Integer Programming. The focus of this chapter is on solution techniques for integer programming models.

Size: px

Start display at page:

Download "Integer Programming. The focus of this chapter is on solution techniques for integer programming models."

Brice Francis
5 years ago
Views:

1 Integer Programming Introduction The general linear programming model depends on the assumption of divisibility. In other words, the decision variables are allowed to take non-negative integer as well as fractional values. However, we quite often face situations where the planning models contain integer valued variables. For instance, trucks in a fleet, generators in a powerhouse, pieces of equipment, investment alternatives and there are a myriad of other examples. In all such cases, an integer solution is desired, which can be easily obtained by rounding off the fractional values of the variables. ut, rounding-off may result in sub-optimal or infeasible solutions. To overcome such difficulties, a different optimization model, which is referred to as integer programming has been developed. Definition Integer (or discrete) programming problem is a type of problem in which some, or all, of the variables are allowed to take only integral values. The focus of this chapter is on solution techniques for integer programming models. In this chapter, we drop the assumption of divisibility. The mathematical model of the problem is as follows: Optimize (maximize or minimize) c j x j a ij x j (, =, ) b i ; i = 1, 2,..., m x j 0; j = 1, 2,..., n x j integer valued; j = 1, 2,..., p n The discreteness stipulation distinguishes an integer from a linear programming problem. If all the variables are restricted to take only integral values (i.e., p = n), the model is called a pure integer programming problem. To the contrary, if some variables are restricted to take only integer values, and the remaining are free to take any non-negative values, then it is called a 1

2 mixed integer programming problem. When the decision variables are required to take value of either zero or one, it is called zero-one programming. The mathematical model of zero-one programming is as follows: Optimize (maximize or minimize) c j x j a ij x j (, =, ) b i ; i = 1, 2,..., m x j = 0 or 1; j = 1, 2,..., n The effective use and applications require, as a first step, the formulation of the model when the problem is presented. In this section, we illustrate the formulation of integer programming problems. The best way to explain a topic is through examples. So consider the following examples. Example 1 You have entered in a treasure cave (with a password KHUL JA SIM SIM) full of three types of valuable stones, amethyst (A), ruby (R), and topaz (T). Each piece of A, R, and T weighs 3, 2, 2 kg., and is known to have a value of 4, 3, 1 crore respectively. You have got a bag that can carry a maximum of 11 kg. Your problem is to decide on how many pieces of each type to carry, within the capacity of your bag, so as to maximize the total value carried. The stones cannot be broken. Solution. We start the formulation exercise by defining the decision variables. 2

3 Let x 1 = Number of amethysts to be carried x 2 = Number of rubies to be carried x 3 = Number of topaz to be carried The objective function here is to maximize the total value carried, which is given by the linear function. Maximize 4x 1 + 3x 2 + x 3 Since one amethyst is of 3 kg, one ruby is of 2 kg, one topaz is of 2 kg, and the capacity of the bag being 11 kg., the constraint can be expressed as 3x 1 + 2x 2 + 2x 3 11 Finally, we note the fact that stones cannot be broken, i.e., the variables have to take discrete values, which may be stated algebraically as x 1, x 2 and x 3 are all non-negative integers. Thus, we have the following formulation: Maximize (4x 1 + 3x 2 + x 3 ) 3x 1 + 2x 2 + 2x 3 11 x 1, x 2 and x 3 are all non-negative integers. Go-No-Go Decisions Example 2 wants to take up four projects. However, because of budget limitations, not all the projects can be selected. It is known that project j has a present value of c j and would require an investment of a jt in period t. The capital available in period t is b t. The problem is to choose projects so as to maximize present value, without violating the budget constraints. Formulate the problem as an I.P. 3

4 Solution. For choice situations of 'yes-no', 'go-no-go' type, where the objective is to determine whether or not a particular activity is to be undertaken, integer binary variables that can take a value of 0 or 1, can be used to represent the decision variables. Here, we find that for each project, we want to find out whether it should be taken up or not, as such we define four decision variables x j (j = 1, 2, 3, 4) corresponding to each project, and x j = 1 if a if project is selected. 0 otherwise Then, the objective function and the constraints can be expressed in terms of the decision variables, to give the required formulation: Maximize c j x j a jt x j b t, for all period t. x j = 1 or 0, for all projects. asically, there are two algorithms to determine the optimal solution for an integer programming problem. One of these is the cutting plane algorithm devised by Gomory and the other is the branch & bound algorithm developed by Land & Doig. The next section concentrates on the cutting plane method. Cutting Plane Algorithm Historically, the first method for solving I.P.P. was the cutting plane method. This method is for the pure integer programming model. The procedure is, first, ignore the integer stipulations, and solve the problem as an ordinary LPP. If the solution satisfies the integer restrictions, then an optimal solution for the original problem is found. Otherwise, at each iteration, additional constraints are added to the original problem. These constraints are added to reduce or cut the solution space in every successive iteration, ruling out the current fractional solution, while ensuring that no integer solution is excluded in the process. The method terminates as soon as an integer-valued is obtained. 4

5 In this method, convergence is guaranteed in a finite number of iterations. To summarize the approach, a series of steps are stated below. Steps 1. Use the simplex method to find an optimal solution of the problem, ignoring the integer condition. 2. Examine the optimal solution. Terminate the iterations if all the basic variables have integer values. Otherwise, construct a Gomory's fractional cut from the row, which has the largest fractional part, and add it to the original set of constraints. Gomory's constraint - Σ f i x j -f I - Σ f i x j + S I = -f I Where: f I - fractional part S I - slack variable In case of a tie, you are at liberty to choose any one arbitrarily. "The difficulty in life is the choice." - George Moore 3. Now, find a new linear programming solution. If the solution thus obtained is integral valued, then this is the required optimal solution of the original I.P.P.; otherwise, return to step 2. Cutting Plane Method 5

6 Example 1 Maximize z = x 1 + 2x 2 2x 2 7 x 1 + x 2 7 2x 1 11 x 1, x 2 are integers 0 Solution. First, solve the above problem by applying the simplex method. After introducing slack variables, the standard form of LPP becomes Maximize z = x 1 + 2x 2 + 0x 3 + 0x 4 + 0x5 2x 1 + x 3 = 7 x 1 + x 2 + x 4 = 7 2x 1 + x 5 = 11 Initial basic feasible solution x 1 = 0, x 2 = 0, and z = 0 x 3 = 7, x 4 = 7, x 5 = 11 Table 1 c j c asic variables x 1 x 2 x 3 x 4 x 5 0 x x x z j c j

7 Key column = x 2 column Minimum (7/2, 7/1) = 7/2 Key row = x 3 row Pivot element = 2 x 3 departs & x 2 enters. Table 2 c j c asic variables x 1 x 2 x 3 x 4 x 5 2 x / /2 0 x / /2 0 x z j c j x 4 departs and x 1 enters. Table 3 c j c asic variables x 1 x 2 x 3 x 4 x 5 2 x / /2 (3 + 1/2) 1 x / /2 (3 + 1/2) 0 x (4 + 0) z j c j 0 0 1/2 1 0 This solution is not acceptable since all the basic variables must be integer valued. Thus, we construct a Gomory's fractional cut to get the desired optimal solution. Choose the largest fractional part under the X column. Here both the fractional parts are same. So either of the two may be used to develop Gomory's constraint. Taking first row as the source row, the corresponding equation is 0x 1 + 1x 2 + 1/2x 3 + 0x 4 + 0x 5 = 3 + 1/2 (1 + 0)x 2 + 1/2x 3 = 3 + 1/2 7

8 After applying the formula we get - 1/2 x 3-1/2-1/2 x 3 + x 6 = - 1/2 x 6 is a slack variable y adding the above equation in Table 3, we get the new table Table 4 c j c asic variables x 1 x 2 x 3 x 4 x 5 x 6 2 x / /2 1 x / /2 0 x x / /2 z j c j 0 0 1/ In the above table, there is a negative value under X column; therefore, apply the dual simplex method. Select the most negative value from X column, i.e., -1/2 Therefore, key row = x 6 row Key Column: Min z 3 - c a 43 Min 1/ /2 Key column = x 3 column x 6 departs & x 3 enters 8

9 Table 5 c j c asic variables x 1 x 2 x 3 x 4 x 5 x 6 2 x x x x z j c j The optimal solution is x 1 = 4, x 2 = 3 z = X 3 = 10 Example 2 Maximize z = x 1 + 4x 2 2x 1 + 4x 2 7 5x 1 + 3x 2 15 x 1, x 2 are integers 0 Solution. First, solve the above problem by applying the simplex method (try it yourself).the final simplex table is presented below. Final Simplex Table c j c asic variables x 1 x 2 x 3 x 4 9

10 4 x 2 1/2 1 1/4 0 7/4 (1 + 3/4) 0 x 4 7/2 0-3/4 1 39/4 (9 + 3/4) z j c j Taking first row as the source row, the corresponding equation is 1/2x 1 + 1x 2 + 1/4x 3 + 0x 4 = 1 + 3/4 1/2x 1 + (1 + 0)x 2 + (1-3/4)x 3 = 1 + 3/4 Gomory's constraint - (1/2x 1-3/4x 3 ) -3/4-1/2x 1 + 3/4x 3 + x 5 = -3/4 Table c j c asic variables x 1 x 2 x 3 x 4 x 5 4 x 2 1/2 1 1/ /4 0 x 4 7/2 0-3/ /4 0 x 5-1/2 0 3/ /4 z j c j Table c j c asic variables x 1 x 2 x 3 x 4 x 5 4 x x / /2 (4+1/2) 1 x / /2 (1+1/2) z j c j 0 0 5/

11 Taking second row as the source row, the corresponding equation is: 0x 1 + 0x 2 + 9/2x 3 + 1x 4 + 7x 5 = 4 + 1/2 or (4 + 1/2)x 3 + (1 + 0)x 4 + (7 + 0)x 5 = 4 + 1/2 Gomory's constraint -1/2x 3-1/2-1/2x 3 + x 6 = -1/2 Table c j c asic variables x 1 x 2 x 3 x 4 x 5 x 6 4 x x / /2 1 x / /2 0 x / /2 z j c j 0 0 5/ In the above table, there is a negative value under X column; therefore, apply the dual simplex method. Final Table c j c asic variables x 1 x 2 x 3 x 4 x 5 x 6 4 x x x x z j c j

12 The optimal solution is x 1 = 3, x 2 = 0 z = X 0 = 3 Graphical Method This section deals with the geometric representation of an integer programming problem. To illustrate the concept of cutting plane method through graphical method, consider again the following problem. Example Maximize z = x 1 + 4x 2 2x 1 + 4x 2 7 5x 1 + 3x 2 15 x 1, x 2 are integers 0 Solution. First, solve the above problem by applying the simplex method. After introducing slack variables, we have 2x 1 + 4x 2 + x 3 = 7 or x 3 = 7-2x 1-4x 2...(i) 5x 1 + 3x 2 + x 4 = 15 or x 4 = 15-5x 1-3x 2...(ii) Gomory's constraint - (1/2x 1-3/4 x 3 ) -3/4...(iii) Substituting the value of x 3 in equation (iii). - 1/2x 1 + 3/4 (7-2x 1-4x 2 ) -3/4-2x 1-3x (iv) Gomory's constraint -1/2x 3-1/2...(v) 12

Substituting the value of x 3 in equation (v) -1/2 (7-2x 1-4x 2 ) -1/2 x 1 + 2x 2 3.

13 Substituting the value of x 3 in equation (v) -1/2 (7-2x 1-4x 2 ) -1/2 x 1 + 2x (vi) The inclusion of two cuts( -2x 1-3x 2-6, x 1 + 2x 2 3) give the new corner point A where x 1 = 3, x 2 = 0 and z = 3 ranch & ound Method The branch & bound method can be used to solve problems containing a few integer valued variables. It can be applied to both mixed & pure integer programming problems. This method partitions the area of feasible solution into smaller parts until an optimal solution is obtained. If the number of variables is large, or if the LP solution to the problem is not optimal, then don't use the ranch & ound method, because the number of iterations required to solve such a problem may be too large. The method, in outline, is: 13

14 Steps 1. First, solve the given problem as an ordinary LPP. 2. Examine the optimal solution. Terminate the iterations if the optimal solution to the LPP satisfies the integer constraints. Otherwise, go to step Divide the problem into two parts. Problem 1: xk [t] max. z = cx ax b xk [t] x 0 Problem 2: xk [t] + 1 max. z = cx ax b xk [t] + 1 x 0 where [t] is the largest integer. 4. Now, solve problem 1 & 2 separately. 5. If for any of the sub-problems, optimal integer solution is obtained, then that problem is not further branched. Otherwise, move to step 3. ranch & ound Method Example Maximize z = x 1 + x 2 3x 1 + 2x 2 12 x 2 2 x 1, x 2 are integers 0 14

15 Solution. First, we solve the above problem by applying the simplex method. After introducing slack variables, we have 3x 1 + 2x 2 + x 3 = 12 x 2 + x 4 = 2 where x 3 and x 4 are slack variables. Initial basic feasible solution x 1 = 0, x 2 = 0, and z = 0 x 3 = 12, x 4 = 2 Final Table c j c asic variables x 1 x 2 x 3 x 4 1 x /3-2/3 8/3 1 x z j c j 0 0 1/3 1/3 x 1 = 8/3, x 2 = 2 Since the solution obtained is not an integer solution, we choose x 1 = 8/3 (2 + 2/3), which has a fractional value and divide the problem into the following two sub-problems (problem - II and problem - III) by adding two new constraints x 1 2 & x 1 3, because [x 1 ] = [2 + 2/3] = 2. Problem - II Maximize z = x 1 + x 2 3x 1 + 2x 2 12 x 2 2 x 1 2 x 1, x 2 0 Introducing slack variables 15

16 3x 1 + 2x 2 + x 5 = 12 x 2 + x 6 = 2 x 1 + x 7 = 2 where x 5, x 6 and x 7 are slack variables. Final Table c j c asic variables x 1 x 2 x 5 x 6 x 7 0 x x x z j c j Optimal solution to problem - II is x 1 = 2, x 2 = 2 Since all the variables in problem - II are integer valued, there is no need to branch this problem further. Problem - III Maximize z = x 1 + x 2 3x 1 + 2x 2 12 x 2 2 x 1 3 x 1, x 2 0 We use the two phase method to solve this problem. Introducing slack, surplus & artificial variables 3x 1 + 2x 2 + x 8 = 12 x 2 + x 9 = 2 x 1 - x 10 + A 1 = 3 16

17 where: x 8 and x 9 are slack variables. x 10 is a surplus variable. A 1 is an artificial variable. Final Table c j c asic variables x 1 x 2 x 8 x 9 x 10 1 x /2 0 3/2 3/2 0 x /2 1-3/2 1/2 1 x z j c j 0 0 1/2 0 1/2 x 1 = 3, x 2 = 3/2 The solution obtained is not integer valued. Since [x 2 ] = [1 + 1/2] = 1, we divide the problem - III into two parts by adding two new constraints x 2 1 & x 2 2. Thus, we form two subproblems (problem - IV and problem - V). Problem - IV Maximize z = x 1 + x 2 3x 1 + 2x 2 12 x 2 2 x 1 3 x 2 1 x 1, x 2 0 We use the two phase method to solve this problem. Introducing slack, surplus & artificial variables 3x 1 + 2x 2 + x 11 = 12 x 2 + x 12 = 2 x 1 - x 13 + A 1 = 3 x 2 + x 14 = 1 17

18 where: x 11, x 12 & x 14 are slack variables. x 13 is a surplus variable. A 1 is an artificial variable. Final Table c j c asic variables x 1 x 2 x 11 x 12 x 13 x 14 0 x / /3 1/3 0 x x / /3 10/3 1 x z j c j 0 0 1/ /3 x 1 = 10/3, x 2 = 1 Problem - V Maximize z = x 1 + x 2 3x 1 + 2x 2 12 x 2 2 x 1 3 x 2 2 x 1, x 2 0 The solution to problem - V is infeasible. So we will not consider this problem. ut the solution to problem to problem - IV is not integer valued. Since [x 1 ] = [3 + 1/3] = 3, we divide the problem - IV into two parts by adding two new constraints x 1 3 & x 1 4. Thus, we form two sub-problems (problem - VI and problem - VII). Problem - VI Maximize z = x 1 + x 2 18

19 3x 1 + 2x 2 12 x 2 2 x 1 3 x 2 1 x 1 3 x 1, x 2 0 Introducing slack, surplus & artificial variables 3x 1 + 2x 2 + x 15 = 12 x 2 + x 16 = 2 x 1 - x 17 + A 1 = 3 x 2 + x 18 = 1 x 1 + x 19 = 3 where: x 15, x 16, x 18 and x 19 are slack variables. x 17 is a surplus variable. A 1 is an artificial variable. Final Table c j c asic variables x 1 x 2 x 15 x 16 x 17 x 18 x 19 0 x x x x x z j c j This problem has multiple optimal solution. As x 2 is zero, it enters into the basis. The optimal solution is x 1 = 3, x 2 = 1 19

20 Since all the variables in problem - VI are integer valued, there is no need to branch this problem further. Problem - VII Maximize z = x 1 + x 2 3x 1 + 2x 2 12 x 2 2 x 1 3 x 2 1 x 1 4 x 1, x 2 0 Introducing slack, surplus & artificial variables 3x 1 + 2x 2 + x 20 = 12 x 2 + x 21 = 2 x 1 - x 22 + A 1 = 3 x 2 + x 23 = 1 x 1 - x 24 + A 2 = 4 where: x 20, x 21 and x 23 are slack variables. x 22 and x 24 are surplus variables. A 1 and A 2 are artificial variables. Final Table c j c asic variables x 1 x 2 x 20 x 21 x 22 x 23 x 24 1 x / /2 0 0 x / /2 2 1 x x / /2 1 0 x

z j c j 0 0 1/2 0 0 0 1/2 The optimal solution is x 1 = 4, x 2 = 0 Since all the variables in problem - VII are integer valued, there is no need

21 z j c j 0 0 1/ /2 The optimal solution is x 1 = 4, x 2 = 0 Since all the variables in problem - VII are integer valued, there is no need to branch this problem further. The history of the complete branch and bound solution is displayed by means of the following tree like diagram. 21

Metode Kuantitatif Bisnis. Week 4 Linear Programming Simplex Method - Minimize

Metode Kuantitatif Bisnis. Week 4 Linear Programming Simplex Method - Minimize Metode Kuantitatif Bisnis Week 4 Linear Programming Simplex Method - Minimize Outlines Solve Linear Programming Model Using Graphic Solution Solve Linear Programming Model Using Simplex Method (Maximize)