we can delete the three extraneous nodes and obtain a partial enumeration tree that we will call an implicit enumeration tree because it is a smaller tree that still contains all the nodes that could be an optimal solution to the problem. One solution method that tries to get as small an implicit enumeration as is possible is the branch and bound method. 5/15/2008 Linear Programming Relaxations of Integer Programming Problems Linear Programs can be written in the form: Min f(x) = cx Subject to D = {x|Ax b, x 0} An Integer Program: Min f(x) = cx Subject to D = {x|Ax b, x 0, x integer} The linear program can be interpreted as a relaxation of the integer program, with f f and D and D as given. 5/15/2008 5/15/2008 Some Immediate Consequences 1. If the linear programming relaxation is infeasible, so is the integer programming (IP) problem. 2. The value of the optimal solution of the IP problem can be no smaller than the optimal value of the integer programming problem. 3. If an optimal solution of the linear programming problem is feasible in the IP problem (that is, if the integer constraints are satisfied, then the linear programming problem solution is optimal for the IP problem. 5/15/2008 This is valuable because: Polynomial time LP algorithms exist, while IP problems are NP Complete. 5/15/2008 4 Moreover: For important special cases of IP, the usual methods of solving LP relaxations (e.g., the simplex method) automatically lead to integer solutions. For example, when A is totally unimodular this happens. Examples include maximum flow problems, assignment problems, and other applications of practical interest. See Chapters 19-21 of [Schriver, 1986] Also remember: Any feasible solution for the IP problem is an upper bound for the IP optimum value! 5/15/2008 5/15/2008 Ways of Pruning an Enumeration Tree Infeasible: If relaxation is infeasible so is the integer problem at the node. Integer: The relaxation satisfies the integer constraints. Fathomed: The current bound is worse than the best known feasible solution. 5/15/2008 This scheme can easily be generalized to mixed integer programming problems Mixed integer optimization problem: Min cx + dy st. Ax + Fy = b x0, y 0, y integer LP relaxation: Min cx + dy st. Ax + Fy = b x0, y 0 5/15/2008 The Relaxation Solution LP of the LP Relaxation For this simple problem wecan solve the LP relaxation by inspection. To do this we rank the 4 Opportunities by their total return in $s per $ invested. Opportunity 1: $ 8,000/$ 6,700 = 1.194 Opportunity 4: $ 4,000/$ 3,400 = 1.176 Opportunity 2: $11,000/$10,000 = 1.100 Opportunity 3: $ 6,000/$ 5,500 = 1.091 5/15/2008 Max 8x1 + 11x2+ 6x3+ 4x4 s.t. 6.7x1 + 10x2+5.5x3+3.4x419 0 xj 1, j=1,,4 5/15/2008 5 Solution of the LP Relaxation We invest in Opportunity 1, all the way to $6,700.Then we invest in the next best Opportunity, 4, for an additional $3,400. Finally, we invest the remaining $8,900 in Opportunity 2. The total return for the relaxation is $8,000 + $4,000 + ($8,900/$10,000)*$11,000 = $21,790. Therefore, we know that we can not get more than this value for any binary solution. LP Relaxation at (1,0,?,?) At this node we have already committed to investing in Opportunity 1, and rejecting Opportunity 2. We have committed $6,700, for a return of $8,000. Investments in Opportunities 3 and 4 are still open. The relaxation is now: Max 8 +6x3+ 4x4 s.t. 5.5x3 + 3.4x4 19-6.7 = 12.3 0 x3 1, 0 x4 1 5/15/2008 5/15/2008 LP Relaxation at (1,0,?,?) (cont.) Since Opportunity 4 outperforms Opportunity 3, we invest an additional $3,400 in Opportunity 4, leaving $8,900 to invest. Our only remaining opportunity is to invest $5,500 of the remaining $8,900 in Opportunity 3. The total return for this relaxation is $8,000 + $4,000 + $6,000 = $18,000. Branch and Bound Our branch and bound solution method relies on 3 items: 1. A list of unpruned leaves of a partial enumeration tree 2. A relaxation value at each of the leaves, and 3. The best feasible integer solution discovered so far. We start the process at the root of the tree (?,?,.?), the relaxation at the root, and if we know a feasible solution we use it, otherwise we set it to (- ) if minimizing (maximizing) 5/15/2008 5/15/2008 Branch and Bound (cont.) The iteration consists of the following steps: 1. Select a leaf from the list of unpruned leaves of the current partial enumeration tree: If the list is empty, the best integer feasible solution we have saved is optimal. 2. Branch: Pick a variable that is not specified and add two leaves to the enumeration tree corresponding to the variable being 1, and the other to the variable being assigned 0. 5/15/2008 5/15/2008 Branch and Bound (cont.) 3. For each of the two solve the relaxation: a) If the relaxation problem is infeasible delete the new node. b) If it is feasible as an integer program compare the value of the relaxation with the current new best integer feasible solution. i. If the relaxation value is better, make it the new best integer feasible solution. Delete each node in the leaf list for which its relaxation value is worse or equal to the new best integer feasible solution. ii. In either case, delete the new node. c) Otherwise, assign the relaxation value to the new leaf. 4. Go to 1. 6 A Simple Capital Budgeting Problem (cont.) We now solve our example using branch and bound. Round brackets () enclose the assignments of values; e.g., (0,?,1,0) indicates that x1 is 0, x2 and x4 are unspecified, and x3 is 1. <R, P> indicates the investment, P, and the total return, R, for the partial assignment. [z] indicates the relaxation value. 5/15/2008 7 An Example from the Text An Example from the Text (cont.) The enumeration tree depicts the final one for a problem with 89 binary variables. The complete enumeration tree would have over 60000 (twenty six zeroes) possible decisions, and 90 levels. The shown tree has 22 levels. 5/15/2008 5/15/2008 Stronger and Weaker Relaxations Relaxations other than LP relaxations can be used. We will give two illustrations: one type that is easier than LP relaxations that may not be as strong; and one that is more difficult, but usually stronger. 5/15/2008 Stronger and Weaker Relaxations (cont.) As we observed in our discussion of duality. Any feasible solution to the dual program is a bound for the primal. Specifically, in the case we have in our example, we are maximizing, ...

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