C&O 370 Deterministic OR Models Winter 2011

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1 C&O 370 Deterministic OR Models Winter 2011 Assignment 1 Due date: Friday Jan. 21, 2011 Assignments are due at the start of class on the due date. Write your name and ID# clearly, and underline your last name. Contents 1 Problem 1: LP Review/Duality/AMPL 10 Marks 2 2 LP Formulation and AMPL; a Transportation Problem - 10 Marks 5 3 LP Formulation Manufacturing 10 Marks 8 1

2 1 Problem 1: LP Review/Duality/AMPL 10 Marks Consider the primal linear programming problem, (P), given in AMPL form. (The file with the AMPL problem is also available here: examp1.mod.) var x1; var x2; var x3; var x4; var x5; var x6; var x7; var x8; minimize Expense: +(-6)*x1+(-9)*x2+(12)*x3+(-14)*x4+(-23)*x5+(5)*x6+(-18)*x7 +(-2)*x8; subject to T1: +(-3)*x1+(5)*x2+(3)*x3+(4)*x4+(-2)*x5+(2)*x6+(9)*x7 <= -6; subject to T2: +(5)*x1+(2)*x2+(15)*x3+(-5)*x4+(6)*x5+(3)*x6+(6)*x7 -x8= 2; subject to T3: +(3)*x1+(0)*x2+(-2)*x3+(3)*x4+(10)*x5+(0)*x6+(4)*x7 <= -44; subject to T4: +(-3)*x1+(6)*x2+(-1)*x3+(5)*x4+(-7)*x5+(-5)*x6+(6)*x7 +(3)*x8 <= -9; subject to xlimit1: x1 >=0; subject to xlimit2: x2 >=0; subject to xlimit3: x3 >=0; subject to xlimit4: x4 <=0; subject to xlimit5: x5 <=0; subject to xlimit6: x6 >=0; subject to xlimit7: x7 <=0; 1. Write down the dual problem (D) of (P). (In the dual problem, the constraints should be a mixture of,, = constraints. The variables should be a mixture of,, and free.) Solution 1.1 The primal problem is: min c T x s.t. Ax ineq b b, x ineq x 0, x R 8, where ineq b, ineq x, represent the appropriate inequalities/equality for the constraints and the variables, respectively. (Note that this problem was generated randomly using the file available here: lpexamp1.m.) Then, the dual program (D) is: max b T y s.t. c ineq x A T y, y ineq b 0, y R 4. Note that the inequalities for constraints go with the inequaities for variables. (With 2

3 an equality constraint corresponding to a free variable.) We get the following: max subject to ( ) y = y y 1, y 3, y 4 0, y 2 free. 2. Write down all the complementary slackness conditions for (P),(D). Solution 1.2 We can write the primal constraint as the equality Ax s = b and the dual constraint as the equality A T y + z = c, where the slack variables s = Ax b ineq b 0, z = c A T y ineq x 0 have the appropriate signs given by their corresponding inequalities. Then the complementary slackness constraints are given by This is equivalent to x T z = s T y = 0. x z = 0 R 7, s y = 0 R 4, where corresponds to the componentwise/elementwise product (also called the Hadamard product). Therefore, all the inequalities can be expressed as x (A T y c) = 0 R 8, (Ax b) y = 0 R 4. 3

4 Explicitly, the complementary slackness conditions for the primal (P) are ( [ 3 ] ) x y ( 6) = 0 ( [5 ] ) x y ( 9) = 0 ( [3 ] ) x y 12 = 0 ( [4 ] ) x y ( 14) = 0 ( [ 2 ] ) x y ( 23) = 0 ( [2 ] ) x y 5 = 0 ( [9 ] ) x y ( 18) = 0 ( [0 ] ) x y ( 2) = 0 and the complementary slackness conditions for the dual (D) are ( [ 3 ] ) y x ( 6) ( [5 ] ) y x 2 ( [ 3 ] ) y x ( 44) ( [ 3 ] ) y x ( 9) = 0 = 0 = 0 = 0 3. Consider the following possible vector of solutions for (P). x = ( ) T First, use the given data and confirm that this vector provides a feasible solution to (P). Then, use part 2 to find a solution to (D) and also, to show (provide the details) whether or not the vector x above is optimal for (P). Solution 1.3 Substituting the given (approximate) solution x into the equality constraint T2 and solving for the unknown x 8 yields x All primal constraints except T1 are active at x. By the complementary slackness conditions for y, we conclude that the first component of the dual variable is y 1 = 0. 4

5 On the other hand, all entries of x except x 1, x 2, x 7, x 8 are approximately equal to 0. The complementary slackness conditions using x imply that the (four) 1, 2, 7, 8 constraints of the dual are active. This yields the linear system of equations y 2 y 3 = y The unique solution to this system yields the optimal dual solution 0 y = 5/ /12 (1) Since this vector has the correct signs, we conclude that we have a certificate of optimality for x. 4. Is the dual optimal solution unique? Is the primal optimal solution unique? Solution 1.4 Uniqueness holds for the dual, since the solution of the equations in (1) is unique. For the primal, the value for y above implies that the last three constraints hold as equality. Moreover, all dual constraints except those corresponding to x 3, x 4, x 5, x 6 are active at y. This implies that we have three equations with four unknowns, i.e. we only conclude that x 3 = x 4 = x 5 = x 6 = 0. We now have to check whether there are any other solutions x with the correct sign pattern. Another optimal solution (to 4 decimals) (confirmation of optimality can be done as above for x) is x = ( ) T. 2 LP Formulation and AMPL; a Transportation Problem - 10 Marks Suppose that there are two canning plants (at Halifax, Winnipeg) and three markets (at Montreal, Toronto, Vancouver). Table 1 provides the data; shipping distances are in thousands of KM, shipping costs are assumed to be \$90.00 per case per thousand KM, and supplies (and demands) are in numbers of cases. 5

6 Markets Montreal Toronto Vancouver Plants Shipping Distances Supplies Halifax Winnipeg Demands Table 1: shipping data for Problem 2 1. Formulate an LP problem for minimizing the transportation cost while meeting customer demand and satisfying the supply constraints. (Your solution should include a description of the sets, the main decision variables, and the constraints.) Solution 2.1 The sets are: set plants; set markets. The decision variables are: x ij - the number of cases to be shipped from plant i to market j. Then the LP becomes: min 1.244x x x x x x 23 objective s.t. 3 j=1 x 1j 350 supply constraint 3 j=1 x 2j 600 supply constraint 2 i=1 x i1 325 demand constraint 2 i=1 x i2 300 demand constraint 2 i=1 x i3 275 demand constraint x ij 0, i, j Note that the supply is 950 cases while the demand is only 900 cases. For transportation problems one often sets up a dummy market with an appropriate demand to make the supply and demand equal. In that case, one can replace the inequality constraints with equality constraints. 2. Solve the LP using the AMPL software. Submit a printed version of your LP model (including any data files), and a log of your session on AMPL that shows (i) the optimal value, and (ii) an optimal solution. Solution 2.2 The file trans.mod is: set Plant; set Market; # set of plants # set of markets param Supply {i in Plant}; param Demand {j in Market}; param ShipDist {i in Plant, j in Market}; # number of cases # number of cases # thousand KMs 6

7 var x {Plant, Market} >= 0; # cases to ship from each plant to each market minimize TransDist: sum{ i in Plant, j in Market } ShipDist[i,j] * x[i,j]; subject to PlantSupply {i in Plant}: sum{ j in Market } x[i,j] <= Supply[i]; subject to MarketDemand {j in Market}: sum{ i in Plant } x[i,j] >= Demand[j]; The file trans.dat is: set Plant := Halifax Winnipeg; set Market := Montreal, Toronto, Vancouver; param: Supply := Halifax 350 Winnipeg 600 ; param: Demand := Montreal 325 Toronto 300 Vancouver 275 ; param ShipDist: Montreal Toronto Vancouver := Halifax Winnipeg ; The output from AMPL is: ILOG AMPL , licensed to "university-waterloo, canada". AMPL Version (SunOS 5.9) ampl: model trans.mod ampl: data trans.dat ampl: solve; ILOG CPLEX , licensed to "university-waterloo, canada", options: e m b q CPLEX : optimal solution; objective dual simplex iterations (0 in phase I) ampl: display x; x := Halifax Montreal 325 Halifax Toronto 25 Halifax Vancouver 0 7

8 Winnipeg Montreal 0 Winnipeg Toronto 275 Winnipeg Vancouver 275 ; ampl: display PlantSupply; PlantSupply [*] := Halifax Winnipeg 0 ; ampl: display MarketDemand; MarketDemand [*] := Montreal Toronto Vancouver ; ampl: quit; 3. Repeat part 2 but replace the first number of supplies (350) and the first number of demands (325) using the two three digit numbers formed from: the first 3 digits of your student ID, and the second 3 digits of your student ID. The larger of these two numbers replaces the supply and the smaller number replaces the demand. 3 LP Formulation Manufacturing 10 Marks A liquor company produces and sells two kinds of liquor: blended whiskey and bourbon. The company purchases intermediate products in bulk, purifies them by repeated distillation, mixes them, and bottles the final product under their own brand names. In the past, the firm has always been able to sell all that it produced. The firm has been limited by its machine capacity and available cash. The bourbon requires 3 machine hours per bottle while, due to additional blending requirements, the blended whiskey requires 4 hours of machine time per bottle. There are 20,000 machine hours available in the current production period. The direct operating costs, which are principally for labor and materials, are \$3.00 per bottle of bourbon and \$2.00 per bottle of blended whiskey. The working capital available to finance labor and material is \$4000; however, 45% of the bourbon sales revenues and 30% of the blended-whiskey sales revenues from production in the current period will be collected during the current period and be available to finance operations. The selling price to the distributor is \$6 per bottle of bourbon and \$5.40 per bottle of blended whiskey. 8

9 1. Formulate a linear program that maximizes contribution subject to limitations on machine capacity and working capital. (Your solution should include a description of the sets, the main decision variables, and the constraints.) Solution 3.1 The sets are the (two) types of whiskey and the (two) types of constraints (constraints on machine hours and available cash). The variables are the number of bottles of each time of whiskey. We denote the number of bottles for the blended whiskey and the bourbon by x W and x B, respectively. max 5.4x W + 6x B objective s.t. 4x W + 3x B machine hours constraint 2x W + 3x B (.3)(5.4)x W + (.45)(6)x B available cash constraint x W, x B 0 Note that the second constraint is equivalent to.38x W +.3x B The slope of the level line for the objective function is: slope obj = 5.4/6 =.9; while the slopes for the constraints are slope hrs = 4/3 and slope cash =.1407, respectively. 2. What is the optimal production mix to schedule? Solution 3.2 From looking at the graph of the feasible set and the level lines of the objective function, we see that only the first constraint is active and that the optimal solution (production mix) is x W = 0, x B = 20000/3. 3. Can the selling prices change without changing the optimal production mix? Solution 3.3 As mentioned above, the slopes slope obj = 5.4/6 =.9 > slope hrs = 4/3. Therefore, we can change the selling price as long as the inequality of the slopes does not change, i.e. as long as slope obj = ( W )/(6 + B ) slope hrs = 4/3. The intervals for the objective coefficients are: < c W 8 and 4.05 c B < Suppose that the company could spend \$400 to repair some machinery and increase its available machine hours by 2000 hours. Should the investment be made? Solution 3.4 The increase in the income would be \$2000(6)/3, as the constraint for available cash is not violated. Therefore, it is worthwhile to ivest the \$ What interest rate could the company afford to pay to borrow funds to finance its operations during the current period? Solution 3.5 The cash constraint is not active. Therefore, it does not pay to borrow funds. 9

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