A queueing system can be described as customers arriving for service, waiting for service if it is not immediate, and if having waited for service,

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1 Queuig System

2 A queueig system ca be described as customers arrivig for service, waitig for service if it is ot immediate, ad if havig waited for service, leavig the service after beig served.

3 The basic queuig System Queuig System Populatio of prospective customers Arrivig Customers Etered customers C C C queue Service mechaism Customer i service Served Customers

4 Basic Elemets of a Queuig System (i) Populatio of prospective customers (ii) Arrival Patter or probability distributio of iter arrival times (iii) Queue (iv) Queue disciplie or service disciplie (v) Service Mechaism (vi)probability distributio of customer service times (vii) State of the System

5 Populatio of prospective customers This is a iput source from which customers requirig service are geerated over time. Size of populatio: If arrival rate of customer to the queueig system is affected by o. of customer i the queueig system the populatio is assumed as fiite otherwise ifiite.

6 Arrival patter of customers We assume customers arrive i a Poisso process. This is same as assumig the iterval times are expoetial radom variables with the same parameter (i.e., the same parameter which give i Poisso Process ) Queue: A queue is characterized by the maximum permissible umber m of customers that it ca cotai. A queue is called ifiite or fiite if this umber is ifiite of fiite.

7 Queue Disciplie The rule for selectig the ext the customer from the queue for sevice FCFS First Come First Serve LCFS Last Come First Serve SIRO Service i Radom Order Sometimes o the certai priority basis Service Mechaism A service statio may have oe or several servers arraged i parallel. Aother service mechaism may cosists of several service facilities i a series.

8 Service Time Distributio Service times for differet customers will be assumed to be idepedet ad expoetial radom variables. State of the System: The state of the system at a istat of time refers to the total umber of customers i the system at that istat which icludes those who are uder service ad also those who are i queue.

9 Termiology ad Notatios N = the umber of the customers i the system at ay istat. Q= the umber of the customers i the queue at ay istat. P (T)=P(N(T)=) =P(there are -customers i the system at time T) P =P(N=) =P(there are -customers i the system at ay istat ), =0,1,2,. Note: N ad Q are radom variables ad R N =R Q ={0,1,2, }

10 s = the umber of servers workig i parallel i the queuig system. = mea arrival rate whe the state of the system is. = mea service rate whe the state of the system is (which is differet from mea service time). Note: Relatio betwee mea service rate ad mea service time Mea service time =1/(mea service rate)

11 L = Expected umber of customers i the system at ay istat = E(N) P 0 L q = expected umber of customers i the queue at ay istat = E(Q)

12 Now E( Q) kp( Q k) k 0 k 0 kp( N k s) ( s) P( N ) s s ( s) P

13 Server utilizatio = s Note that: Server utilizatio ad sigle server system. both are same for I geeral, both are differet otatios or termiology.

14 W W q = Mea waitig time of a customer i the system at ay istat. = Mea waitig time of a customer i the queue at ay istat. Relatio betwee W ad If the mea service time is Wq 1 the 1 W Wq

15 Kedall s Notatio: A queuig system ca be expressed as (a/b/c): (d/e/f) a Iter-arrival time distributio b Service time distributio c No. of servers d Queue disciplie e System capacity f size of the populatio We use the symbol M for Expoetial distributio ( which ca be used for a ad b )

16 Example: ( M/ M/ 4) :( FCFS / 20 / ) Iter arrival time is expoetial, service time is also expoetial for each of the 4 servers, queue disciplie is first come first serve basic, the maximum umber of customers allowed to the system is 20 (4 i service ad 16 i queue), ad the populatio size is ifiite.

17 Little s Law: These relatioships do ot hold whe the maximum customers allowed ito the system is fiite. I case fiite, the modified little s law as follows: where, called effective arrival rate q q L W L W q q L W L W 0 P

18 I each queueig problem, the parameters P, L, L q, W ad W q to be determied. The mai problem is to calculate P. So we eed to have some method to determie P. Birth ad Death Process Birth is a etry ito the system ad death is a departure after service completio. Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 3 / 10

19 BIRTH AND DEATH PROCESS Assumptios 1. Give that N(T ) =, the coditioal probability distributio of the time util the ext birth is expoetial with parameter λ, = 0, 1, 2, 2. Give that N(T ) =, the coditioal probability distributio of the time util the ext death is expoetial with parameter µ, = 1, 2, with µ 0 = 0. 3 P(more tha oe evet occurs i iterval (T, T + T )) = o( T ) 4 Oly oe birth or oe death ca occur at a time. Also births ad deaths are idepedet. Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 4 / 10

20 FROM THE ABOVE 4 ASSUMPTIONS 1. P(o birth i (T, T + T ) N(T ) = ) = 1 λ T + o( T ). 2. P(o death i (T, T + T ) N(T ) = ) = 1 µ T + o( T ). 3. P(exactly oe birth i (T, T + T ) N(T ) = ) = λ T + o( T ). 4. P(exactly oe death i (T, T + T ) N(T ) = ) = µ T + o( T ). 5. P(two or more births/deaths i (T, T + T ) N(T ) = ) = o( T ). 6. P(o births ad o deaths i (T, T + T ) N(T ) = ) = 1 (λ + µ ) T + o( T ). Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 5 / 10

21 BIRTH AND DEATH PROCESS Let A be the evet that exactly oe birth ad o death i (T, T + T ). Let B be the evet that exactly oe death ad o birth i (T, T + T ). Let C be the evet that o birth ad o death i (T, T + T ). The P(A N(T ) = ) = λ T + o( T ). P(B N(T ) = ) = µ T + o( T ). P(C N(T ) = ) = 1 (λ + µ ) T + o( T ). Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 6 / 10

22 BIRTH AND DEATH PROCESS Sice P (T ) = P(N(T ) = ), which implies P (T + T ) = P(N(T + T ) = ). Now P (T + T ) = P(N(T ) = & C or N(T ) = + 1 & B or N(T ) = 1 & A) + o( T ) P (T + T ) = P(N(T ) = & C) + P(N(T ) = + 1 & B) + P(N(T ) = 1 & A) + o( T ) P (T + T ) = P(C N(T ) = )P (T ) + P(B N(T ) = + 1)P +1 (T ) + P(A N(T ) = 1)P 1 (T ) + o( T ) Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 7 / 10

23 BIRTH AND DEATH PROCESS P (T + T ) = {1 (λ + µ ) T }P (T ) + µ +1 P +1 (T ) T + λ 1 P 1 (T ) T + o( T ) P(T + T ) P(T ) T = (λ + µ )P (T ) + µ +1 P +1 (T ) + λ 1 P 1 (T ) + o( T ) T dp(t ) dt = (λ + µ )P (T ) + µ +1 P +1 (T ) + λ 1 P 1 (T ) The solutios of above equatios are very difficult to obtai aalytically, except for some very special cases. Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 8 / 10

24 SPECIAL CASE(PURE BIRTH): Pure Birth process (i.e. µ = 0) with λ = λ, P 0 (0) = 1, ad P j (0) = 0 for j 0. The the above equatio becomes dp (T ) dt = λp (T ) + λp 1 (T ), for 1 ad dp 0 (T ) dt = λp 0 (T ). The solutio i this case will be P (T ) = (λt ) e λt!, 0 ad T > 0. Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 9 / 10

25 SPECIAL CASE(STEADY STATE/STATIONARY SYSTEM): If lim T P (T ) = P ad lim T dp (T ) dt = 0 for all 0,we have 0 = (λ + µ )P + µ +1 P +1 + λ 1 P 1 for 1 ad P +1 = 0 = µ 1 P 1 λ 0 P 0. λ µ +1 P for 0 ad P = λ 0λ 1 λ 1 µ 1 µ 2 µ P 0 for 1. ) = 1. As =1 P = 1 which implies P 0 ( 1 + λ 0 µ 1 + λ 0λ 1 µ 1 µ 2 + Thus, the steady state probabilities exist if the series S = 1 + C 1 + C 2 + <, where C = λ 0λ 1 λ 1 µ 1 µ 2 µ The P 0 = 1/S ad P ca be determied for ay. Dr. J. K. Sahoo (BITS) Lecture 3 Sectio-1 10 / 10

26 Queueig Models: Model 1. ( M/M/1 ) :( FCFS/ / ) with, 0,1,2,..., 1,2,... Now ad λ C = µ = ρ, 1,2, S = 1+ C + C +... if Thus for existece of steady state probabilities, we must have 1.

27 0 P = 1 S = 1- ρ Also, P = CP 0 = ρ (1- ρ), = 0,1,2,... which is geometric distributio with probability of success 1- ρ Hece, the steady state umber of customers follows geometric distributio. Performace Measures: L 1 L E( N), W Wq W, Lq Wq ( ) ( ) s

28 Model 2. ( M/M/1 ) : ( FCFS/m/) with This gives S, for 0,1,..., m1 for all. 0, for m m1 1 if 1 or 1 ad S m 1, m m

29 Therefore, , 1 1, 1 (1 ),, 0,1,..., 1 1,, 0,1,..., 1 m m P m ad m P m m

30 m m1 P P (1 P ) m 0 0 Now, if, L E( N ) P m the 0 0 m1 m m1 m1 0 1 m1 0 (1 ) (1 ) ( m 1) 1 1 m 1 m If, the L m 1 2 m 1 L L W, Wq W (1 P ) ad L W q m q 1

31 Model 3. ( M/M/s ) :( FCFS/ / ) with, s, 0,1,2,... ad s, s 1 1 Therefore, C, for 1,2,...,( s 1).!! 1 ad C, for s. s s s! S 2 s 1 s s 1 s s ! ( s 1)! s! ss! s s! s 1! s! 1 s if s.

32 !! 1, 0,1,..., ( 1).!,.! s s s Now P S s s P s ad P P s s s For existece of steady state probabilities, we must have. s

33 s k q ( ) s k k s k 0 k 0 s s L s P kp k P! 0 s k s 1 P 0 k P0 1 k s! k 0 s ss! s Lq W q 1 W W q L W 2

34 Model 4. ( M/M/s ) : ( FCFS/m/) with, for 0,1,..., m 1, for s ad 0, for m s, for s 2 2 s s Thus, C, C,..., C 2 2 s! s m ad Cs 1,..., Cm ms s! s s s! Also S C C C, m

35 We get S 2 s1 s ms ms 2 ( s 1)! s! s s ms1 1 s1 s s, 1 0! s! 1 s s s 1 s ad S ( m s 1), if 1! s! s 0

36 Therefore, P 0 ms1 1 s1 s s, if 1! s! 0 1 s s 1 s 1 s ( m s 1), if 1! s! s 0 1

37 P0, for 0,1,..., s 1! Also, P P0, for s, s 1,..., m s s s! 0, for m 1, m 2... Now L ( s) P kp q k s s k0 ms k 0 k sk k s s! P 0 P ( m s) ss! s s s s1 2 ms1 0 2 ms1

38 , (1 ) 1,, m q q q Also P L Thus W W W L W ) ( 1) ( 1! s s s m s s m P ss L s m s m s q

39 Model 5. ( M /M /s) : ( FCFS/K/K ) This Model, the limited source model i which there are oly K customers, is called the machie repair model, the machie iferece model or the cyclic queue model. K- total umber of machies workig i parallel. s- Number of mechaics (servers) ad s K. Assumptios: 1. The time util breakdow of a machie has expoetial distributio with parameter ad the machie s fuctio idepedetly. 2. The service time of each machie is expoetially distributed with parameter ad service times are idepedet.

40 Let there be machies i the repair shop (system). ( K ) machie s are fuctioig. Let T be the time util ext machie arrives. T mi{ T1, T2,..., T( K ) }, T where is the time util breakdow of the machie. T i has expoetial distributio with parameter The arrival occurs i poisso process with mea ( K ), whe there are -customers i the system. ( K ), 0,1,..., K 1 0, K i th ( K).

41 Now, 1,2,..., s 1 s, s ad K K K K K C1 K, C C 2 ( 1) ( 1) K!, 1,2,..., s 1!( K )! K!, s, s 1,... K s s s!( K )! 2

42 S s1 0 K K! K! s!( K )! s s!( K )! s P 0 1 S K! P0, 1, 2,..., s 1!( K )! K! P P0, s, s 1,..., K s s s!( K )! 0, K

43 L K 0 P, K P ( K ) P ( K ) P ( K L) K W L, 1 Wq W, L q W q.

44 Problem 1: O a average, 5 customers reach the barber s shop every hour. Determie the probability that exactly 2 customers will reach i a 30 miute period, assumig that the arrivals follow Poisso distributio. As: 5, t 0.5 P( X 2) ( t) 2 e t 2!

45 Problem 2: I a barber s shop with a sigle barber, there are three chairs. Out of these, oe chair is for hair cuttig ad the other two for waitig iside the shop. The customers arrive accordig to Poisso distributio with average 2 customers per hour. The service time is expoetial with mea 20 miutes per customers. If all the chairs are occupied, the ay customer that comes goes elsewhere for hair cuttig. Fid the probability that (i) a customer will get a chair to sit. (ii) a customer will get immediate service. (iii) a customer will wait outside the shop or go elsewhere for hair cuttig. Also fid L ad W.

46 Solutio: (i) 2, 3, P P P 1P m=3, 2. 3 (ii) P 0 (iii) P( 3) L P P P P 2 3, L 2(1 P), W.

47 Problem 3: Trais arrive at the yard expoetially every 15 miutes ad the service time is also expoetial with a mea of 30 miutes. If the lie capacity of the yard is limited to 4 trais, fid (a) The probability that the yard is empty. (b) The average umber of trais i the system. As: (a) P (b) L

48 Problem 4: Problem VII.3 Page 7.70 A telephoe exchage has two log distace operators. Durig the peak hours, log distace calls arrive i a Poisso process at the mea rate of 15 per hour. The legth of service o these calls is approximately expoetial with mea 5 miutes. (a) Fid the probability that a subscriber will wait for his log distace call durig the peak hours? (b) Fid the mea waitig time i the queue?

49 Solutio: (M/M/2) : (FCFS//) model with = 15, = 12, = 5/4<2 P ! s!(1 / s) 2 0 s1 s 2 P 1 = P 0 =15/52 Thus the probability that a ew subscriber will wait for the log distace call = Probability that the umber of customers i the system 2=1 - P 0 - P 1 = 0.48

50 L q s P 2 0 ss!(1 / s) W q Lq hr 3.2mi

51 Problem 5: Problem VII.5 Page 7.70 At a service statio, customers arrive i a Poisso process at the rate of 15 per hr. The service time of a customer is expoetial with mea 6 miutes. A server has to be paid Rs. 4/- per hour ad the cost icurred ad bore by the service statio due to the presece of a customer i the system is Rs. 10/- per hr per customer. (a) What is the miimum umber of servers eeded for steady state to exist? (b) What is the optimum umber of servers eeded to miimize the expected cost per hr? (c) If the service statio charges Rs. 20/- from each served customer (irrespective of the time devoted to servig him), what is the optimum umber of servers eeded to maximize the per hr et reveue of the service statio?

52 Solutio: Solutio We have a (M/M/s) : (FCFS//) model with = 15, = 10, =1.5 Thus for steady state to exist we must have < s or s 2. Case (i) s = 2. P 0 1, 7 L q 9, 3 8 L Expected Total cost per hour = 10 L + 4s = Case (ii) s=3. Expected Total cost per hour=10*66/38+4*3=29.36 Case (ii) s 4. Expected Total cost per hour 31. Thus optimum umber of servers =3.

53 Problem 6: Arrival rate of telephoe calls at a telephoe booth is accordig to Poisso distributio, with a average time of 9 miutes betwee two cosecutive arrivals. The legth of a telephoe call is assumed to be expoetially distributed with mea 3 miutes. (a) Determie the probability that a perso arrivig at the booth will have to wait. (b) Fid the average queue legth that forms from time to time. (c) Fid the fractio of a day that the phoe will be i use. As. (a) 0.33,(b) 0.166, (c) 0.33

54 Problem 7. I a cliic, patiets arrive i Poisso patter at a average rate of 15 patiets per hour. The waitig room ca accommodate maximum 14 patiets. If the doctor takes o a average 6 miutes per patiet ad assumig expoetial service time, fid (i) the effective arrival rate. (ii) the probability that a arrivig patiet will ot wait. (iii) the probability that a arrivig patiet will fid a vacat seat.

55 Solutio 7. We have (M/M/1) : (FCFS/15/) with = 15, = 10, =1.5. (i) the effective arrival rate=15(1- P 1 5 ). (i) (iii) P 0. P ( 1 4 ) 1 P. 1 5

56 Problem 8. I a car repairig shop, there are two mechaics for repairig the cars. The cars come i a Poisso patter o a average 3 cars per day. The service time for ay mechaic is expoetially distributed with mea 4 hours. Assumig 8 hours workig day, fid (a) the hours of the day for which all the mechaics are busy. (b) the probabilities that both the mechaics are idle, oe mechaic is idle. (c) the expected umber of cars i the queue. (d) the expected waitig time of a car i queue.

57 Solutio 8. We have (M/M/2) : (FCFS/ /) with = 3, = 2 per day ad =1.5. (a) P ( 2 ) 1 P P. (b) P(both the mechaics are idle)= P(oe mechaic is idle)= 0 1 P. 0 P 1 0. (c) the expected umber of cars i the queue= (d) the expected waitig time of a car i queue= P 2 L W q. q

58 Problem 9: Problem VII.15 Page 7.73 Trucks arrive at a service statio i a Poisso process at a rate of 10 per hour. The service statio has two servers i parallel, each with a mea service rate of 10 trucks per hour. Service times are expoetial. There is space for oly two trucks beig served ad for two i waitig. If all spaces are filled up, the trucks will go to aother service statio.

59 (a) (i) What is the expected umber of trucks i the system? (ii) What fractio of customers are lost? (iii) What is the per hour effective arrival rate of the service statio? (b) Aswer all the four questios above if 3 spaces are provided for waitig. (c) If the provisio of the third waitig space costs a retal of Rs. 5/- per hour ad if the reveue from the service of a truck is Rs. 10/-, will the provisio of the 3 rd waitig space be profitable for the service statio?

60 Solutio: We thus have a (M/M/2):(FCFS/4/) model with = 10 per hour ad = 10 per hour. Thus = 1. P ! 2! 2 2! P 1 = P 0 = P 0 =8/23, P 2 = 2 P 0 /2 =4/23, P 3 = 3 P 0 /4 =2/23, P 4 = 4 P 0 /8 = 1/23

61 (i) L = P P P P 4 = 26/23 (ii) Fractio of customers lost = P 4 = 1/23 (iii) Effective arrival rate =(1- P 4 ) = 220/23=9.57 (b) If there is oe more waitig space, the we have a (M/M/2):(FCFS/5/) model. P ! 2! 2 2! 4 2!

62 P 1 = P 0 = P 0 =16/47, P 2 = 2 P 0 /2 =8/47, P 3 = 3 P 0 /4 =4/47, P 4 = 4 P 0 /8 = 2/47, P 5 =1/47. L=57/47, 460 (1 P5 ) c) Extra trucks serviced per hour by providig extra waitig space = = The extra reveue eared per hour = Rs. 2.2 which is less tha the retal cost Rs Thus, the ew arragemet is ot profitable to the compay.

63 Problem 10: VII.14 A factory has 4 machies. The time util breakdow of a machie is expoetial with mea 10 hours. The time to repair a machie is expoetial with mea 5 hours. A repairma costs Rs. 15/- per hour. The cost of lost time of a ioperative machie is Rs. 30/- per hour per machie. Should the factory hire oe repairma or two repairme? Justify.

64 Solutio: This is a machie repair model with K = 4, = 1/10, = 1/5. Thus =1/2. Case (i) s=1 = (4-), =0, 1, 2, 3 ad 0 elsewhere. =, = 1, 2, P 1 = 4 P 0 = 2 P 0, P 2 = 12 2 P 0 = 3 P 0 P 3 = 24 3 P 0 = 3 P 0, P 4 = 24 4 P 0 = 3/2P P0, P1, P2, P3, P

65 L= P 1 +2P 2 +3P 3 +4P 4 = 46/21 ad the expected cost for the firm = 30 L+15=80.71 Case (ii) s=2. = (4-), =0, 1, 2, 3 ad 0 elsewhere. 1 = ad = 2 for = 2,3, P 1 = 4 P 0 = 2 P 0, P 2 = 6 2 P 0 = 3/2 P 0 P 3 = 6 3 P 0 = 3/4 P 0, P 4 = 3 4 P 0 = 3/8P 0

66 ad P P, P, P, P L=14/9 ad the expected cost for the firm=30l+30=76.67 Thus the firm will hire two repairme.

67 Problem VII.8 (Page 7.71) A bak has two tellers workig o savigs accouts. The first teller hadles withdrawals oly ad the secod teller hadles deposits oly. The service time for both deposits ad withdrawals are expoetial with mea 3 miutes per customer. Depositors arrive i a Poisso fashio with a mea rate of 16 per hour, while withdrawers arrive i a Poisso fashio with a mea rate of 14 per hour. What would be the effect o the average waitig time i the system for depositors ad withdrawers if each teller could hadle both deposits ad withdrawals? What would be the effect if this could be accomplished oly by icreasig the mea service time to 3.5 miutes? Solutio Withdrawals Deposits W/D & DEP W/D&DEP /7 7/10 4/5 3/2 7/4 P 0 3/10 1/5 1/7 1/15 L q 49/30 16/5 27/14 343/60 L 7/3 4 24/7 112/15 W 1/6 =10 mts 1/4 = 15 mts 4/35 = 6.86 mts 112/450 =14.93 mts

First come, first served (FCFS) Batch

First come, first served (FCFS) Batch Queuig Theory Prelimiaries A flow of customers comig towards the service facility forms a queue o accout of lack of capacity to serve them all at a time. RK Jaa Some Examples: Persos waitig at doctor s