Chapter 7 Integer Programming Models

1. Chapter 7Integer Programming Models

2. In manufacturing, products are often indivisible, so a production plan that calls for fractional output is not acceptable. There are also many situations that require logical decisions of the form yes/no, go/no go, assign / don't assign. Designers faced with selecting from a finite set of alternatives, schedulers seeking the optimal sequence of activities, or transportation planners searching for minimum cost vehicle routes all face discrete decision problems.

3. A manufacturer is planning to construct new buildings at four local sites designated 1,2, 3, and 4. At each site, there are three possible building designs labeled A, B, and C. There is also the option of not using a site. The problem is to select the optimal combination of building sites and building designs. Preliminary studies have determined the required investment and net annual income for each of the 12 options. SITE SELECTION EXAMPLE

4. The company has an investment budgetof $100 million ($100M). The goal is to maximize total annual income without exceeding the investment budget .

5. Let I = {A, B, C}, J= (1,2,3,4) be the set of site options. Also, denote by pij the annual net income and by aij the investment required for the design/site combination i,j. As a first try, you propose the following model for finding the maximum net annual income.

6. the optimal solution is yA1= yA3= yB3=yB4=yC1=1 with all other values of yij equal to zero and z = 40. Of the available budget, $99M is used.

7. You seem to have omitted some of the logic of the problem, because two designs are built on the same site—that is, Al and Cl, and also A3 and B3, are all in the solution. In addition, your supervisor now realizes that you were not alerted to several other logical restrictions imposed by the owners and architects—i.e., (a) site 2 must have a building; (b) design A can be used at sites 1,2, and 3 only if it is also selected for site 4; (c) at most two of the designs may be included in the plans.

8. The following additional constraints are needed to guarantee a feasible solution. Site 2 must have a building: There can be at most one building at each of the other sites:

9. Design A can be used at sites 1,2, and 3 only if it is also selected for site 4: To formulate the constraints associated with design selection,three new binary variables are introduced. Let Wi=

10. At most two designs may be used: WA + WB+ WC 2 Finally, the yij and wi variables must be tied together: the optimal solution is yA1 = yA4 = yB2 = yB3 = wA =wB = 1 with all other variables equal to zero and z = 37. All the budget is spent.

11. Logical Constraints Bound Constraints: yi=0 or 1 for i=1,2,…,n The n decisions are mutually exclusive: y1 + y2 +…+yn 1 At most k in the subset may be chosen: y1 + y2 +…+yn k At least k in the subset must be chosen: y1 + y2 +…+yn k Exactly k must be chosen: y1 + y2 +…+yn = k

12. Implication Constraints Let w be a binary variable that corresponds to a decision implied by one or more related decisions. • Decision w is implied if any one of the other n decision variables has a value of l: y1 + y2 +…+yn nw

13. • Decision w is implied if all of the other decision variables are 1: y1 + y2 +…+yn n-1+w • Decision w is implied if at least k of the other n decision variables are 1: y1 + y2 +…+yn (k-1)+[n-(k-1)]w

14. Binary Variable Implied by a Real Variable Consider the binary variable y representing the decisionwhether or not to build a facility and the real variable x representing the number of products produced by the facility. A logical constraint that restricts the variable x to be 0 unless y is 1 is where u is an upper bound on x.

15. Maximize z = subject to bi ,i=1, . . .m xj 0 and integer, j = 1,...,p, and xj > 0, j = p+ l,...,n GENERAL CONSIDERATIONS Model Structure

16. The first p variables are restricted to integer values, whereas the remaining n - p variables can assume any nonnegative real values. When all the variables are constrained to the values 0 and 1, this restriction allows us to represent binary decisions such as yes/no and gives rise to what is called a binary or 0-1 programming problem.

17. Simple Transformations Converting a General IP to a 0-1 Model Although it is somewhat inefficient from a computational point of view, it is possible to use only binary variables in IP models. A simple transformation replaces a bounded general integer variable with a weighted sum of several binary variables. Let x be an integer variable bounded by zero from below and by u from above. Let t be the smallest integer such mat 2t+1 > u.

18. x= y1 +2 y2 +4y3…+2tyt+1= yj= 0or 1, j =1 ,..., t + l where 2t u < 2t+1 For example, if x is bounded by 15, then t = 3 (because 24 > 15 23). The transformation is x =y1 + 2y2 + 4y3 + 8y4.

19. Discrete Decision Values and Functions Suppose that x can take on only one of a finite set of values—i.e., x D = {d1, d2,..., dr}. We can model this situation as x= d1y1 +d2 y2 + d3y3…+ dryr y1 + y2 +…+ yr= l yj=0 or l , j=1,...,r Ex: x {5, 8, 13, 20}

20. Complemented Variables For some algorithms, it is necessary to have all positive signs in the objective function. When only binary variables appear in the model, replacing the corresponding variable with its complement can reverse a negative coefficient. The complement of binary variable xj is 1- xj which is also binary. Ex. Z=x1-4x2+5x3

21. SYSTEM DESIGN WITH FIXED CHARGES Fixed Charge Model In many situations, undertaking an activity means that a fixed charge or setup cost is incurred in addition to the variable cost associated with the level of the activity. In the telecommunications network design problem, there is a tradeoff between construction costs and operating costs for a given demand. In a manufacturing problem, there is usually a fixed charge for setting up a machine and a variable cost for each item produced. The same is true for building and operating most facilities.

22. To construct a model for the network design problem, let xkbe the flow on proposed link k A'. The total cost for link k can be modeled by the following nonlinear concave function. where fkis the fixed cost coefficient and ck is the variable cost coefficient. LP formulations cannot handle this kind of function directly.

23. When hk(xk) is to be minimized and fk > 0, the following transformation allows us to formulate the objective as a linear function. hk(xk)=fkyk+ckxk xk 0 , yk = 0 or 1 xk  uyk (u is the upper bound ofxk)

24. KO(i) is the set of arcs originating at node i and KT(i) is the set of arcs terminating at node i, we have C1: C2: C3: C4:

25. All of the data for the model are included in Figure 7.1. Constraint C2 represents the implication that if xk is greater than zero, yk must be 1.

27. FACILITY LOCATION PROBLEM A logistics company wants to set up a distribution network in a new region of the country. There are five possible locations for warehouses and five customer locations that use the commodities supplied by the warehouses. Table 7.2 displays the data defining the problem, including unit shipping costs between potential warehouse sites and customers, fixed and variable costs for constructing the warehouses, customer demands, and warehouse capacities.

28. The goal is to select warehouse sites and sizes and to establish a shipping pattern between warehouses and customers that minimizes total shipping costs as well as the amortized cost of construction .All demands must be met and the capacity at each facility must not be exceeded.

29. Let us say that m potential sites for warehouses have been identified and that the locations and demands of the n customers are known. Let dj be the demand for customer j. The shipping cost between each potential warehouse site i and each customer j has been estimated as cij The cost of establishing a warehouse at location i consists of a fixed cost fi and a variable cost vi per unit of warehouse capacity. The maximum capacity at warehouse site i is ui

30. Define the following variables. zi = size of warehouse at location i xij = amount of product shipped from warehouse i to customer j The mathematical programming model is as follows. All demands must be met: Supply must not be exceeded:

31. Shipping from a location implies that the warehouse has been built: Nonnegative shipments: Nonnegative size: Integrality:

34. Uncapacitated Facility Location Problem It is assumed that there is no limit on the size of the warehouses. Although the same mathematical programming model applies with ui , set at an arbitrarily large value, we present a slightly different formulation that allows for a more efficient solution technique. Let xij = proportion of demand j satisfied by warehouse i

35. Because capacity is unlimited, it can be shown that it is optimal to meet the demand of each customer from a single warehouse. As such, the unit transportation cost and the variable facility cost can be combined with the demand to obtain the cost coefficient associated with the new variable xij. This is the cost of supplying the entire demand of customer j from warehouse i.

36. The new model is as follows. Minimize All demands must be met: Shipping from a location implies that the warehouse has been built Simple bounds: Integrality: yi = 0 or 1, i = 1, . . . , m

37. For the second constraint, it is necessary to multiply the yi variable on the RHS by n to allow for the extreme case in which all customers are serviced by warehouse i. In an alternative formulation, these m implication constraints representing the potential warehouse sites are replaced by mn implication constraints each representing the relation between an individual transportation link and a site: xij yi , i=1,...,m ; j=1,....n

38. Although this is inefficient from a modeling point of view because we have increased the number of constraints by a factor of n, IP algorithms may now be able to find solutions more quickly. This increase in efficiency results from the fact that if we relax the integrality requirement on all yi, the feasible region associated with the expanded model is much tighter than that of the original.

39. Covering Problem Consider a microelectronics company that would like to manufacture six new products. Initial cost estimates indicate that the equipment needed to make any of the products is very expensive, and that to make each product individually would probably require too much investment. It is possible, however, to produce composite devices that through different inter­connections can perform the functions of two or more of the products. In fact, a very complex device can be constructed that would have the same functionality as all six products, but it would require the use ofunproven technology and so has been ruled out.

40. After studying the design and manufacturing issues, the company's engineers have come up with 14 options, each performing a subset of functions, formanagement to consider. The first six options are the products themselves. To define the problem mathematically, let cj be the equipment cost for device j and let the column vector Aj represent the set of functions that the device performs. For instance, the vector A14 = (1, 0,0,1,1,0)T indicates that device 14 can be used for products 1,4, and 5. The problem is to find the set of devices with the minimumtotal equipment cost that can perform the functions of all six products.

41. For convenience, Figure 7.6 shows the 14 vectors A 1, A2,..., A14 arrayed in a 6×14 matrix in which each row corresponds to a particular function.

43. Assume that there are, in general, n devices or options, and let For x the model is Minimize cx subject to Ax e xj = 0 or l, j =1,..., n where e = (1...., 1)T is an n–dimensional vector of 1's.

44. We must select devices to cover all functions. The ith row of A identifies the set of devices that includes the ith function. The corresponding constraint ensures that at least one of the devices selected will perform the function. In the optimal solution for this example, x5= x7= x13=1, and all other variables are zero, with cost z = 62. Each function is covered by this solution, and function 2 is covered twice.

45. Partitioning Model If we now add the restriction that each function must be included once and only once in the selected devices , the partitioning problem results. The general formulation is as follows. Minimize cx subject to Ax = e xj = 0 or l, j =1,..., n For this problem, the subsets cannot overlap, and thus the solution obtained for the covering model is not feasible because function 2 is covered twice. Solving this problem, we find that x1=x3= x5= x13=1 and that all other variables are zero, with cost z = 63.

46. Traveling Salesman Problem Starting at an arbitrary location, we would like to find a tour that visits each node exactly once and returns to the original location. The goal is to minimize the sum of the lengths of the selectedarcs. Let n be the number of nodes in the network and let cij be the length of the arc passing from i to j. When it is infeasible to go from i to j either the coefficient cij is set to an arbitrarily large number or the corresponding variable is omitted from the formulation.

47. The matrix shown in Figure 7-7 gives the distance between each pair of nodes. The dashes along the diagonal rule out self-loops. Also, the data imply an asymmetric problem structure, because the distance from node i to node j is generally not equal to the distance from node j to node i.

49. A subtour is a sequence of unique nodes that starts and ends at the same location but visits only a subset of the n nodes. A solution containing subtours is not feasible. Define the following decision variables:

50. When i =j, xij does not exist so it is not included in the model. We now give the mathematical programming formulation of the asymmetric TSP. Tour length: Minimize z = Cl: Exactly one successor for each node: C2: Exactly one predecessor for each node: C3:Subtour elimination: C4: Integrality. xij = 0 or l, i≠j = l , . . ., n