Foundations of Operations Research (Prof. Roberto Cordone) September 11th, 2014 Time available: 2 hours and 30 minutes

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1 Foundations of Operations Research (Prof. Roberto Cordone) September 11th, 2014 Time available: 2 hours and 30 minutes Note: Report all computational passages (unmotivated results will be ignored). Exercise 1 - A catering service must organize two coffee-breaks, for different customers. They have already prepared 30 cookies, whose price is 1.5 Euros each, and 50 fruit tarts, whose price is 1 Euro each. The catering service is free to mix the two kinds of desserts, without exceeding the budget of the two customers, which is 30 Euros for the first one and 40 Euros for the second. In addition, the first customer wants the cookies to be at least twice as many as the tarts, whereas the second wants a balanced distribution, with a difference not larger than 5 units between cookies and tarts. Give a mathematical programming formulation for the problem of maximizing the revenue of the catering service (without solving it). Exercise 2 - The following AMPL model describes the problem of maximizing the profit of an ice-cream producer. set F set R # Flavours # Resources set F := Cream Lemon Chocolate set R := Milk Sugar WorkTime param a{r} # Resource availability param c{r} # Unitary resource cost param d{f} # Flavour demand param pi # Price of the ice-cream param k{r,f} # Amount of resource # required for flavour var x{f} >= 0 maximize f : pi * sum{f in F} x[f] - sum{i in R} c[r] * k[r,f] * x[p] subject to Resource{r in R} : sum{f in F} k[r,f] * x[f] <= a[r] subject to Demand{f in F} : x[f] <= d[f] end param a := Milk 50 # litres Sugar 100 # kgs WorkTime 900 # hours param c := Milk 0.5 # Euros/litre Sugar 0.8 # Euros/kg WorkTime 1/60 # Euros/hour param d := Cream 13 # Euros/litre Lemon 10 # Euros/kg Chocolate 15 # Euros/hour param pi := 5 # Euros/kg param k := Cream Lemon Chocolate Milk Sugar WorkTime end The corresponding output is:

2 f = x [*] := Cream Lemon Chocolate x.rc [*] := Cream Lemon Chocolate Resource.dual [*] := Milk Sugar WorkTime Resource.slack [*] := Milk Sugar WorkTime Demand.dual [*] := Cream Lemon Chocolate Demand.slack [*] := Cream Lemon Chocolate Answer the following management questions, motivating your answers: 1. What is the production level of Chocolate? 2. What amount of each resource is left after production? 3. If the production cost of Chocolate should decrease by 0.5 Euros/kg, would the optimal production levels change? 4. What is the largest cost that it would be profitable to pay for additional WorkTime? Exercise 3 - Define the minimum spanning tree problem. Describe Prim s algorithm to solve the problem. Discuss the complexity of Prim s algorithm as a function of the graph size, specifically describing the data structures that allow to improve it with respect to the most trivial implementation. Exercise 4 - Given the graph with the following cost matrix, find the shortest paths from node 5 to all the other nodes Which algorithm should be used, and why? What is the complexity of this algorithm with respect to the graph size?

3 Exercise 5 - Given an undirected graph and two vertices s and t, define the concept of s t cut. Define the maximum flow problem. Prove that, given every feasible solution of the maximum flow problem, the flow across every s t cut is not larger than the capacity of the cut. Exercise 6 - Define the Linear Programming problem. Solve graphically the following instance of the problem: maxz = x 1 +2x 2 2x 1 +2x 2 1 +x 1 2x x 1 +2x 2 5 x 1,x 2 0 Exercise 7 - Write the dual of the following Linear Programming problem: maxz = 3x 1 x 2 x 1 x 2 2 x 1 +2x 3 1 x 1 0 x 2 R x 3 0 Apply the complementary slackness conditions to find the dual basic solution corresponding to x 1 = 0, x 2 = 2, x 3 = 1/2. Based on the two corresponding solutions, can you state whether the primal one is optimal? Is it optimal? Exercise 8 - Given the following Integer Linear Programming problem minz = x 1 +x x 1 2x 2 +x 3 = 9 2 3x 1 +x 2 +x 4 = 3 x 1,x 2,x 3,x 4 N solve its continuous relaxation with the simplex method. Is it integer? If it is not, find a Gomory cut in integer and fractionary form. Add the Gomory cut to the optimal tableau and reoptimize. Repeat until the solution becomes integer.

4 Solutions Please, write to me if you find mistakes in the solutions: that will be to the common good. Exercise 1 - The following is one of the possible models. The decision variables x c1, x c2, x t1, x t2 determine the number of cookies and tarts sold to the first and second customer. They are all integer and nonnegative. The objective is the total revenue of the catering service: maxf = 1.5[Euros/cookie] (x c1 +x c2 )[cookies]+1.0[euros/tart] (x t1 +x t2 )[tarts] The cookies and tarts sold should not exceed the available number. x c1 [cookie]+x c2 [cookie] 30[cookie] x t1 [tarts]+x t2 [tarts] 50[tarts] The money required from each customer should not exceed the corresponding budget. 1.5[Euros/cookie] x c1 [cookies]+1.0[euros/tart] x t1 [tarts] 30[Euros] 1.5[Euros/cookie] x c2 [cookies]+1.0[euros/tart] x t2 [tarts] 40[Euros] For the first customer, the cookies should be at least twice as many as the tarts for the second one, both possible differences between the two numbers (cookies minus tarts and tarts minus cookies) should not exceed 5. x c1 2x t1 x c2 x t2 5 x t2 x c2 5 All variables should be nonnegative integers. x c1,x c2,x t1,x t2 N As a rule of thumb, I subtracted 1 point for not defining the variables, and for reverting the sign of the constraints ( at most instead of at least ), 0.5 points for minor mistakes. Exercise 2-1. The production level of Chocolate is x[chocolate] = Milk and WorkTime are completely exhausted there are Resource.slack[Sugar] = kgs left of Sugar. 3. The Chocolate ice-cream is not produced because its reduced cost is negative (x.rc[chocolate] = ), meaning that, if one should force the production of one (small) unit of Chocolate the objective function (profit) would decrease. Directly reducing the cost of Chocolate by 0.5 Euros would increase the profit by the same amount. Therefore, the production would become profitable, and the optimal production levels would indeed change.

5 4. Increasing the availability of WorkTime by one (small) unit would increase the profit by Resource.dual[WorkTime] = Euros. Since WorkTime is measured in hours, the largest profitable cost of additional WorkTime is Euros/hour (assuming that the number of hours is not so large to affect the optimal basis. The value of each question is 1 point. Exercise 3 - For the MST, see lesson 4, slide 4 (1 point). For Prim s algorithm, see lesson 4, slides 10, 19, 20 (2 point). For its complexity, see lesson 4, slides (1 point). Exercise 4 - The graph has only nonnegative cost arcs, and it has circuits. Both Dijkstra s and Floyd-Warshall s algorithms could be used, but the former is more efficient, and the problem requires only the shortest paths from one node to the others. So, Dijkstra s algorithm is the correct choice (1 point). The complexity of Dijkstra s algorithm is O ( n 2), where n is the number of nodes of the graph (1 point). Application of the algorithm (2 points): Step 0: l j π j Step 1: mark node 5 and update labels. l j π j Step 2: mark node 3 and update labels. l j π j Step 3: mark node 1 and update labels. l j π j Step 4: mark node 4 and update labels. Step 5: mark node 2 and terminate. l j π j

6 Exercise 5 - For the concept of s t cut, see lesson 8, slide 7 (1 point). For the maximum flow problem, see lesson 8, slide 4 (1.5 point). For the proof, see lesson 8 slide 9 (1.5 points). Exercise 6 - For the definition of Linear Programming, see lesson 10, slide 6 (1 point). The optimal solution is x = (1,3/2) with z = 2 (3 points). Exercise 7 - The dual problem is (1 point): minw = 2y 4 +y 5 y 4 +y 5 3 y 4 = 1 2y 5 0 y 4 0 y 5 0 The complementary slackness conditions x i y i = 0 for i = 1,...,5 are (1 point): x 1 (y 4 +y 5 3) = 0 x 2 (y 4 1) = 0 x 3 (2y 5 ) = 0 (2 x 1 +x 2 ) y 4 = 0 (x 1 +2x 3 1) y 5 = 0 The dual basic solution corresponding to x 1 = 0, x 2 = 2, x 3 = 1/2 is (1 point) 0 (y 4 +y 5 3) = 0 2 (y 4 1) = 0 { y 4 = 1 1/2 (2y 5 ) = 0 y 5 = 0 (2 0 2) y 4 = 0 (0+2 1/2 1) y 5 = 0 The dual solution is unfeasible, because it does not satisfy the first constraint y 4 + y 5 3. Therefore, the primal solution is not optimal (1 point). Exercise 8 - The starting tableau of the problem is z x 1 x 2 x 3 x 4 z /2 7/ The current solution is already optimal, but it is not integer (1 point).

7 There is a single row with a fractionary right-hand-side, that can be used to generate a Gomory cut, that is row 1. The Gomory cut in integer form is (0.5 points) 3x 1 2x 2 +x 3 4 The Gomory cut in fractionary form is (0.5 points) The augmented tableau becomes 1 2 x z x 1 x 2 x 3 x 4 x 5 z /2 7/ /2-1/ Applying the dual simplex method on pivot element a 31 (only negative row, only negative element) yields (2 points): z x 1 x 2 x 3 x 4 x 5 z which is integer: x = (1,0,1,0) with z = 1.