IE 5112 Final Exam 2010

Transcription

1 IE 5112 Final Exam There are six cities in Kilroy County. The county must decide where to build fire stations. The county wants to build as few fire stations as possible while ensuring that there is at least one fire station within 15 minutes driving time of each city. Driving times between cities are shown in the table below. Formulate an integer programming problem that will determine how many fire stations are required and in which cities they should be located. To From City 1 City 2 City 3 City 4 City 5 City 6 City City City City City City Solution. Let x i = { 1 if a fire station is built in city i, 0 otherwise. To model the requirement that there is at least one fire station within 15 minutes of city 1, we require that x 1 +x 2 1. The full problem formulation is minimize 6 i=1 x i subject to x 1 + x 2 1 x 1 + x 2 + x 6 1 x 3 + x 4 1 x 3 + x 4 + x 5 1 x 4 + x 5 + x 6 1 x 2 + x 5 + x 6 1 x i {0,1} i = 1,...,6

2 2. Subzero clothing company is capable of manufacturing three types of clothing: vests, jackets, and pants. Each type of clothing requires the appropriate machinery. This machinery can be rented at the following rates: vest machinery costs $200 per week, jacket machinery costs $150 per week, and pants machinery costs $100 per week. Each type of clothing also requires the amounts of labor and material shown in the table below. Each week 150 hours of labor and 160 square yards of material are available. The per-unit variable cost and the per-unit selling price of each clothing type are shown in the table. Assume that Subzero can sell all of the clothing that it manufactures. Also, note that the cost of renting machinery for a particular clothing type is only incurred if that clothing type is produced. Formulate an integer programming problem to determine the production levels of each clothing type so that overall profit is maximized. Clothing type Labor (hours) Material (sq. yds) Sale price Variable cost Vests 3 4 $12 $6 Jackets 2 3 $8 $4 Pants 6 4 $15 $8 Solution. Letthe decisionvariablesx 1,x 2,x 3 be thenumberofvests, jackets, andpants, respectively, to produce. Because the machinery rental costs are only incurred if clothing of that particular type is produced, this is a fixed-charge problem. Introduce auxiliary binary variables to model the fixed-charge. { 1 if x i > 0 Let y i = i = 1,2,3. 0 otherwise The mathematical formulation of the Integer Programming problem is maximize 12x 1 + 8x x 3 6x 1 4x 2 8x 3 200y 1 150y 2 100y 3 subject to 3x 1 + 2x 2 + 6x x 1 + 3x 2 + 4x x 1 My 1 x 2 My 2 x 3 My 3 where M is a large number, x i 0, x i are integer-valued, y i {0,1}, and i = 1,2,3.

3 3. A certain missile has a guidance control system. Course correction signals form a sequence {X n } where X n is the state of the system after the nth signal is sent and received. The state space for X is 0: No correction required. 1: Minor correction required. 2: Major correction required. 3: Abort and self-destruct. Suppose that {X n } can be modeled as a discrete-time Markov chain with one-step transition matrix /3 1/6 1/ /3 1/6 1/ If the system is initially in state 1, find the probability that the missile will self-destruct. Solution. We want to know f 13, the probability of beging absorbed into state 3, starting from state 1. States 0 and 3 are absorbing, so f 03 = 0 and f 33 = 1. The relationships for for f 13 and f 23 are Solving these equations we obtain f 13 = 1/21. f 13 =p 10 f 03 +p 11 f 13 +p 12 f 23 +p 13 f 33 f 23 =p 20 f 03 +p 21 f 13 +p 22 f 23 +p 23 f 33.

4 4. A coffee shack has one clerk to serve customers. Customers arrive according to a Poisson process at a mean rate of 10 per hour. However, if there is at least one person in line (in queue) then arriving customers may balk. The probability that a customer will balk increases with the length of the queue. In particular, the probability that a customer will balk is n/5 for n = 2,3,4,5. Note that if n = 1, then the probability of balking is zero because there is no one in line (in queue.) In addition, the clerk gets confused when there are people in line, which causes his service rate to decline. In particular, the clerk s service rate is µ n = 15 2n, for n = 1,2,3,4,5. Service times are Exponentially distributed. (a) Construct the rate transition diagram for this continuous-time Markov chain. Solution. The arrival rate is 10 the probability that a customer will not balk. For example, λ 2 = 10(3/5) = 6. λ 0 = 10 λ 1 = 10 λ 2 = 6 λ 3 = 4 λ 4 = µ 1 = 13 µ 2 = 11 µ 3 = 9 µ 4 = 7 µ 5 = 5 (b) Develop the steady-state balance equations. You do not need to solve the equations. Solution. 10π 0 =13π 1 (10+13)π 1 =10π 0 +11π 2 (11+6)π 2 =10π 1 +9π 3 (9+4)π 3 =6π 2 +7π 4 (7+2)π 4 =4π 3 +5π 5 5π 5 =2π 4 5 π i =1 i=0

5 5. A local business person is considering building a self-service car wash with some number of stalls. It is estimated that cars will arrive to the car wash at the rate of two per hour. The average time taken to wash a car is estimated to be 20 minutes. Assume that interarrival times and service times are exponentially distributed. Customers in this particular town do not like to wait. If an arriving customer finds that all stalls are in use, then she will drive awayand her business is lost. That is to say, if all stalls are busy then a customer will balk will probability one. Using the estimates given above, determine the minimum number of stalls required so that the long-run probability that a customer will balk is less than.15. Solution. First notice that because the interarrival and service times are distributed exponnential, the system can be modeled as a continuous-time Markov chain. Also notice that there is never a queue, so it is inappropriate to use queueing theory results which assume that there is a queue. We can determine the minimum number of stalls n by computing π n, the steady-state proportion of time that all stalls are busy. We do this for increasing values of n until π n <.15. Suppose that there is one stall. Then the steady-state balance equations are λπ 0 = µπ 1 π 0 +π 1 = 1, where λ = 2 and µ = 3. Solving this system of equations gives π 0 = 3/5 and π 1 = 2/5. So one stall is insufficient. Now suppose that there are two stalls. The steady-state equations are λπ 0 = µπ 1 (λ+µ)π 1 = λπ 0 +2µπ 2 π 0 +π 1 +π 2 = 1, Solving this system of equations gives π 0 = 18/34, π 1 = 12/34, and π 2 = 4/34. So two stalls will suffice.

6 6. Consider a discrete-time Markov chain that has the following one-step transition matrix /5 0 1/ /4 0 1/2 1/ /2 0 1/10 2/ /3 0 1/3 1/3 0 (a) Determine the classes of this Markov chain and, for each class, determine whether it is transient or recurrent. Solution. First some background information: We say that a state j is accessible from another state i if p (n) ij > 0 for some n > 0. If j is accessible from i and i is accessible from j, then we say that i and j communicate. States that communicate with each other are in the same class. Transience and recurrence are class properties. Transient means that upon entering a state, there is a positive probability that the process may never return to the state. Recurrent means that upon entering a state, the process will definitely return to the state if you let it run for a long enough period of time. A special case of recurrence is an absorbing state. Upon entering an absorbing state, the process never leaves. If there is only one class, then we say that the Markov chain is irreducible. For this particular Markov chain, state 3 is absorbing so it is recurrent and forms a class of its own. States {0,1,2,4} communicate and so they form a class. Furthermore states {0,1,2,4} are all transient because it s possible for the process to go to state 3. (b) For each of the classes identified in part 6a, determine the period of the states in that class. Solution. The period of a state is the multiple of time steps in which the process may return to the state. More formally, the period of a state i is an integer t that is determined as follows. gcd{t 1 : P(x t = i x 0 = i) > 0} where gcd means the greatest common divisor. Periodicity is a class property, which means that all states in the same class have the same period. For this Markov chain, the period of state 0 (and hence of states 1, 2, and 4) is gcd{2,4,6,8,...} = 2. State 3 is absorbing and so its period is one. We say that state 3 is aperiodic.