MA231-2020-exam-solutions-corrected2

MA Operational Research Methods Exam Solutions Summer Question (a) Consider the following Markov chain diagram, with transition probabilities on each arc: (i) Explain in plain words what is a recurrent state and what is a transient state. [ marks] Solution : A recurrent state is a state where the probability of returning to this state at some point in time in the future is 1 . A transient state is a state where the probability of returning to it at some point in the future is strictly less than 1 (eventually it will be absorbed to the recurrent states and never return). (ii) For the Markov chain above identify the transient, recurrent and absorbing states. [ marks] Solution : Transient: 1 , 2 , 3 , 4 , 5 ; Recurrent: 6 , 7 ; Absorbing: 6 , 7 . (iii) List the communicating classes. [ marks] Solution : Communicating classes: { 1 } , { 2 } , { 3 } , { 4 } , { 5 } , { 6 } , { 7 } . (iv) What is the probability of terminating in state 6 if the starting state is 3 ? [ marks] © LSE ST /MA Page of

Solution : Let f i = the probability of terminating in state 6 given that the initial state is i . Observing that, starting from 3 , we will never reach state or state , we are not interested in f 1 and f 2 . We have: f 3 = 0 . 5 f 4 + 0 . 1 f 6 + 0 . 4 f 7 f 4 = 0 . 3 f 5 + 0 . 6 f 6 + 0 . 1 f 7 f 5 = 1 f 6 We know that: f 6 = 1 and f 7 = 0 , this gives us: f 3 = 0 . 5 f 4 + 0 . 1 f 4 = 0 . 3 f 5 + 0 . 6 f 5 = 1 from which we easily conclude f 3 = 0 . 55 . (v) Starting from state , what is the expected number of times that arc (3 , 4) will be traversed before the chain terminates in an absorbing state? [ marks] Solution : Since { 3 } is a communication class, and is transient, we will never cross arc (3 , 4) more than once. Hence, the expected number of times we will cross arc (3 , 4) starting from state equals the probability of crossing (3 , 4) . This is equal to the product of the probability of reaching state 3 starting from state 1 and the probability of going from state 3 to state 4 . Note that the probability of reaching state from 1 is 0 . 8 (after one step we reach state with probability , and after that we either go to with probability 0 . 8 , or we get absorbed by 7 with probability 0 . 5 . Hence the expected number of times we cross (3 , 4) is 0 . 8 · p 34 = 0 . 8 · 0 . 5 = 0 . 4 . (b) Consider the following transition matrix for a Markov chain with two states, 1 and 2 : P = 0 . 4 0 . 6 0 . 2 0 . 8 (i) Given that the chain is in state 2 at time t , what is the probability that it will be in state 2 at time t + 3 ? [ marks] Solution : We want P 3 22 : P 3 22 = ( nd row of P 2 ) × ( nd column of P ) We need P 2 : P 2 = 0 . 28 0 . 62 0 . 24 0 . 76 Thus, P 3 22 = (0 . 24 0 . 76) × (0 . 6 0 . 8) = 0 . 752 (ii) In the long run, what is the proportion of time that the chain is in state 1 ? [ marks] © LSE ST /MA Page of

A B C G D E F t s 0 5 5 5 5 5 8 5 4 10 3 0 0 5 5 5 10 13 17 20 The gure illustrates the earliest start time of each activity at the top of each node, as well as the critical path, represented by the boldfaced arcs. The minimum duration of the project is 20 , de ned by the critical path s - A - B - C - G - t . (iii) Determine the free oat and the total oat of each activity. [6 marks] Solution : The only activities with a non-zero free oat are E , with a free oat of 17 - (10+5) = 2 , and F with a free oat of 17 - (5 + 10) = 2 . TO determine the total oats, we run the CPA algorithm in reverse. The next gure shows the network with the latest start time of each activity. A B C G D E F t s 0 5 5 5 5 5 8 5 4 10 3 0 0 7 5 7 12 13 17 20 Task Earliest Latest Free Total start start oat oat A B C D E F G (b) A patisserie, which has a franchise from a large chain, in any given day bakes its signature cakes until am. Each cake costs the shop £ to make. Orders from restaurants arrive starting from am through the morning, and deliveries are made in the afternoon. Demand of signature cakes is exponentially distributed with mean . If not enough cakes are baked to satisfy demand, the shop can purchase more cakes from another shop from the chain, at a cost of £ . If too many cakes are baked in one day, in the evening the shop will sell them, at a loss, for £6 each, and they are certain to sell all the excess. How many signature cakes should the shop bake in every given day? (Recall that the exponential distribution with mean μ has probability density function μ - 1 e - x/μ ) and cumulative density function 1 - e - x/μ .) [8 marks] © LSE ST /MA Page of

Solution : This is a newsvendor problem with unit purchase cost c = 10 , shortage cost p = 12 , and holding cost h = - 6 . We need to nd S * such that F ( S * ) = p - c p + h = 1 3 . That is 1 - e - S * 100 = 1 3 . Reordering and taking the logarithm we get - S * 100 = ln 2 3 , therefore S * = 100 ln 3 2 = 40 . 5 . © LSE ST /MA Page of

[ point] When n = 2 , Tom needs to decide on water, where V 2 = 4 and U 2 = 5 . So he cannot take food with him unless s = 4 , 5 , 6 , 7 , f (2 , s ) = max x { 5 x + f (1 , s - 4 x ) } s x = 0 x = 1 x * f (2 , s ) f (1 , 0) = 0 – f (1 , 1) = 0 – f (1 , 2) = 0 – f (1 , 3) = 4 – f (1 , 4) = 4 +f( , ) = f (1 , 5) = 4 +f( , ) = 6 f (1 , 6) = 4 +f( , ) = f (1 , 7) = 4 +f( , ) = [ points] When n = 3 we have V 3 = 2 and U 3 = 6 and the state is s = 7 : s x = 0 x = 1 x * f (3 , s ) f (2 , 7) = 9 6 + f (2 , 5) = 11 Thus, Tom should put the clothes and the water but not the food (b) In a small local airport, check-in is performed by a single airline employee. Before the morning ight, an average of passengers join the check-in queue every six minutes. The airline of cer takes, on average, seconds during the check-in process for each passenger. Assume that interarrival times and service times are both exponentially distributed. (i) What is the expected number of passengers in the check-in system? [ marks] (ii) How long, on average, does a passenger stay in the queue? [ marks] Solution : Consider a minute as the time unit. We have λ = 5 / 6 , μ = 3 / 2 and ρ = 5 / 9 . The expected number of costumers in the check-in system is equal to ρ/ (1 - ρ ) = (5 / 9) / (4 / 9) = 1 . 25 . The expected queueing time is equal to λ μ ( μ - λ ) = 5 / 6 3 / 2 · (3 / 2 - 5 / 6) = 5 / 6 minutes or 50 seconds. Question (a) Players A and B play a game of cards. Each starts the game holding 3 cards: Ace, Two and Three. First, A selects one of the cards and puts it on the table (facedown) without showing it to B. Then, B does the same. An Ace is worth point, a Two is worth points and a Three is worth points. When they turn their cards, if they are different then the player with the card that has the highest points collects in £s the difference in points. (For example, if A has an Ace and B has a Three, then B gets £ from A.) If the cards are the same, then for two Three’s, A collects £ from B, for two Two’s, nothing happens, and for two Aces, B collects £ from A. (i) Write the game in strategic form. [ marks] © LSE ST /MA Page of

Solution : The strategic form of the game is as follows, where A is the row player and B is column player: ( points) B 1 B 2 B 3 A 1 - - - - A 2 + - - A 3 + + + + * + + * + (ii) Find each player’s optimal strategies. [ marks] Solution : The game has a solution in pure strategies: ( A 3 , B 2 ) . So, A should play Three and B should play Two. (iii) To make the game fair, how much should each player pay/receive to enter the game? [ marks] Solution : The value of the game is +1 , which is unfair to B . Thus to make the game fair: A should pay £ to enter and B should receive £ to enter. (b) Consider the following strategic form of a zero-sum game between row player R and column player C (as always, the entries represent the payoffs to the row player). C C C C C C6 R - R -6 R - R - - - R - (i) Reduce the game matrix using dominance. [ marks] Solution : We reduce the matrix by dominance: R 3 dominates R 1 , R 2 (strictly); R 5 dom- inates R 4 (strictly). Then: C 2 dominates C 5 ; C 4 dominates C 3 . The resulting matrix is: C 1 C 2 C 4 C 6 R 3 - R 5 - (ii) Write down the linear programming formulation of the reduced game from player R’s per- spective. [ marks] Solution : The linear programming formulation from R ’s perspective is as follows: max u u ≤ 4 x 3 u ≤ 3 x 3 + x 5 u ≤ - 4 x 3 + 2 x 5 u ≤ 5 x 3 - 2 x 5 1 = x 3 + x 5 x 3 , x 5 ≥ 0 © LSE ST /MA Page 8 of

Question Consider the problem of nding a minimum cost diet containing at least units of protein and units of carbohydrates. Four foods are available with the following nutritional content and costs. Food Protein/kilo Carbohydrate/kilo Cost in £/kilo 8 8 (a) Formulate the minimum cost diet problem as a linear programming problem. [ marks] Solution : min 18 x 1 +8 x 2 +13 x 3 +12 x 4 2 x 1 +3 x 2 +4 x 3 +5 x 4 ≥ 70 5 x 1 +4 x 2 +3 x 3 +2 x 4 ≥ 90 x 1 , . . . , x 4 ≥ 0 (b) Write down the dual of the problem you formulated in part (a), and solve it by drawing a - dimensional diagram. Find the corresponding solution to the primal problem. [6 marks] Solution : max 70 y 1 +90 y 2 2 y 1 +5 y 2 ≤ 18 3 y 1 +4 y 2 ≤ 8 4 y 1 +3 y 2 ≤ 13 5 y 1 +2 y 2 ≤ 12 y 1 , y 2 ≥ 0 x 1 1 2 3 4 x 2 1 2 3 4 The optimal solution is a the intersection of constraints 2 and 4 , given by the solutions to the equations 3 y 1 +4 y 2 = 8 5 y 1 +2 y 2 = 12 which is y * = ( 16 7 , 2 7 ) . The corresponding primal solution is obtained by solving the system 3 x 2 +5 x 4 = 70 4 x 2 +2 x 4 = 90 © LSE ST /MA Page of

which gives the point ¯ x = (0 , 5 7 , 0 , 155 7 ) . Since ¯ x is nonnegative, it is feasible, which proves that ¯ x and ¯ y are optimal primal and dual solutions. (c) Use your solution to answer the following questions. (i) There are two foods that will not be used in any minimum cost solution, irrespective of the quantity of proteins or carbohydrates needed in a diet. Which ones? Why? [ marks] Solution : The dual constraints relative to x 1 and x 3 are redundant, hence they can never be de ning in any basic feasible solution. It follows that x 1 and x 3 will be 0 in every optimal solution, irrespective of what the objective function is. Hence foods and are never used in a minimum cost solution. (ii) Due to your new-found passion for body building, your protein requirement has doubled. How much does your daily food cost increase? [ marks] Solution : The RHS of the rst constraint increases by 70 . We can check that this is within the allowable increase, since the solution to the system 3 x 2 +5 x 4 = 140 4 x 2 +2 x 4 = 90 is x 2 = 85 / 7 , x 4 = 145 / 7 , which is nonnegative. Hence the cost of the diet increases by 70 · ¯ y 1 = 160 . (iii) It turns out that food is “really good for you”, so you should have at least one unit of it. By how much does the minimum cost of a diet increase due to this new constraint? (This is with respect to the original problem, not with the doubling of the protein requirement at point (ii).) [ marks] Solution : If we introduce the constraint x 1 ≥ 1 , at the optimum we will have x 1 = 1 , since the optimal value is currently x 1 = 0 . Hence we will need to solve the LP min 8 x 2 +13 x 3 +12 x 4 3 x 2 +4 x 3 +5 x 4 ≥ 68 4 x 2 +3 x 3 +2 x 4 ≥ 85 x 2 , . . . , x 4 ≥ 0 Note that the feasible region of the dual is unchanged, since the dual constraint corre- sponding to x 1 was redundant, and that the solution ¯ y is still optimal, since the solution to the system 3 x 2 +5 x 4 = 68 4 x 2 +2 x 4 = 85 is nonnegative. It follows that the value of the solution increases by 18 - 2¯ y 1 - 5¯ y 2 = 12 . (d) More generally, suppose you have n foods to choose from, j = 1 , . . . , n , each with unit cost c j , and m nutrients, i = 1 , . . . , m , that you need in your diet, each with minimum requirement b i . Let a ij be the content of nutrient i in one unit of food j . (i) Write down an LP to determine a minimum-cost diet satisfying the nutritional requirements. [ marks] © LSE ST /MA Page of

Question 6 Biko, a manufacturer of frames for high-end road bicycles, produces two types of frames, the “speedy” and the “ultimate”. The table below shows the usage of resources, as well as their sales price. Speedy Ultimate Labour hours hours Carbon ber 6 kg kg Price £6 £ The resources available are , hours of labour and , units of carbon ber . At most , “speedy” and , “ultimate” can be sold. Furthermore, Biko can purchase “speedy” and “ultimate” frames for £ and £ each, respectively, from a third party producer. (a) Formulate as a linear program the problem of producing “speedy” and “ultimate” frames in order to maximize revenue. (The LP should use variables.) [6 marks] Solution : Let x 1 : number of “speedy” frames produced, x 2 : number of “ultimate” frames produced, x 3 : number of third party “speedy” frames, x 4 : number of third party “ultimate” frames. max 640 x 1 +1100 x 2 +100 x 3 +100 x 4 2 x 1 +4 x 2 ≤ 90000 6 x 1 +10 x 2 ≤ 29000 x 1 + x 3 ≤ 50000 x 2 + x 4 ≤ 50000 x 1 , . . . , x 4 ≥ 0 (b) Below we report the Excel sensitivity report relative to the optimal solution of the problem. An- swer the following questions. [ marks] (i) Biko intends to change the price of the “speedy” frame they produce (while leaving the retail price of the third party “speedy” unchanged at £6 ). What is the highest and lowest price they can set without changing the optimal solution? Solution : The lowest price they can set is £6 per “speedy”. The price can be increase arbitrarily without changing the optimal solution. (ii) How much should the price of the “ultimate” frame produced by Biko increase, before it becomes pro table to produce them? Solution : They should increase it by £8 . (iii) The third party producer would like to increase the price they charge Biko on their “ultimate” frames. How much can they increase it before loosing Biko as a customer? Solution : They can increase it up to £ 8 . After that the pro t Biko makes on third party ultimates is less than £ /unit, which is below the lower bound indicated in the solution. (iv) Suppose Biko could buy the “ultimate” frame for £ 6 from yet another producer. What would be the optimal solution? © LSE ST /MA Page of

Solution : This is equivalent to increasing the pro t of variable x 4 by £ /unit. This is within the ranges (since the upper-bound for x 4 is +Inf.). Therefore the optimal solution does not change, but the optimal value increases by £ 40 · 50 , 000 = £ 2 , 000 , 000 . (v) How much would Biko be willing to pay for more kilograms of carbon ber? And for more hours of labour? Solution : The carbon ber constraint is ineffective, therefore Biko would not be willing to spend anything to increase the availability. The dual value of the labour constraint is , and is within the range for the labour constraint. It follows that Biko would be willing to spend up to £ 270 · 100 = £ 27 , 000 for extra hours of labour. (vi) How much should Biko be willing to spend in advertisement, at most, in order to boost demand of the “ultimate” frame by , units? Solution : The dual value of the constraint relative to ultimates demand is . The al- lowable increase for the constraint is +in nity. This implies that, if the RHS increases by , units, the pro t increases by £ , , .Biko should therefore spend no more than £ , . (vii) Biko is considering starting producing mountain bikes frames, and needs to price them. Each frame would require kg of carbon ber and hours of labour. How much should each mountain bike frame be priced in order to be pro table? Solution : If we introduce a new variable x 5 indicating the number of mountain bike frames produced, and denote by p 5 its price, the dual constraint relative to the new variable x 5 is 12 y 1 + 5 y 2 ≥ p 5 . Therefore the previous optimal solution remains optimal (i.e. x 5 = 0 in the optimal solu- tion) as long as p 5 ≤ 12 · 0+5 · 270 = 1 , 350 . Thus we need to set a price of at least £ , for the mountain bike frames in order for them to be pro table. Variable Cells Final Reduced Objective Allowable Allowable Name Value Cost Coef cient Increase Decrease Speedy 6 + ∞ Ultimate -8 8 + ∞ Third party speedy Third party ultimate + ∞ 8 Constraints Final Shadow Constraint Allowable Allowable Name Value Price R.H. Side Increase Decrease Labour 6666.66666 Carbon + ∞ Demand speedy + ∞ Demand ultimate + ∞ END OF PAPER © LSE ST /MA Page of