Unit 4 simulation and queing theory(m/m/1)

Monte carlo simulation and queuing theory problems

1. Introduction to simulation- MONTE CARLO SIMULATION

2. System: The physical process of interest Model: Mathematical representation of the system Models are a fundamental tool of science, engineering, business, etc. Models always have limits of credibility Simulation: A type of model where the computer is used to imitate the behavior of the system Monte Carlo simulation: Simulation that makes use of internally generated (pseudo) random numbers Random Number : Random numbers  are  numbers  that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is not impossible to predict future values based on past or present ones.  2 Basics

3. 3 Ways to Study System Focus of course System Experiment w/ actual system Experiment w/ model of system Physical Model Mathematical Model Analytical Model Simulation Model

4. “The Monte Carlo method is a numerical solution to a problem that models objects interacting with other objects . A Monte Carlo simulation is a model used to predict the probability of different outcomes when the intervention of random variables is present. Monte Carlo simulations help to explain the impact of risk and uncertainty in prediction and forecasting models. A variety of fields utilize Monte Carlo simulations, including finance, engineering, supply chain, and science. The basis of a Monte Carlo simulation involves assigning multiple values to an uncertain variable to achieve multiple results and then to average the results to obtain an estimate. It represents an attempt to model nature through direct simulation of the essential dynamics of the system in question. In this sense the Monte Carlo method is essentially simple in its approach. 4

5. Business and finance are plagued by random variables, Monte Carlo simulations have a vast array of potential applications in these fields. Monte Carlo Method: A Monte Carlo simulation takes the variable that has uncertainty and assigns it a random value. The model is then run and a result is provided. This process is repeated again and again while assigning the variable in question with many different values. Once the simulation is complete, the results are averaged together to provide an estimate. 5

6. let’s consider a simple system with simple inputs: 6 As A, B, C and D are always the same, the output will always be the same and it can be easily calculated Imagine that input A has a range of possible values – the output will also be variable. And when there are many more possible inputs and all of them have a range of possible values, the output is not that simple to calculate. That’s where you need to use Monte Carlo simulation.

7. Steps in monte carlo simulation: Step 1:Clearly define the problem. Step 2:Construct the appropriate model. Step 3:Prepare the model for experimentation. Step 4:Using step 1 to 3,experiment with the model. Step 5:Summarise and examine the results obtained in step 4. Step 5:Evaluate the results of the simulation. 7

8. A manufacturing company keeps stock of a special product. Previous experience indicates the daily demand as given below 8 Daily demand 5 10 15 20 25 30 probability 0.01 0.20 0.15 0.50 0.12 0.02 Simulate the demand for the next 10 days. Also find the daily average demand for that product on the basis of simulated data. Consider the following random numbers: 82,96,18,96,20,84,56,11,52,03

9. Solution: Step 1:Generate tag values 9 Daily demands Probability Cumulative probability Tag values(Random num range) 5 0.01 0.01 00-00 10 0.20 0.21 01-20 15 0.15 0.36 21-35 20 0.50 0.86 36-85 25 0.12 0.98 86-97 30 0.02 1.00 98-99 Step 2: Simulate for 10 days Days Random num Daily demand 1 82 20 2 96 25 3 18 10 4 96 25 5 20 10 6 84 20 7 56 20 8 11 10 9 52 20 10 03 10 Average demand=(20+25+10+25+10+20+20+10+20+10)/10 =170/10=17 units/day

10. 2)A tourist car operator finds that during the past few months the cars use has varied so much that the cost of maintaining the car varied considerably. During the past 200 days the demand for the car fluctuated as below 10 Trips per week Frequency 16 1 24 2 30 3 60 4 40 5 30 Using random numbers 82,96,18,96,20,84,56,11,52,03, simulate the demand for 10 week period

11. Solution: Step 1:Generate tag values 11 Trips/week frequency Probability Cumulative probability Tag values 16 16/200=0.08 0.08 00-07 1 24 24/200=0.12 0.20 08-19 2 30 30/200=0.15 0.35 20-34 3 60 60/200=0.30 0.65 35-64 4 40 40/200=0.20 0.85 65-84 5 30 30/200=0.15 1.00 85-99 Frequency-Number of occurrences, Total num of occurrences(16+24+30+60+40+30)=200 Step 2: Simulation for next 10 week Weeks Random Num Trips/week 1 82 4 2 96 5 3 18 1 4 96 5 5 20 2 6 84 4 7 56 3 8 11 1 9 52 3 10 03 Avg trips/week=28/10=2.8≈3 trips/week

12. For a particular shop the daily demand of an item is given as follows, Use random numbers 25,39,65,76,12,05,73,89,19,49.Find the average daily demand. Daily demand 5 10 15 20 25 30 Probability 0.01 0.20 0.15 0.50 0.12 0.02 Solution: Generate tag values 12 Daily demand Probability Cumulative probability Tag values 0.01 0.01 00 10 0.20 0.21 01-20 20 0.15 0.36 21-35 30 0.50 0.86 36-85 40 0.12 0.98 86-97 50 0.02 1.00 98-99

13. Step 2: Simulation for 10 days 13 Days Random num Daily demand 1 25 20 2 39 30 3 65 30 4 76 30 5 12 10 6 05 10 7 73 30 8 89 40 9 19 10 10 49 30 Avg daily demand= 240/10=24

14. An automobile company manufactures around 150 scooters.Daily production varies from 146 to 154,the probability distribution is given below. Step 1:Generate tag values for production/day Step 1:Generate tag values for production/day 14 Production/day 146 147 148 149 150 151 152 153 154 probability 0.04 0.09 0.12 0.14 0.11 0.10 0.20 0.12 0.08 The finished scooters are transported in a lorry accomodading150 scooters. using the following random numbers 80,81,76,75,64,43,18,26,10,12,65,68,69,61,57 simulate 1)Average number of scooters waiting in the factory 2)Average number of empty space in the lorry

15. Step 1:Generate tag values for production/day 15 Production/day probability Cumulative probability Tag values 146 0.04 0.04 00-03 147 0.09 0.13 04-12 148 0.12 0.25 13-24 149 0.14 0.39 25-38 150 0.11 0.50 39-49 151 0.10 0.60 50-59 152 0.20 0.80 60-79 153 0.12 0.92 80-91 154 0.08 1.00 92-99 Step 2: Simulate for 15 days to get avg no of waiting scooters and empty space, lorry can accommodate 150 scooters Days Random num Production/day No of scooters waiting No of empty space in lorry 1 80 153 3 - 2 81 153 3 - 3 76 152 2 - 4 75 152 2 - 5 64 152 2 - 6 43 150 - - 7 18 148 - 2 8 26 149 - 1 9 10 147 - 3 10 12 147 - 3 11 65 152 2 - 12 68 152 2 - 13 69 152 2 - 14 61 152 2 - 15 57 151 1 -       Total=21 Total=9 Avg no of scooters waiting= 21/15=1.4 Avg No of space in the lorry= 9/15=0.6

16. An automobile production line turns out about 100 cars/day, but deviation occur owing to many causes.Production of cars are described by the probability distribution given below. 16 Production/day 95 96 97 98 99 100 101 102 103 104 105 probability 0.03 0.05 0.07 0.10 0.15 0.20 0.15 0.10 0.07 0.05 0.03 Finished cars are transported across the bay at the end of each day by ferry. If ferry has space for only 101 cars, what will be the average number of cars waiting to be shipped and what will be the average number of empty space on ship? Simulate the production of cars for next 15 days, consider the random numbers 97,02,80,66,96,55,50,29,58,51,04,86,24,39,47.

17. Step 1:generate tag values 17 Production/day Probabillity Cumulative probability Tag values 95 0.03 0.03 00-02 96 0.05 0.08 03-07 97 0.07 0.15 08-14 98 0.10 0.25 15-24 99 0.15 0.40 25-39 100 0.20 0.60 40-59 101 0.15 0.75 60-74 102 0.10 0.85 75-84 103 0.07 0.92 85-91 104 0.05 0.97 92-96 105 0.03 1.00 97-99

18. Step 2:Simulate for 15 days, ferry can transport 101 cars 18 Days Random numbers Productions/day No of cars waiting Empty space in the ship 1 97 105 |105-101|=4 - 2 02 95 - (101-95) =6 3 80 102 1   4 66 101 - - 5 96 104 3 - 6 55 100 - 1 7 50 100 - 1 8 29 99 - 2 9 58 100 - 1 10 51 100 - 1 11 04 96 - 5 12 86 103 2 - 13 24 98 - 3 14 39 99 - 2 15 47 100 - 1       Total=10 Total=23 Avg num of cars waiting=10/15 Avg empty space in the ship=23/15

19. Strong is a dentist who schedules all her patients for 30 minutes appointment. Some of the patients take more or less than 30min depending on the type of dental works to be done. The following summary shows the various categories of work, their frequency and the time actually needed to complete the work 19 Category Filling crown cleaning extracting checkup Time required 45 60 15 45 15 Number of patients 40 15 15 10 20 Simulate the dentist clinic for 4 hrs and find out the avg waiting time for the patients as well as the idleness of doctor.Assume that the patients show up at the clinic at exactly scheduled time. Arrival time starts at 8AM.Use the following random number for handling the same 40,82,11,34,25,66,19,79

20. category Time required No of patients(Frequency) probability Cumulative probability Tag values Filling 45 40 0.40 0.40 00-39 Crown 60 15 0.15 0.55 40-54 Cleaning 15 15 0.15 0.70 55-69 Extracting 45 10 0.10 0.80 70-79 Checkup 15 20 0.20 1.00 80-99     Total=100       20 Random num Category Time required (min) Arrival time of patients Service time Start time End time waiting time for patients(min) Idleness of doctor 40 crown 60 8.00 8.00 9.00 - 82 checkup 15 8.30 9.00 9.15 30(9-8.30) - 11 Filling 45 9.00 9.15 10.00 15 - 34 Filling 45 9.30 10.00 10.45 30 - 25 Filling 45 10.00 10.45 11.30 45 - 66 Cleaning 15 10.30 11.30 11.45 60 - 19 Filling 45 11.00 11.45 12.30 45 - 79 Extracting 45 11.30 12.30 1.15 60 - Step 1:find the cumulative probability and tag values Step 2:Simulate for 4 hrs Avg waiting time for patients=(30+15+30+45+60+45+60)/8=285/8=35.62 min≈36min Waiting time for patients=(start time of service-arrival time)

21. Bright Bakery keeps stock of a popular brand of cake. Previous experience indicates the daily demand as given below: Consider the following sequence of random numbers; 48, 78, 19, 51, 56, 77, 15, 14, 68,09. Using this sequence simulate the demand for the next 10 days. Find out the stock situation if the owner of the bakery decides to make 30 cakes every day. Also estimate the daily average demand for the cakes on the basis of simulated data. 21 Daily demand 10 20 30 40 50 Probability 0.01 0.20 0.15 0.50 0.12 0.02

22. Daily demand Probability Cumulative probability Tag values 0.01 0.01 00 10 0.20 0.21 01-20 20 0.15 0.36 21-35 30 0.50 0.86 36-85 40 0.12 0.98 86-97 50 0.02 1.00 98-99 22 Step 1:find the cumulative probability and tag values Step 2:Simulate for 10 days, make 30 cakes every day Days Random num Daily Demand Stock condition 1 48 30 - 2 78 30 - 3 19 10 20 4 51 30 - 5 56 30 - 6 77 30 - 7 15 10 20 8 14 10 20 9 68 30 - 10 09 10 20 Avg daily demand=220/10=22

23. An airport entry has only one check-in counter, customers arrive at this counter at random from 1 to 6 min,each interarrival time has same probability of occurrence’s, probability for service time is given below. Service time(ST) 1 2 3 4 5 6 Probability 0.10 0.20 0.25 0.30 0.10 0.05 Develop a simulation table for 10 customers. Consider the random number for inter arrival time(IAT) as 13,27,15,48,9,22,53,35,2 and RN for ST as 84,10,74,53,17,91,79,67,38,89. 1.Calculate avg ST, avg time between arrival, waiting time of customer, avg time spent by customer in system,avg waiting time of customers those who wait.

24. step 1:tag values for service time Service time probability Cumulative probability Tag values 1 0.10 0.10 00-09 2 0.20 0.30 10-29 3 0.25 0.55 30-54 4 0.30 0.85 55-84 5 0.10 0.95 85-94 6 0.05 1.00 95-99 Step 2: tag values for IAT, equal probability of occurance [out of 6 option any one can occur, 1/6=0.166666667=0.17(approx.)] Inter arrival time probability Cumulative probability Tag values 1 1/6=0.17 0.17 00-16 2 0.17 0.34 17-33 3 0.17 0.51 34-50 4 0.17 0.68 51-67 5 0.17 0.85 68-84 6 0.17 1.00 85-99

25. Simulation for 10 customers Customer no RN for IAT IAT AT RN for ST Service time Service time WT for customer Time spent by customer in system begin end 1 - - 84 4 4 - (4-0)=4 2 13 1 1 10 2 4 6 (4-1)=3 (6-1)=5 3 27 2 3 74 4 6 10 (6-3)=3 (10-3)=7 4 15 1 4 53 3 10 13 6 (13-4)=9 5 48 3 7 17 2 13 15 6 (15-7)=8 6 9 1 8 91 5 15 20 7 (20-8)=12 7 22 2 10 79 4 20 24 10 (24-10)=14 8 53 4 14 67 4 24 28 10 (28-14)=14 9 35 3 17 38 3 28 31 11 (31-17)=14 10 2 1 18 89 5 31 36 13 (36-18)=18     Total=18     Total=36     Total=69 Total=105

26. Time spent by customer in the system= (service end time-arrival time) Avg waiting time=total WT/total customers =69/10=6.9min Avg time spent by the customer in the system=105/10=10.5 Avg service time=36/10=3.6≈4min Avg time between arrival=18/10=1.8≈2min Avg waiting time of customers those who wait=69/9=7.66=total waiting time of customers/number of customers waiting

27. A Grocery store has only one checkout counter counter , customers arrive at this counter at random from 1 to 8 min,each interarrival time has same probability of occurance.The probability for service time is given below Service time(ST) 1 2 3 4 5 6 probability 0.10 0.20 0.30 0.25 0.10 0.05 Simulate the arrival for 10 customers. Consider the random number for inter arrival time(IAT) as 913,727,015,948,309,922,753,235,302 and RN for ST as 84,10,74,53,17,91,79,67,89,38. 1.Calculate avg WT, avg ST,avg time between arrival, avg time spent by customer in queue, avg waiting time of customers those who wait.

28. Step1: Generate tag vales for service time , Service time(ST) probability Cumulative probability Tag values 1 0.10 0.10 00-09 2 0.20 0.30 10-29 3 0.30 0.60 30-59 4 0.25 0.85 60-84 5 0.10 0.95 85-94 6 0.05 1.00 95-99 Step 2: Generate tag values for IAT, Given that P=1/8=0.125 Inter arrival time probability Cumulative probability Tag values 1 0.125 0.125 000-124 2 0.125 0.250 125-249 3 0.125 0.375 250-374 4 0.125 0.500 375-499 5 0.125 0.625 500-624 6 0.125 0.750 625-749 7 0.125 0.875 750-874 8 0.125 1.000 875-999

29. Step 3:simulation for 10 customers Customer no RN for IAT IAT AT RN for ST ST Service time Time in Queue Time spent in system Idle time for server Begin End 1 - - 84 4 4 - (4-0)=4 - 2 913 8 8 10 2 8 10 - (10-8)=2 (8-4)=4 3 727 6 14 74 4 14 18 - 4 (14-10)=4 4 15 1 15 53 3 18 21 (18-15)=3 6 - 5 948 8 23 17 2 23 25 - 2 2 6 309 3 26 91 5 26 31 - 5 1 7 922 8 34 79 4 34 38 - 4 3 8 753 7 41 67 4 41 45 - 4 3 9 235 2 43 89 5 45 50 (45-43)=2 7 - 10 302 3 46 38 3 50 53 (50-46)=4 7 - Total   46     36     9 45 17

30. Avg ST=36/10=3.6min Avg time between arrival=46/10=4.6 min Avg time spent by customer in the queue(WT)=9/10=0.9min Avg time spent by customer in the system=45/10=4.5min Avg idle time of server=17/10 Avg waiting time of those who wait=9/3=3min

31. Ex 4:Consider a store with one checkout counter.prepare a simulation table and find the avg waiting time of customers in queue,avd idle time of server,avg service time.consider the following data. IAT 3 2 6 4 4 5 8 7 Sevice time 4 5 5 8 4 6 2 3 4

32. Customer no IAT Arrival time ST Service time WT Idle time of server Time spent in system Begin End 1 - 4 4 - - 4 2 3 3 5 4 9 1 - 6 3 2 5 5 9 14 4 - 9 4 6 11 8 14 22 3 - 11 5 4 15 4 22 26 7 - 11 6 4 19 6 26 32 7 - 13 7 5 24 2 32 34 8 - 10 8 8 32 3 34 37 2 - 5 9 7 39 4 39 43 - 2 4       41     Total=32   73 Since the inter arrival time and service time are specified in the problem, simulation for 9 customers is given in the following table Avg waiting time of those who wait=32/7=4.57 Avg service time=41/9=4.55 Avg Waiting Time=32/9 Idle time of server=2/9=0.22 Avg time spent in the system=73/9=8.1

33. Simulation of Queueing Systems Simulation is often used in the analysis of queueing models. A simple but typical queueing model Queueing models provide the analyst with a powerful tool for designing and evaluating the performance of queueing systems .  queuing situation involves two parts. Someone or something that requests a service—usually referred to as the customer, job, or request. Someone or something that completes or delivers the services—usually referred to as the server. Typical measures of system performance, Server utilization, length of waiting lines, and delays of customers. For relatively simple systems: compute mathematically For realistic models of complex systems: simulation is usually required

34. A queueing system is described by Calling population Arrival rate Service mechanism System capacity Queueing discipline

35. Calling population Calling population : the population of potential customers, may be assumed to be finite or infinite. Finite population model : if arrival rate depends on the number of customers being served and waiting, e.g., model of corporate jet, if it is being repaired, the repair arrival rate becomes zero. Infinite population model : if arrival rate is not affected by the number of customers being served and waiting, e.g., systems with large population of potential customers.

36. System Capacity System Capacity: A limit on the number of customers that may be in the waiting line or system. Limited capacity , e.g., an automatic car wash only has room for 10,cars to wait in line to enter the mechanism. If system is full no customers are accepted anymore Unlimited capacity , e.g., concert ticket sales with no limit on the number of people allowed to wait to purchase tickets.

37. Arrival Process For infinite-population models : • In terms of interarrival times of successive customers. • Arrival types: Random arrivals : interarrival times usually characterized by a probability distribution. Scheduled arrivals : interarrival times can be constant or constant plus or minus a small random amount to represent early or late arrivals. • Example: patients to a physician or scheduled airline flight arrivals to an airport At least one customer is assumed to always be present,so the server is never idle, e.g., sufficient raw material for a machine.

38. Queue Behavior and Queue Discipline Queue behavior : the actions of customers while in a queue waiting for service to begin, for example: leave when they see that the line is too long, leave after being in the line when its moving too slowly , move from one line to a shorter line Most queuing formula assume that all arrivals stay until service is completed Balking refers to customers who do not join the queue Reneging refers to customers who join the queue but give up and leave before completing service. Jockeying : refers to waiting customers move from one queue to another. Queue discipline : the logical ordering of customers in a queue that determines which customer is chosen for service when a server becomes free, for example: • First-in-first-out (FIFO) • Last-in-first-out (LIFO) • Service in random order (SIRO) • Shortest processing time first (SPT) • Service according to priority (PR)

39. Service Times and Service Mechanism In queuing system,  service time  is defines as the  time  required to serve a custom. May be constant or random. The  service mechanism  is the way that customers receive  service  once they are selected from the front of a  queue . service mechanism: a description of the resources needed for service to begin how long the service will take (the  service time distribution ) the number of servers available whether the servers are in series (each server has a separate queue) or in parallel (one queue for all servers) whether pre-emption is allowed (a server can stop processing a customer to deal with another "emergency" customer) Assuming that the service times for customers are independent and do not depend upon the arrival process is common.  Service mechanism in a queuing system is characterized by Server’s behavior .

40. A = Arrival distribution (M for Poisson, D for deterministic, and G for general) B = Service time distribution (M for exponential, D for deterministic, and G for general) S = number of servers c=calling population d=service discipline 40 Kendall’s Notation A / B / s :c/d

41. An M/M/1 queue is a stochastic process whose  state space(set of all possible combination of system)  is the set {0,1,2,3,...} where the value corresponds to the number of customers in the system, including any currently in service. Arrivals occur at rate λ according to a  Poisson process  and move the process from state  i  to  i  + 1. Service times have an  exponential distribution  with rate parameter μ in the M/M/1 queue, where 1/μ is the mean service time. A single server serves customers one at a time from the front of the queue, according to a  first-come, first-served  discipline. When the service is complete the customer leaves the queue and the number of customers in the system reduces by one. The buffer is of infinite size, so there is no limit on the number of customers it can contain. 41 Single Server Queuing System (M/M/1): (α/ FCFS)

42. ρ = utilization factor (probability of all servers being busy) L q = average number in the queue L or Ls = average number in the system W q = average waiting time in the queue W or Ws = average time in the system P = probability of 0 customers in system P n = probability of exactly n customers in system λ =Arrival rate μ =Service rate 42 Measuring Queue Performance

43. 43

44. 1. Average server utilization or traffic density 2. Average number of customers waiting In queue In the system 3. Average time customer is waiting In the queue In the system 4. Probability of 0 customers in system 5. Probability of exactly n customers in system = 6.Average length of non empty queue 7.Probabili ty that the number of customers is greater than k,   44 Operating Characteristics for M/M/1 Queue

45. Ex 1:The arrival rate of customers at a petrol bunk follows a poisson distribution with a rate of 27 per hour. The petrol bunk has only one unit of service. The service rate at the petrol bunk has exponential distribution with rate of 36 per hour. Determine the following. 1)What is probability of having zero customers in the system? What is probability of having 6 customers in the system? What is probability of having 10 customers in the system? The values of L s , L q , W s , W q 45

46. Solution: Given: Arrival rate λ =27per hour Service rate μ=36/hr Then utilization factor ρ = λ / μ=27/36=0.75 The probability of having 0 customers in system P0= ρ (1- ρ )=1- ρ =1-0.75=0.25 The probability of having 6 customers in the system P6= ρ 6 (1- ρ )=0.0178(1-0.75)=0.0445 The probability of having 10 customers in the system P10= ρ 10 (1- ρ )=0.0563(1-0.75)=0.0141 Ls= ρ /(1- ρ ) =0.75/(1-0.75)=3 customers Lq = ρ 2 /(1- ρ ) =0.75 2 / /(1-0.75)=2.25 customers Ws =1/(μ- λ ) =1/(36-27)=0.011 hr Wq = ρ /(μ- λ ) =0.75/(36-27)=0.0833 hr 46

47. Ex 2:At one man book binding centre, customers arrive according to poisson distribution with arrival rate of 4 per hour and the book binding time is exponentially distributed with a mean of 12 minutes. Find out the following. The average number of customers in the book binding centre The average number of customers waiting for book binding. The percentage of time arrival can walk in straight without having to wait. The percentage of customers who have to wait before getting into book binders table. 47

48. Solution: Given: Arrival rate λ =4/60=1/15 min Service rate μ=1/12 min Then utilization factor ρ = λ / μ=12/15=0.8 The average number of customers in the system(customers in the queue and book binding center ) Ls= ρ /(1- ρ ) =0.8/1-0.8=4 customers The average number of customers waiting for book binding Lq = ρ 2 /(1- ρ ) =(0.8) 2 /(1-0.8)=3.2 customers The percentage of times arrival can walk straight into book binders table without waiting is server utilization= ρ %=0.8*100=80% The percentage of customers who wait before getting into book binders table=(1- ρ )%=(1-0.8)%=20% 48

49. Ex 3:A person repairing wrist watches observes that the time spent on the wrist watches has an exponential distribution with mean 20 minutes. If the wrist watches are repaired in the order in which they come in and their arrival is poisson distribution with an average rate of 15 for 8hr day, what is repairman's expected idle time each day? On an average, how many jobs are ahead of wrist watch just brought in? 49

50. Solution: Given: Arrival rate λ =15/(8*60)=1/32 units/min Service rate μ=1/20 units/min Then utilization factor ρ = λ / μ=20/32=5/8 The number of jobs ahead of the wrist watch just brought in=average number of jobs in the system Ls= ρ /(1- ρ ) =(5/8)/(1-(5/8))=5/3 The number of hours for which the repairman remains busy in an 8hr day=8* ρ =8*(5/8)=5hr Hence the time for which repairman remains idle in an 8hr day=8-5=3hrs 50

51. Example 4:In a s supermarket, the arrival rate of customer is 10 every 30 minutes following poisson process. The average time taken by a cashier to list and calculate the customers purchase is 2.5minutes following exponential distribution. A)What is the probability that the queue length exceeds 6. B)What is the expected time spent by customer in system. 51

52. Solution: Given: Arrival rate λ =10/30 min=1/3 customers/min Service rate μ=1/2.5 customers/min Then utilization factor ρ = λ / μ=3/2.5 A)Probability that queue length is > 6= ρ 6 =0.3348 B) Expected time spent by the customer in the system Ws =1/(μ- λ ) =1/(0.4-0.33)=14.28minutes 52

53. Example 5:In a public telephone booth the arrival are at the rate 15 per hour. A call on an average takes 3 minute. If there is just one phone, find A)The expected number of callers in the booth at any time. B)The probability of time the booth is expected to be idle. 53

54. Solution: Given: Arrival rate λ =15/60 calls/min=1/4 Service rate μ=1/3 per min Then utilization factor ρ = λ / μ=3/4 The expected number of callers in the booth at any time(Average length of non empty queue) Ln= μ/(μ- λ )=4 callers Probability of busy server is ρ ,Hence the probability of server being idle is (booth expected to be idle)(1- ρ ) =1-(3/4)=1/4=0.25 54

55. Example 6:Customers arrive at random at one window drive in a bunk according to poisson distribution with 10 per hr. Service time per customer exponential with mean 5 min. The space in front of the window including that for serviced car can accommodate a maximum of 3 cars 1)what is the probability that an arriving customer can drive directly to the space in the front of the window? 2)what is the probability that an arriving customer will have to wait outside the indicated space? 3)How long is an arriving customer expected to wait before starting service? 55

56. Solution: Given: Arrival rate λ =10/hr=10/60=1/6 cars/min Service rate μ=1/5 cars per min Then utilization factor ρ = 5/6 1)p +p 1 +p 2 =p + ρ p o + ρ 2 p =p (1+ ρ + ρ 2 ) where p o =(1- ρ ) =0.42 2)Probability that arriving customer wait(1-0.42)=0.58 3)Time for which arriving customers wait before getting service( average waiting time in queue) Wq = =0.417   56

57. In a railway marshalling yard, goods train arrive at the rate of 30 trains per day. Assuming that service time follow exponential distribution with a mean of 36 minutes.calculate the following. 1)The mean queue size The probability that queue size exceeds 10 57

58. Solution: Given: Arrival rate λ =30/(24*60) trains/min =1/48 trains/min Service rate μ=1/36 trains/min Then utilization factor ρ = 36/48=0.75 1) Lq = =(0.75) 2 /(1-0.75)=2.25 trains 2)p(>10)=(0.75) 10 =0.056   58

59. ‰An airport runway for arrivals only has ‰Arriving aircraft join a single queue for the runway.The ‰Exponentially distributed service time with a rate µ = 27 arrivals / hour (As you computed in PS1.) ‰Poisson arrivals with a rate λ = 20 arrivals / hour 59 An Example of M/M/1 Queue or infinity/FCFS

60. Now suppose we are in holidays and the arrival rate increases λ = 25 arrivals / hour ‰ How will the quantities of the queueing system change? 60

61. Now suppose we have a bad weather and the service rate decreases µ = 22 arrivals / hour How will the quantities of the queueing system change? 61

62. Codification  is used to properly classify equipment's, raw materials, components and spares to suit the particular needs of any organisation.  Codification  is helpful to prevent duplication and multiplicity of stores and the mistakes which are caused by the normal practice of describing the material. Objectives of Codification Bringing all similar items together. To enable putting up of any future item in its proper place. To classify an item according to its characteristics. To give a unique code number to each item to avoid duplication and ambiguity. 62

63. A  database  is a data structure that stores organized information. Most  databases  contain multiple tables, which may each include several different fields. For  example , a company  database  may include tables for products, employees, and financial records. Databases are used  just about everywhere including banks, retail, websites and warehouses. Banks use  databases  to keep track of customer accounts, balances and deposits. Retail stores can use  databases  to store prices, customer information, sales information and quantity on hand. 63