YORK UNIVERSITY FACULTY OF ARTS DEPARTMENT OF MATHEMATICS AND STATISTICS Instructor : Dr. Igor Poliakov MATH 4570 6.0, YEAR 2006-07 APPLIED OPTIMIZATION (TEST #4 ) (SOLUTIONS) March 29, 2007 Name (print) (Family) (Given) Student I.D. Instructions: 1. This is a closed-book Test. Calculators are permitted. 2. Show all significant steps. No marks will be given for the answer alone. 3. Solve the problems in the spaces provided. If you need more space, use the back of the page (indicate this fact on the original page). 4. YOU ARE ALLOWED TO USE AID SHEET. 5. USE PEN TO WRITE THE FINAL ANSWER. 6. DRAW A BOX AROUND YOUR FINAL ANSWERS. Test # 4 contains 7 questions on 8 pages. Read the instructions carefully and sign below: (Signature) QUESTION 1 2 3 4 5 6 7 Total MARKS 8 6 6 6 10 4 6 46 SCORE Page 1 of 8
1. [8] The time between arrivals of busses follows an exponential distribution, with a mean of 60 minutes. a) What is the probability that exactly four buses will arrive during the next 2 hours? b) That at least two buses will arrive during the next 2 hours? c) That no buses will arrive during the next 2 hours d) A bus has just arrived. What is the probability that it will be between 30 and 90 minutes before the next bus arrives? a. Number of buses arriving in next two hours is Poisson with mean 2. Thus from (7) P(4 buses in next two hours) = e -2 (2) 4 /4! =.09 b. P(at least 2 buses) = 1 - P(0 buses) - P(1 bus) = 1 - e -2-2e -2 =.59 c. P(0 buses) = e -2 (2) 0 /0! = e -2 =.14 d. Interarrival time has density (in hours) e -t. Thus we seek 3/2 3/2 e -t dt = [-e -t ] = e -.5 - e -1.5 =.38 1/2 1/2 Page 2 of 8
2. [ 6] My home uses two light bulbs. On average, a light bulb lasts for 22 days (exponentially distributed). When a light bulb burns out, it takes an average of 2 days (exponentially distributed) before I replace the bulb. a) Formulate a three-state birth-death model of this situation. b) Determine the fraction of the time that both light bulbs are working c) Determine the fraction of the time that no light bulbs are working. a. Let the state be the number of working bulbs. Possible states are 0, 1, and 2. Birth = a bulb is repaired while Death = bulb burns out. Then the birth-death parameters are as follows: λ 0 = 1/2 + 1/2 = 1 µ 0 = 0 λ 1 = 1/2 µ 1 = 1/22 λ 2 = 0 µ 2 = 1/22 + 1/22 = 1/11 Steady state probabilities may be found from π 0 = π 1 /22, π 1 /2 + π 1 /22 = π 0 + π 2 /11, π 2 /11 = π 1 /2, π 0 + π 1 + π 2 = 1 or π 1 = 22π 0' π 2 = 121π 0. Thus π 0 (1 + 22 + 121) = 1 or π 0 = 1/144, π 1 = 11/72, π 2 = 121/144. b. π 2 = 121/144 c. π 0 = 1/144 Page 3 of 8
3. [6] A fast-food restaurant has one drive-through window. An average of 40 customers per hour arrive at the window. It takes an average of 1 minute to serve a customer. Assume that interarrival and service times are exponential. a) On the average, how many customers are waiting in line? b) On average, how long does a customer spend at the restaurant (from time of arrival to time service is completed)? c) What fraction of the time are more than 3 cars waiting for service (this includes the car (if any) at the window)? a. L q = 40 2 /(60-40)60 = 1.33 customers b. W = 1/(60-40) = 1/20 hour = 3 minutes c. 1 - π 0 - π 1 - π 2 - π 3 = 1-1/3-2/9-4/27-8/81 = 16/81. Page 4 of 8
4. [6] An average of 40 cars per hour (interarrival times are exponentially distributed) are tempted to use the drive-in window at the Hot Dog King Restaurant. If a total of more than 4 cars are in line ( including the car at the window) a car will not enter the line. It takes an average of 4 minutes (exponentially distributed) to serve a car. a) What is the average number of cars waiting for the drive-in window (not including a car at the window)? b) On the average, how many cars will be served per hour? c) I have just joined the line at the drive-in window. On the average, how long will it be before I have received my food? We have an M/M/1/4 system with µ = 15 cars/hour and λ = 40 cars/hour. a. L q = L - L s ρ = 40/15 = 2.67 2.67 [1-5(2.67) 4 + 4(2.67) 5 ] L = ------------------ = 3.44 (1-2.67 5 ) (1-2.67) 1-2.67 π 0 = ------- =.012 L s = 1 - π 0 =.99 1-2.67 5 L q = 3.44 -.99 = 2.45 b. We seek λ(1 - π 4 ) = 40(1 -.61) = 15.6 cars/hour (π 4 = (2.67) 4 (.012) =.61) L 3.44 c. We seek W = -------- = --------- =.22 hours λ(1 - π 4 ) 40(1 -.61) Page 5 of 8
5.[10] An average of 90 patrons per hour arrive at a hotel lobby (interarrival times are exponential), waiting to check in. At present, there are 5 clerks, and patrons are waiting in a single line for the first available clerk. The average time for a clerk to service a patron is 3 minutes (exponentially distributed). Clerk earn $10 per hour, and the hotel assesses a waiting time cost of $20 for each hour that a patron waiting in line (see Table on the last page). a)[4] b)[6] Compute the expected cost per hour of the current system. The hotel is considering replacing one clerk with an Automatic Clerk Machine (ACM). Management estimates that 20% of all patrons will use an ACM. An ACM takes an average of 1 minute to service a patron. It costs $48 per day ( 1 day = 8 hours) to operate an ACM. Should the hotel install the ACM? Assume that all customers who are willing to use the ACM wait in a single queue. a. M/M/5 system with λ = 90 customers/hr and μ = 20 customers/hour. P(j 5) =.76. Then W q = P(j 5)/(100-90) =.076 hours. Expected Cost Per Hour = 10(5) + 20(90)W q = 50 + 1800(.076) = $186.80. b. With ACM we have an M/M/1 (the ACM) having λ = 18 customers per hour and μ = 60 customers per hour and an M/M/4 having λ = 72 customers/hour and μ = 20 customers/hour. For ACM W q =18/(60(60-18)) =.0071 hours and Expected ACM Cost Per Hour = 6 + 20(18)W q = $8.57. For M/M/4 system P(j 4) =.79 so W q =.79/8 =.0988 and Expected Cost Per Hour = 10(4) + 20(72)(.0988) = $182.27 Thus with ACM total hourly expected cost is 182.27 + 8.57 = 190.83. Thus do not use ACM. Page 6 of 8
6. a) [2] Each week, the Columbus Record Club attracts 100 new members. Members remain members for an average of one year ( 1 year = 52 weeks). On the average, how many members will the record club have? b) [2] The State U doctoral program in business admits an average of 25 doctoral students each year. If a doctoral student spends an average of 4 years in residence at State U, how many doctoral students would one expect to find there? a) λ = 100 members/week W = 52 weeks. Then L = λw = 5200 members. b) L = average number of doctoral students on campus 1/µ = W = 4 years so L = 25(4) = 100 students. Page 7 of 8
7.[6] Consider an M / G /1/ GD / / queuing system in which an average of 10 arrivals occur each hour. Suppose that each customer s service time follows an Erlang distribution, with rate parameter 1 customer per minute and shape parameter 4. a) Find the expected number of customers waiting in line. b) Find the expected time that a customer will spend in the system. c) what fraction of the time will the server be idle? From (9) 1/µ = Mean Service Time = 4/60 hours and Variance of Service Time = 4/60 2 = 1/900 hours 2. Then λ = 10 customers/hr. and ρ = 10(4/60) = 2/3 (10) 2 (1/900) + (2/3) 2 a. L q = ------------------------- = 5/6 customers. 2(1-2/3) b. W q = L q /λ = (5/6)/10 = 1/12 hour, and W = W q = (1/µ) = 1/12 + 1/15 =.15 hours. c. π 0 = 1-ρ = 1/3 of the time that the server is idle. THE END Total marks are 54 Page 8 of 8