Non-Markovian Queues and Queue Networks1 MISRIMAL NAVAJEE MUNOTH JAIN ENGINEERING COLLEGE DEPARTMENT OF MATHEMATICS PROBABILITY AND QUEUING THEORY (MA 2262) SEMESTER-IV QUESTION BANK – V UNIT– V – NON-MARKOVIAN QUEUES AND QUEUE NETWORKS PART-A Problem: 1 Write down Pollaczek - Khinchine formula. Solution: i) Average number of Customers in the system ( ) 2 2 2 2 1 ì o p p p + = + ÷ where ì p u = . ii) Average queue length ( ) 2 2 2 2 1 ì o p p + = ÷ Problem: 2 Write the steady-state equations (flow balance equations) for a two-station sequential Queue with blocking. Solution: 0,0 2 0,1 p p ì u = 1 1,0 2 1,1 0,0 p p p u u ì = + 2 0,1 1 1,0 2 ,1 ( ) b p p p ì u u u + = + 1 2 1,1 0,1 ( ) p p u u ì + = 2 ,1 1 1,1 b p p u u = Problem: 3 Write the flow-balance equations of open Jackson networks. Solution: 0,1,2,... 1 k j j i ij j k i r P ì ì = = = + ¯ where P ij is the probability that a departure from server i joins the Queue at server j. Problem: 4 Write the flow-balance equations of closed Jackson networks. Solution: 0,1,2,... 1 k j i ij j k i P ì ì = = = ¯ Non-Markovian Queues and Queue Networks 2 where P ij is the probability that a departure from server i joins the Queue at server j. Problem: 5 State the characteristics of Jackson networks. Solution: (a) Arrivals from outside through node i follow a Poisson process with mean arrival rate r i . (b) Service times at each channel at node i are independent and exponentially distributed with parameter i u (c) The probability that a customer who has completed service at node i will go next to node j (Routing probability) is P ij , i= 1, 2, 3, ….k and j = 0, 1, 2, 3, …k. and P i0 denotes the probability that a customer will leave the system from node i. P ij is independent of the state of the system. If λ j is the total arrival rate of customers to node j, 0,1,2,... 1 k j j i ij j k i r P ì ì = = = + ¯ Problem: 6 Distinguish between open Jackson networks and closed Jackson networks Solution: We have, 0,1,2,... 1 k j j i ij j k i r P ì ì = = = + ¯ In all general cases where, r i ≠ 0 for any i or P i0 ≠ 0for any i, the Jackson networks are referred to as open Jackson networks. In the case of closed Jackson networks, r i = 0 for all i and P i0 = 0for all i. PART-B Problem: 7 Derive Pollaczek - Khinchine relation for ( ) ( ) / /1 : / M G GD · Solution: In order to determine the mean queue length L q and mean waiting time Wq for this system the following technique is used. Let ( ) f t = Probability distribution of service time t with mean ( ) E t and Variance ( ) V t . n =no. of customers in the system just after a customer departs. t = service time of the customer following the one already departed. 1 n = The number of customers left behind the next departing customer. 1 0 n k if n = = ( ) 1 0 n k if n = ÷ + > Where k = 0,1,2….is the number of arrivals during the service time. Alternatively of 1 0 if n o = = 0 0 if n = > 1 1 (1) n n k o = ÷ + + ÷ Non-Markovian Queues and Queue Networks 3 Then taking expectation ( ) ( ) ( ) ( ) 1 1 E n E n E E k o = + + ÷ Since ( ) ( ) 1 E n E n = in steady state ( ) ( ) 1 E E k o = ÷ Squaring (1) both sides ( ) ( ) 2 2 1 1 n n k o = + ÷ + ( ) ( ) ( ) 2 2 2 1 2 1 2 2 1 . n k n k n k o o o = + ÷ + ÷ + + + ÷ ( ) ( ) 2 2 2 1 2 1 2 2 2 n k n k n k o o o o = + ÷ + ÷ + + + ÷ Since o can take values 0 or 1only 2 0 and n o o o = = ( ) 2 1 2 2 2 1 2 1 2 2 n n k k n k k o o o = + ÷ + + ÷ + + ÷ ( ) ( ) 2 2 2 1 2 1 2 1 n k n k k k o = + + + + ÷ ÷ + ( ) ( ) 2 2 1 2 2 1 2 1 2 1 n k n n k k k o ÷ = ÷ + + ÷ ÷ + Taking expectation on both sides ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 (2 ) 1 2 1 E n E k E n E n E k E E k E k o 2 ÷ = ÷ + + ÷ ÷ + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 1 2 1 E k E k E E k E n E k o ÷ + ÷ + = ÷ ( ) ( ) 1 E E k o = ÷ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 1 1 2 1 E k E k E k E k E n E k ÷ + ÷ ÷ + = ÷ . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 E k E k E k E n E k + ÷ = ÷ Now is order to determine ( ) E n , the values of ( ) ( ) 2 and E k E k are to be computed. Since the arrivals follow Poisson distribution ( ) ( ) ( ) 0 / E k E k t f t dt · = í Now ( ) ( ) ( ) ( ) 2 2 / , / E k t t E k t t t ì ì ì = = + ( ) ( ) ( ) 2 2 / / / V k t E k t E k t = ÷ ( ) ( ) ( ) 0 E k t f t dt E t ì ì · = = í ( ) ( ) ( ) 2 2 0 / E k E k t f t dt · = í ( ) ( ) 2 0 t t f t dt ì ì · = + í ( ) ( ) ( ) ( ) ( ) 2 2 2 2 E t E t V t E t E t ì ì ì ì = + = + + ( ) ( ) ( ) ( ) 2 2 2 E k V t E t E t ì ì = + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 V t E t E t E t E t E n E t ì ì ì ì ì + + + ÷ = ÷ Non-Markovian Queues and Queue Networks 4 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 V t E t E t E t E t ì ì ì ì ì + + ÷ = ÷ ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 V t E t E t E t E t ì ì ì ì ì + + ÷ = ÷ ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 E t V t E t E t ì ì ì ì + = + ÷ ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 s E t V t L E n E t E t ì ì ì + = = + ÷ ì s s L W = ( ) 1/ where is the rate of servive. E t u u = Then ( ) 2 2 2 2 2 2 (1/ ) ( ) / 2 1 / 2(1 ) s L ì u o p ì o ì u p ì u p + + = + = + ÷ ÷ where 2 / and ( ) V t ì u p o = = 2 2 2 ( ) 2(1 ) q s L L p ì o p p + = ÷ = ÷ Problem: 8 A one man barber shop takes exactly 25 minutes to complete one haircut. If customers arrive at the barber shop in a Poisson fashion at an average rate of one every 40 minutes, how long on the average a customer spends in the shop? Also find the average time a customer must wait for service. Solution: Since the service time T is a constant=25 minutes, T follows a probability distribution with E(T)=25, Var(T)=0. 1 ì = for every 40 minutes 1 40 = per minutes By Pollaczek-Khinchine formula, ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 var 2 1 1 0 25 25 40 1 40 2 1 .25 40 1 .625 25 25 625 4 5 25 55 1600 15 40 40 1600 3 8 48 48 2 40 s T E T L E T E T ì ì + = + ÷ | | + | \ . = + | | ÷ | \ . = + = + × = + = | | | \ . By Little formula, Non-Markovian Queues and Queue Networks 5 55 55 48 40 1 48 40 s S L W ì = = = × 45.8 = Minutes ( ) ( ) 1 1 45.8 45.8 25 q s W W E T E T u u = ÷ = ÷ = = ÷ 20.8 = Minutes Hence, a customer has to spend 45.8 minutes in the shop and has to wait for service for 20.8 minutes on the average. Problem: 9 Suppose a one person tailor shop is in business of making men’s suits. Each suit requires four district tasks to be performed before it is completed. Assume all four tasks must be completed on each suit before another is started. The time to perform each task has an exponential distribution with a mean of 2 hr. If orders for a suit come at the average rate 5.5 per week (assume an 8hr day, 6 day week), how long can a customer expect to wait to have a suit made? Solution: Given: 5.5 ì = per week 5.5 6 8 = × orders/hr =0.1149 orders/hr The server time for each task 1 ku = where 4 k = i.e., 1 2 . 4 hr u = = (given) 1 8 hr u = =0.125 orders / hr. 0.1149 0.9192 0.125 ì p u = = = Since the service time T is constant then 2 0 o = Here the average service rate u is the average service rate to complete a suit. The expected waiting time of a customer (in the system) is Non-Markovian Queues and Queue Networks 6 ( ) 2 2 2 1 2 1 s W ì o p ì p u + = + ÷ Problem: 10 In a heavy machine shop, the overhead crane is 75% utilized. Time study observations gave the average slinging time as 10.5 minutes with a standard deviation of 8.8 minutes. What is the average calling rate for the services of the crane and what is the average delay in getting service? If the average service time is cut to 8.0 minutes, with a standard deviation of 6.0 minutes, how much reduction will occur, on average, in the delay of getting served? Solution: This is a ( ) ( ) / /1 : / M G FIFO · Process Given: Utilization rate 75 3 75% 100 4 = = = i.e., 3 4 p = Average service time E(T)=10.5 min ( ) 1 1 10.5 E T u = = We know that ì p u = ì pu ¬ = 3 1 0.0714min 4 10.5 ì | || | = = | | \ .\ . i.e., average calling rate for the services of the crane 0.0714min 4.286 ì = = hour To find the average delay in getting service Here 3 8.8, 0.0714, 0.75 4 o ì p = = = = Average delay in getting service ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 0.0714 8.8 0.75 26.815 min 2 1 2 0.0714 1 0.75 q W ì o p ì p + + = = = ÷ ÷ i.e., The average service time is cut to 8.3 minutes then ( ) ( ) 1 1 min 8.0 0.0714min 0.0714 0.5712 1/ 8 6min E T u ì ì p u o = = = = = = = Non-Markovian Queues and Queue Networks 7 Average delay in getting service ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 0.0714 6 0.5712 2 1 2 0.0714 1 0.5712 8.326 min q W ì o p ì p + + = = ÷ ÷ = The Average waiting time has a reduction of 26.8-8.32=18.5 minutes ( ) ( )( ) 2 0 0.9192 1 53.42 2 0.1149 0.0808 0.125 + = + = hrs =1.113 Weeks . Problem: 11 A car manufacturing plant uses one big crane for loading cars into a truck. Cars arrive for loading by the crane according to a Poisson distribution with a mean of 5 cars per hour. Given that the service time for all cars is constant and equal to 6 minutes determine , , S q S q L L W and W . Solution: The given problem is in ( ) ( ) / /1 : / M G FIFO · model. 5cars/hr ì = cars/hr. The service time T is constant with mean | | 1 6 E T u = = min Since the service time T is constant then var[T]= 2 0 o = 1 1 1 6 min 6 . 60 10 10 5 1 10 2 hr hr hr hr u u ì p u = = × = = = = = Using k p ÷ formula, we have ( ) ( ) ( ) 2 2 2 2 1 0 1/ 4 1 3 0.75 / 2 2 1 1/ 2 4 3 1 4 2 0.25 / s q s L car hr L L car hr ì o p p p ì u + = + ÷ + = + = = ÷ = ÷ = ÷ = ( ) ( ) 3/ 4 3 9 1/ 5 20 s s L W hr ì = = = = minutes Non-Markovian Queues and Queue Networks 8 ( ) ( ) 1/ 4 1 3 1/ 5 20 q q L W hr ì = = = = minutes. Problem: 12 A car wish facility operates with only one bay. Cars arrive according to a Poisson distribution with a mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy. The parking lot is large enough to accommodate any number of cars. Find the average number of cars waiting in the packing lot, if the time for washing and cleaning a car follows. (a) Uniform distribution between 8 and 12 minutes (b) A normal distribution with mean 12 minutes and S.D.3minutes. (c) A discrete distribution with values equal to 4,8 and 15 minutes and corresponding probabilities 0.2,0.6 and 0.2. Solution: Mean = 4 ì = / hour = 4 60 per minute 1 15 = per minute Let T be a continuous random variable. Then, E(T) = mean of the uniform distribution ( ) ( ) ( ) 1 1 1 8 12 20 10 2 2 2 a b = + = + = = Var (T) ( ) 2 1 12 b a = ÷ ( ) ( ) ( ) 2 2 1 1 1 4 12 8 4 16 12 12 12 3 = ÷ = = = By the Pollaczek-Kninchine formula, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 var . 2 1 1 4 10 1 15 3 .10 1 15 2 1 .10 15 1 4 100 2 225 3 2 3 2 1 3 1 304 . 2 2 1 304 3 2 152 302 225 3 . . 1 3 3 225 3 2 3 225 225 2 3 s T E T L E T E T ì ì ì + = + ÷ | | + | \ . = + | | ÷ | \ . + = + | | ÷ | \ . = + = + = + = | | | \ . Non-Markovian Queues and Queue Networks 9 1.342 = Cars ~ Car By Little’s formula, 1 15 1.342 1 10 q s L L ì u = ÷ = ÷ 0.675 = cars ~ 1 car (b) 1 , 15 ì = ( ) 12 E T = min, and Var(T)=9 ( ) 1 1 12 E T u = = By the Pollaczek-Kninchine formula, ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 var . 2 1 1 1 9 12 .153 1 4 4 153 15 15 225 .12 . 1 3 15 5 5 225 6 2 1 .12 2 15 15 s T E T L E T E T ì ì ì + = + ÷ | | + | \ . = + = + = + | | | | ÷ | | \ . \ . 4 153 2.5 5 90 = + = Cars. By Little’s formula, 1 12 15 2.5 2.5 2.5 0.8 1.7 2 1 15 12 q s L L Cars Cars ì u = ÷ = ÷ = ÷ = ÷ = ~ (c) The service time T follows the discrete distribution given below. T : 4 8 15 P(T) : 0.2 0.6 0.2 ( ) ( ) ( ) 4 0.2 8 0.6 15 0.2 0.8 4.8 3 TP T E T = = × + × + × = + + ¯ 8.6 = minutes Non-Markovian Queues and Queue Networks 10 ( ) ( ) ( ) ( ) 2 2 4 0.2 8 0.6 15 0.2 0.8 4.8 3 8.6minutes T P T 2 E T = = + × + × = + + = ¯ ( ) ( ) ( ) ( ) 2 2 2 var 86.6 8.6 12.64 E T E T T = ÷ = ÷ = BY Pollaczek-Khinchine formula, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 var . 2 1 1 12.64 8.6 1 225 8.6 1 15 2 1 8.6 15 s T E T L E T E T ì ì ì + = + ÷ + = × + ÷ × | | 1 12.64 73.96 86.6 15 225 0.573 0.573 1.024 1 8.6 225 12.8 2 1 15 Car + = + = + × = ~ ÷ By Little formula, q s L L ì u = ÷ ( ) 1 1 5 1.024 1 86 E T u = ÷ = 8.6 1.025 0.45 5 Car = ÷ = Problem: 13 An automatic car wash facility operates with only one bay; Cars arrive according to a Poisson process with mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy. If the service times for all cars is constant and equal to 10 minutes. Determine , , , q q s q L L W W . Solution: 4 ì = per hour 4 60 = per minute 1 15 = per minute Given service time ‘T’ is a constant. ’T’ follows a distribution with E(T)=10 and Var(T)=0. Non-Markovian Queues and Queue Networks 11 By Pollaczek - Khinchine formula, ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 var . 2 1 1 0 10 1 15 10 1 15 2 1 .10 15 100 10 10 100 15 20 4 225 15 10 15 15 225 10 15 3 2 15 s T E T L E T E T ì ì ì + = + ÷ | | + | \ . = × + | | ÷ | \ . = + = + × = = ÷ | | | \ . By Little formula, 4 15 3 s S L W ì = = × 20 = minutes ( ) ( ) 1 1 20 20 10 q s W W E T E T u u = ÷ = ÷ = = ÷ 10 = minutes. i.e., a customer has to spend 20 minutes in the system and 10 minutes in the queue. . 1 10 10 15 15 q q L W ì = = × = Problem: 14 In a big factory, there are a large number of operating machines and two sequential repair shops, which do the service of the damaged machines exponentially with respective rates of 1/hour and 2/hour. If the cumulative failure rate of all the machines in the factory is 0.5 / hour, find (i) the probability that both repair shops are idle, (ii) the average number of machines in the service section of the factory and (ii) The average repair time of a machine. Solution: Given 0.5 ì = / hour 1 2 = per hour 1 u = 1 per hour 2 u = 2 per hour Non-Markovian Queues and Queue Networks 12 The situation in this problem is comparable with 2-stage Tandem queue with single server at each state. (i) P(both the service stations are idle) ( ) 0 0 1 1 2 2 0 0 0, 0 . 1 . . 1 1/ 2 1/ 2 1/ 2 1/ 2 1 1 1 1 1 1 1 1 1 3 3 1 2 4 2 4 8 P ì ì ì ì u u u u = | | | | | | | | = ÷ ÷ | | | | \ . \ . \ . \ . | | | || | | | = ÷ ÷ | | | | \ . \ .\ . \ . | || | | | = ÷ = = | | | \ .\ . \ . (ii) The average number of machines in service 1 2 1 1/ 2 1/ 2 1/ 2 1 4 2 1 1 1 1/ 2 3/ 2 3 3 1 2 2 2 ì ì u ì u ì = + ÷ ÷ = + = + = + = ÷ ÷ (iii) The average repair time 1 2 1 1 1 1 1 1 8 1 1 1 3 3 1 2 2 2 2 2 u ì u ì = + ÷ ÷ = + = + = | | | | ÷ ÷ | | \ . \ . Problem: 15 A TVS company in Chennai containing a repair section shared by a large number of machines has 2 sequential stations with respective service rates of 3 per hour and 4 per hour. The cumulative failure rate of all the machines is 1 per hour. Assuming that the system behavior can be approximated by the above 2-stage tendon queue, find (i) the probability that booth the service stations are idle (free) (ii) the average repair time including the waiting time. (iii) the bottleneck of the repair facility. Solution: STEP-1: Model Identification The current situation comes under the sequence queue model, Since any number of machines can be repaired, each station comes under the model ( ) ( ) / /1 : / M M FCFS · STEP 2: Given Data Cumulative failure rate 1 ì = Service rate of station I 1 3 u = Non-Markovian Queues and Queue Networks 13 Service rate of station II 2 4 u = STEP 3: To find the following (1) The probability that both the servive stations are idle. (2) The average repair time and waiting time (3) The bottleneck of the repair facility. STEP 4: Required Computations (i) P( m customers in the I station and n customers in the II station 1 1 2 2 1 1 2 2 0 0 00 1 1 1 1 1 1 2 3 1 1 1 3 3 4 4 3 4 2 m n mn P P ì ì ì ì u u u u | | | || | | | = ÷ ÷ | | | | \ . \ .\ . \ . | | | || | | | = ÷ ÷ = × = | | | | \ . \ .\ . \ . (ii) The average number of machines in service at the system(both the stations) 1 2 1 1 1 1 5 3 1 4 1 2 3 6 s L ì ì u ì u ì = + ÷ ÷ = + = + = ÷ ÷ = 1 machine (approximately) Average repair time (including waiting time) 1 2 1 1 u ì u ì = + ÷ ÷ 5 6 = hours =50 minutes. (iii) Since 1 2 1 3 ì ì u u = > 1 4 = We get the service station 1 is the bottleneck of the repair facility. Problem: 16 In the Airport reservation section of a city junction, there is enough space for the customers to assemble, form a queue and fill up the reservation forms. There are 5 reservation counters in front of which also there is enough space for the customers to wait. Customers arrive at the reservation counter section at the rate of 40 per hour and takes one minute on the average to fill up the forms. Each reservation clerk takes 5 minutes on the average to complete the business of a customer in an exponential manner.(i) Find the probability that a customer has to wait to get the service in the reservation counter section (ii) Find the total waiting time for a customer in the entire reservation section. Assume that only those who have the filled up reservation forms will be allowed into the counter section. Solution: Non-Markovian Queues and Queue Networks 14 The queuing system in the form-filling portion is only a M/M/1 model, since each customers is served by himself or herself (viz., one server) For this system, 40 ì = / hour and 60 u = /hour. ( ) 40 40 2 60 40 20 s E N ì u ì = = = = ÷ ÷ ( ) 1 1 20 s E W u ì = = ÷ hour or 3 min The queuing system in the reservation counter section is an M/M/s model with 40 ì = /hour, 12 u = /hour and 5 s = [Since the output of the M/M/1 system in the same as the input of that system, by Burke’s theorem and the output of this system (namely, 40/hour) becomes the input of the M/M/s system] (i) P(a customer has to wait in the counter section) ( ) 2 0 . , .... 1 1 P s s ì u ì u | | | \ . = | | ÷ | \ . Where ( ) ( ) ( ) 1 1 0 0 1 5 4 1 0 10 1 2 3 4 / 1 . 1 10 1 10 3 . 3 5! 1 2/ 3 100000 1 10 1 10 1 10 1 10 1 10 243 1 0! 3 1 3 2 3 3 3 4 3 120 r s s r r r P r s s r ì u ì u ì u ÷ ÷ = ÷ ÷ = ¦ ¹ ¦ ¦ | | ¦ ¦ = + ´ ` | | | \ . ¦ ¦ ÷ | ¦ ¦ \ . ¹ ) | | | | | \ . + | ÷ \ . | | | | | | | | | | = + + + + + | | | | | \ . \ . \ . \ . \ . ¯ ¯ 1 3 ÷ | | | \ . 1 10 100 1000 1 100000 100000 3 1 3 18 162 24 81 243 120 ÷ | || | = + + + + + + | | \ .\ . | | | | 1 1 1 10 100 1000 1 10000 1 10.288 3 18 162 24 81 21.206 10.288 31.494 0.032 ÷ ÷ ÷ = + + + + + = + = = Non-Markovian Queues and Queue Networks 15 ( ) 1 ¬ Required probability ( ) 5 10 0.032 3 2 5! 1 3 | | | \ . = | | ÷ | \ . ( ) ( ) | | ( ) 1 0 2 100000 0.032 243 0.3292 1 120 3 / 1 ! 1 s s E N P ss s ì u ì u ì u + | | | \ . = = | | | \ . = + | | ÷ | \ . ( )( ) ( ) ( )( ) ( ) | | 6 10 1 10 3 0.032 2 5 5! 3 1 3 1000000 1 10 729 0.032 1 5 120 3 3 1 131.69 3.3333 0.2195 3.3333 3.5528 600 | | | \ . = + | | ÷ | \ . | | | \ . = + | | | \ . = + = + = ( ) | | ( ) 1 1 3.5528 40 0.0888 s s E W E N hour ì = = = 5.329 = minutes Total waiting time of a customer in the entire reservation room =6+5.329=11.3292 minutes. Problem: 17 The last two things that are done to a car before its manufacture is complete are installing the engine and putting on the tires. An average of 54 cars per hour arrives, requiring these two tasks. One worker is available to install the engine and can service an average of 60 cars per hour. After the engine is installed, the car goes to the tire station and waits for its tires to be attached. Three workers serve at the tire station. Each works on one car at a time and can put tires on a car in an average of 3 minutes. Both inter arrival times and service times are exponential. (i) Determine the mean queue length at each work station. (ii) Determine the total expected time a car spends waiting for service. Solution: 1 1 2 2 54, 1, 60, 3, 20 s s ì u u = = = = = Non-Markovian Queues and Queue Networks 16 Since 1 2 3 and ì u ì u < < , neither queue will “below up” and Jackson’s theorem is applicable. For stage 1 (engine) ( ) ( ) 1 1 2 2 54 0.90 60 0.90 0.81 8.1 1 1 0.90 0.1 1 1 8.1 0.15 54 q q q s L cars W L hour ì p u p p ì = = = = = = = ÷ ÷ = = = For stage 2 (tires) ( ) ( ) ( )( ) ( ) 2 2 54 54 0.90 3 20 60 3 0.83 0.83 0.90 7.47 1 0.90 1 1 7.47 0.138 54 q q q s j L cars W L hour ì p u p ì = = = = > = = = ÷ = = = Thus, the total expected time a car spends waiting for engine installation an tires is 0.15+0.138=0.288 hour. Problem: 18 For a 2- stage (service point) sequential queue model with blockage, compute the average number of customers in system and the average time that a customer has to spend in the system, if 1 ì = , 1 2 2 1 and u u = = Solution: Given 1 2 1, 2, 1 ì u u = = = . The balanced equations are State Rate that the process leaves (0,0) 00 2 01 P P ì u = (1) (1,0) 1 00 00 2 11 P P P u ì u = + (2) (0,1) ( ) 2 01 1 10 1 1 b P P P ì u u u + = + (3) (1,1) ( ) 1 2 11 01 P P u u ì + = (4) (b,1) 2 1 1 11 b P P u u = (5) 00 10 01 11 1 1 b P P P P P + + + + = (6) (1) 00 01 P P ¬ = (7) (2) 10 00 11 2P P P ¬ = + (8) (3) 01 10 1 2 2 b P P P ¬ = + (9) (4) 11 01 3P P ¬ = (10) (5) 1 11 2 Pb P ¬ = (11) Non-Markovian Queues and Queue Networks 17 (7) & (10) 00 01 11 3 P P P ¬ = = (12) (12) & (1) 1 00 01 11 3 3 2 b P P P P ¬ = = = (13) (8) 10 00 00 1 2 3 P P P ¬ = + 10 00 4 2 3 P P = 10 00 2 3 P P = (14) (6) 00 10 01 11 1 1 b P P P P P ¬ + + + + = 00 00 00 00 00 2 1 2 1 3 3 3 P P P P P + + + + = by (13) & (14) 00 2 1 2 1 1 1 3 3 3 P + + + + = 00 00 5 2 1 3 11 1 3 P P + = = (13) 01 3 11 P ¬ = 11 00 1 1 3 1 3 3 11 11 P P | | = = = | \ . i.e., (14) 10 2 3 2 . ., 3 11 11 P i e | | ¬ = = | \ . (13) 1 00 2 3 b P P ¬ = 2 3 3 11 | | = | \ . 1 2 11 b P = ( ) 01 10 11 1 2 3 2 1 2 2 11 11 11 11 5 3 5 6 2 1 11 11 11 11 b L P P P P = + + + | | = + + + | \ . | | = + = + = | \ . 00 3 11 P = 11 1 11 P = 10 2 11 P = Non-Markovian Queues and Queue Networks 18 ( ) ( ) ( ) ( ) 01 10 11 1 00 01 00 01 2 1 1 11 3 3 6/11 6 1 11 11 b P P P P L W P P P P ì ì + + + = = + + = = = | | + | \ . Problem: 19 For a 2-stage (service point) sequential queue model with blockage, compute s L and 1 2 , 1, 1 2 S W if and ì u u = = = . Solution: Given: 1 2 1, 1, 2 ì u u = = = The balanced equations are ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 00 2 01 1 10 00 2 11 2 01 1 10 2 1 1 2 11 01 2 1 1 11 00 10 01 11 1 00 01 10 00 11 01 10 1 11 0, 0 ..... 1 1, 0 ..... 2 0,1 ..... 3 1,1 ..... 4 ,1 ..... 5 1 ..... 6 1 2 ..... 7 2 ..... 8 3 2 ..... 9 3 b b b P P P P P P P P P P b P P P P P P P P P P P P P P Pb P P ì u u ì u ì u u u u u ì u u = = + + = + + = = + + + + = ¬ = = + = + = ( ) ( ) ( ) ( ) ( ) 01 1 11 00 01 11 ..... 10 2 ..... 11 7 & 10 2 6 ..... 12 b P P P P P = ¬ = = ( ) ( ) ( ) ( ) ( ) 00 01 11 1 10 00 00 10 00 12 & 11 2 6 12 ..... 13 1 8 3 4 ..... 14 3 b P P P P P P P P P ¬ = = = ¬ = + = ( ) ( ) ( ) 00 00 00 11 00 00 00 4 1 1 1 6 1 13 & 14 3 2 6 12 4 1 1 1 1 1 3 2 6 12 37 1 12 P P P P P by P P ¬ + + + + = + + + + = = 00 12 37 P = Non-Markovian Queues and Queue Networks 19 ( ) 01 00 1 13 2 1 12 2 37 P P = | | = | \ . ( ) 11 00 1 13 6 1 12 6 37 P P = | | = | \ . ( ) 1 00 1 13 12 1 12 12 37 b P P = | | = | \ . ( ) 10 00 4 14 3 4 12 3 37 P P = | | = | \ . ( ) 01 10 11 1 2 6 6 2 1 22 3 28 2 2 37 37 37 37 37 37 37 b L P P P P = + + + | | | | = + + + = + = | | \ . \ . ( ) ( ) ( ) ( ) 00 01 28/ 37 28/ 37 14 12 6 18/ 37 9 1 37 37 L W P P ì = = = = + | | + | \ . Problem: 20 There are two salesmen in a shop, one in charge of receiving payment and the other in charge of delivering the items. Due to limited availability of space, only one customer is allowed to enter the shop, that too when the clerk is free. The customer who has finished his job 01 6 37 P = 11 2 37 P = 1 1 37 b P = 10 16 37 P = Non-Markovian Queues and Queue Networks 20 has to wait there until the delivery section becomes free. It customers arrive in accordance with a Poisson process at rate 1 and the service times to two clerks are independent and have exponential rates of 1 and 3, find (i) the proportion of customers who enter the ration shop (ii) the average number of customers in the shop and (iii) the average amount of time that an entering customer spends in the shop. Solution: Given : 1 2 1, 1, 3 ì u u = = = The balanced equations are ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 00 2 01 1 10 00 2 11 2 01 1 10 2 1 1 2 11 01 2 1 1 11 00 10 01 11 1 00 01 10 00 11 01 10 1 11 0, 0 ..... 1 1, 0 ..... 2 0,1 ..... 3 1,1 ..... 4 ,1 ..... 5 1 ..... 6 1 3 ..... 7 3 ..... 8 4 3 ..... 9 4 b b b P P P P P P P P P P b P P P P P P P P P P P P P P Pb P P ì u u ì u ì u u u u u ì u u = = + + = + + = = + + + + = ¬ = = + = + = ( ) ( ) ( ) ( ) ( ) 01 1 11 00 01 11 ..... 10 3 ..... 11 7 & 10 3 12 ..... 12 b P P P P P = ¬ = = ( ) ( ) ( ) ( ) ( ) 00 01 11 1 10 00 00 10 00 12 & 11 3 12 36 ..... 13 1 8 4 5 ..... 14 4 b P P P P P P P P P ¬ = = = ¬ = + = ( ) ( ) ( ) 00 00 00 00 00 00 00 00 5 1 1 1 6 1 13 & 14 4 3 12 36 5 1 1 1 1 1 4 3 12 36 97 1 36 36 97 P P P P P by P P P ¬ + + + + = + + + + = = = Non-Markovian Queues and Queue Networks 21 ( ) ( ) 01 00 11 00 1 00 10 00 1 1 36 12 13 3 3 97 97 1 1 36 3 12 12 97 97 1 1 36 1 36 36 97 97 5 5 36 45 14 4 4 97 97 b P P P P P P P P | | = = = | \ . | | = = = | \ . | | = = = | \ . | | ¬ = = = | \ . The proportion of customers entering the shop 00 01 36 12 48 97 97 97 P P = + = + = (ii) ( ) 01 10 11 1 2 b L P P P P = + + + 12 45 3 1 57 8 2 97 97 97 97 97 97 | | = + + + = + | \ . 65 97 = (iii) ( ) 00 01 65/ 67 36 12 1 97 97 L W P P ì = = + | | + | \ . 65 97 65 97 48 48 | || | | | = = | | | \ .\ . \ . Problem: 21 Consider a system of two servers where customers from outside the system arrive at server 1 at a Poisson rate 4 and at server 2 at a Poisson rate 5. The service rates 1 and 2 are respectively 8 and 10. A customer upon completion of service at server 1 is equally likely to go to server 2 or to leave the system (i.e.,P 11 =0,P 12 =1/2); whereas a departure from server 2 will go 25 percent of the time to server 1 and will depart the system otherwise (i.e.,P 21 =1/4,P 22 =0). Determine the limiting probabilities, L and W. Solution: The total arrival rates to servers 1 and 2 call them 1 2 and ì ì - Can be obtained from equation 1 1 k j i ij i r P ì ì = = + ¯ That is, we have 2 1 1 1 11 2 21 1 2 11 21 4 4 1 1 4 0 0, 4 4 i i i P P P P P ì ì ì ì ì = = + = + + = + + = = ¯ Non-Markovian Queues and Queue Networks 22 ( ) | | ( ) 1 2 2 2 2 1 1 12 2 22 1 12 22 2 1 1 4 ...... 1 4 5 5 1 5 0 1/ 2, 0 4 1 5 ...... 2 2 i i i P P P P P ì ì ì ì ì ì ì ì ì = = + = + = + + = + + = = = + ¯ ( ) 1 1 1 1 1 1 1 1 1 4 5 4 2 5 1 4 4 8 1 21 8 4 7 21 8 4 ì ì ì ì ì ì ¬ = + + = + + ÷ = = ( )( ) ( ) 1 2 21 8 3 2 6 4 7 1 5 6 5 3 8 2 ì ì | || | = = = | | \ .\ . = + = + = Given 1 2 1 2 1 2 6 3 8 4 8, 0, , 8 4 10 5 ì ì u u u u = = = = = = Hence P[n at server 1, m at server 2] ( ) | | 1 2 1 1 2 2 1 1 2 2 1 2 1 2 1 1 2 2 1 1 3 3 4 4 3 1 16 1 3 1 1 4 4 5 5 4 4 25 5 125 6 8 6 8 3 4 7 8 6 10 8 2 2 1 1 7 4 5 9 7 9 n n q q L W L ì ì ì ì u u u u ì ì u ì u ì ì ì | | = ÷ ÷ | \ . | | | | | | | | | || || || | = ÷ ÷ = = | | | | | | | | \ . \ . \ . \ .\ .\ .\ . \ . = + ÷ ÷ = + = + = + = ÷ ÷ = = = + = Non-Markovian Queues and Queue Networks 23 Problem: 22 In a departmental store there are 2 sections, namely, grocery section and perishable (vegetables and fruits) section. Customers from outside arrive at the G-Section according to a Poisson process at a mean rate of 10/hour and they reach the P-section at a mean rate of 2/hour. The service times at both the sections are exponentially distributed with parameters 15 and 12 respectively. On finishing the job in the G-section, a customer is equally likely to go the P- section or to leave the store, whereas a customer on finishing his job in the P-section will go to the G-section with probability 0.25 and leave the store otherwise. Assuming that there is only one salesman in each section, find the probability that there are 3 customers in the G-section and 2 customers in the P-section. Find also the average number of customers in the store and the average waiting time of a customer in the store. Solution: This system given is a Jackson’s open queuing system. 1 ì ÷total arrival rates of 1 S 2 ì ÷total arrival rates of 2 S Jackson’s flow balance equations for this open model are 1 2 1 1 11 21 1 1 2 21 , 1, 2,..... 10 10 1 1 10 0 4 4 k j j i ij i i i i i i r P j k P P P P ì ì ì ì ì ì ì ì = = = + = = + = + + = + + = ¯ ¯ | | ( ) | | ( ) | | 1 2 11 2 2 2 1 12 2 22 1 2 1 22 1 1 1 1 1 1 2 1 10 0 ....... 1 4 2 2 1 2 0 ....... 2 2 1 1 2 1 10 2 10 4 2 4 8 7 42 8 4 7 42 2 42 2 12 7 1 2 12 8 2 i i i P P P P P ì ì ì ì ì ì ì ì ì ì ì ì ì ì ì = = + = = + = + + = + = = + + = + + = = × = = = + = ¯ Given 1 2 1 2 1 2 12 4 8 2 15, 12, , 15 5 12 3 ì ì u u u u = = = = = = Non-Markovian Queues and Queue Networks 24 P(n 1 customers in S 1 and S 2 customers in S 2 ) ( ) ( ) ( ) ( ) 1 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 2 2 , 1 1 4 4 2 2 3, 2 1 1 5 5 3 3 64 1 4 1 256 0.0152 125 5 9 3 16875 12 8 4 2 6 15 12 12 8 1 1 1 6 12 2 n n P n n P L E W L ì ì ì ì u u u u ì ì u ì u ì ì | | | | | | | | = ÷ ÷ | | | | \ . \ . \ . \ . | | | || | | | = ÷ ÷ | | | | \ . \ .\ . \ . | || || || | = = = | | | | \ .\ .\ .\ . = + = + = + = ÷ ÷ ÷ ÷ = = = | | 10 2 12 ì = + = 1 2 = hours 1 60 2 = × minutes 30 = minutes. Problem: 23 Consider two servers. An average of 8 customers per hour arrive from outside at server 1 and an average of 17 customers per hour arrive from outside at server 2. Inter arrival times are exponential. Server 1 can serve at an exponential rate of 20 customers per hour and server 2 can serve at an exponential rate of 30 customers per hour. After completing service at server 1, half of the customers leave the system, and half go to server 2. After completing service at server 2, 3 4 of the customers complete service, and 1 4 return to server1. (i) What fraction of the time is server 1 idle? (ii) Find the expected number of customers at each server. (iii) Find the average time a customer spends in the system. (iv) How would the answers to parts (i) –(iii) change if server 2 could serve only an average of 20 customers per hour? Solution: Given 1 8 r = 2 12 21 11 22 17 1/ 2 1/ 4 0 r P P P P = = = = = Jackson’s flow balance equations for this open model are | | 1 2 1 1 1 1 1 11 2 21 2 11 , 1, 2..., 8 1 8 0 0 4 k j j j ij i i i i r P j k r P P P P ì ì ì ì ì ì ì = = = + = = + = + + = + + = ¯ ¯ Non-Markovian Queues and Queue Networks 25 ( ) | | ( ) ( ) 1 2 2 2 2 2 1 1 12 2 22 1 22 2 1 1 1 1 1 8 ..... 1 4 17 1 17 0 0 2 1 17 ..... 2 2 1 1 1 8 17 4 2 17 1 8 4 8 i i i r P P P P ì ì ì ì ì ì ì ì ì ì ì ì = = + = + = + + | | = + + = | \ . = + ¬ = + + = + + ¯ ( )( ) ( ) ( ) 1 1 1 1 2 1 49 8 4 7 49 8 4 49 8 7 2 14 4 7 1 2 17 14 17 7 24 2 ì ì ì ì ì ÷ = = | || | = = = | | \ .\ . ¬ = + = + = (i) Server 1 may be treated as an M/M/1/GP/ ·/ · system with 14, 20 0.7 ì ì u p u = = = = 0 1 1 0.7 0.3 P p = ÷ = ÷ = Thus server 1 is idle 30% of the time. (ii) 1 2 1 1 2 2 L ì ì u ì u ì = + ÷ ÷ 14 24 20 14 30 24 14 24 7 19 4 6 6 3 3 = + ÷ ÷ = + = + = (iii) | | 1 2 1 1 19 19 8 17 25 25 3 75 W L r r ì ì | | = = = = + = + = | \ . 2 2 20 24 u ì = < = . So no steady state solution exists. Non-Markovian Queues and Queue Networks where Pij is the probability that a departure from server i joins the Queue at server j. Problem: 5 State the characteristics of Jackson networks. Solution: (a) Arrivals from outside through node i follow a Poisson process with mean arrival rate ri. (b) Service times at each channel at node i are independent and exponentially distributed with parameter i (c) The probability that a customer who has completed service at node i will go next to node j (Routing probability) is Pij , i= 1, 2, 3, ….k and j = 0, 1, 2, 3, …k. and Pi0 denotes the probability that a customer will leave the system from node i. Pij is independent of the state of the system. If λj is the total arrival rate of customers to node j, j rj i Pij i 1 k j 0,1,2,...k Problem: 6 Distinguish between open Jackson networks and closed Jackson networks Solution: We have, j rj i Pij i 1 k j 0,1,2,...k In all general cases where, ri ≠ 0 for any i or Pi0 ≠ 0for any i, the Jackson networks are referred to as open Jackson networks. In the case of closed Jackson networks, ri = 0 for all i and Pi0 = 0for all i. PART-B Problem: 7 Derive Pollaczek - Khinchine relation for M / G /1 : / GD Solution: In order to determine the mean queue length Lq and mean waiting time Wq for this system the following technique is used. Let f t = Probability distribution of service time t with mean E t and Variance V t . n no. of customers in the system just after a customer departs. t service time of the customer following the one already departed. n1 The number of customers left behind the next departing customer. n1 k if n 0 if n0 n 1 k Where k = 0,1,2….is the number of arrivals during the service time. Alternatively of 1 if n 0 0 if n0 n1 n 1 k (1) 2 E n E k 2 2 E k 1 E k 2 E k 1 1 2 1 E k E k 2 E k 2E 2 k 2 1 E k Now is order to determine E n . the values of E k and E k 2 are to be computed. Since the arrivals follow Poisson distribution E k E k / t f t dt 0 Now E k / t t . 2 2 2 n 2 k 1 2n k 1 2 2n 2k 2 Since can take values 0 or 1only and 2 2 n 0 n1 n 2 k 2 2k 1 2n k 1 2k 2 n2 k 2 2n k 1 2k 1 2k 1 2n 1 k n 2 n1 k 2 2k 1 2k 1 2 Taking expectation on both sides 2 E n 1 E k E n 2 E n1 E k 2 E (2 E k ) 1 2 E k 1 E n E k 2 2 E k E 2 E k 1 1 2 1 E k E 1 E k E n .Non-Markovian Queues and Queue Networks Then taking expectation Squaring (1) both sides E n1 E n E E k 1 Since E n1 E n in steady state E 1 E k 1 2 n n k 1 n 2 k 1 2n k 1 2 2n 2 k 1 . E k 2 / t t t 2 V k / t E k 2 / t E2 k / t E k t f t dt E t 0 2 E k 2 E k 2 / t f t dt t t f t dt 0 0 E t 2 2 E t 2 E k 2 2 V t E 2 t E t V t E 2 t E t E n 2 V t E 2 t E t E t 2 2 E 2 t 2 1 E t 3 . If customers arrive at the barber shop in a Poisson fashion at an average rate of one every 40 minutes. how long on the average a customer spends in the shop? Also find the average time a customer must wait for service.625 25 1600 25 625 4 5 25 55 40 40 1600 3 8 48 48 15 2 40 By Little formula. 2 2 var T E T Ls E T 2 1 E T 1 0 252 25 40 1 40 2 1 . Solution: Since the service time T is a constant=25 minutes.Non-Markovian Queues and Queue Networks 2V t 2 E 2 t 2 E t 2 2 E 2 t 2V t 2 E 2 t 2 E t 1 E t 2 1 E t 2 1 E t 2 E 2 t 2V t E t 2 1 E t Ls E n E t E (t ) 1/ 2 E 2 t V t 2 1 E t Ws Ls where is the rate of servive. 1 for every 40 minutes 1 per minutes 40 By Pollaczek-Khinchine formula.25 40 1 . Var(T)=0. Then Ls / 2 (1/ )2 2 2 1 / ( 2 2 2 ) where / and V (t ) 2 2(1 ) Lq Ls ( 2 2 2 ) 2(1 ) Problem: 8 A one man barber shop takes exactly 25 minutes to complete one haircut. T follows a probability distribution with E(T)=25. 4 2 . Each suit requires four district tasks to be performed before it is completed.5 per week 5.1149 0. The expected waiting time of a customer (in the system) is 5 .. Assume all four tasks must be completed on each suit before another is started. If orders for a suit come at the average rate 5. 1 2 hr.125 orders / hr. a customer has to spend 45. (given) 4 1 hr 8 =0.8 minutes on the average.5 orders/hr 68 =0.8 Minutes 1 Wq Ws 45.1149 orders/hr 1 The server time for each task where k 4 k i.8 E T 1 E T 45.9192 0. 6 day week). The time to perform each task has an exponential distribution with a mean of 2 hr.8 25 20.5 per week (assume an 8hr day. 0.Non-Markovian Queues and Queue Networks 55 Ls 48 55 WS 40 1 48 40 45.8 minutes in the shop and has to wait for service for 20.125 Since the service time T is constant then 2 0 Here the average service rate is the average service rate to complete a suit. how long can a customer expect to wait to have a suit made? Solution: Given: 5. Problem: 9 Suppose a one person tailor shop is in business of making men’s suits.e.8 Minutes Hence. 75 4 2 2 Average delay in getting service 2 2 2 0. with a standard deviation of 6. The average service time is cut to 8.Non-Markovian Queues and Queue Networks Ws 2 2 2 1 2 1 Problem: 10 In a heavy machine shop.0 0. 3 0. on average.e. 75 3 100 4 3 4 Average service time E(T)=10.0714.0714 min 4 10. how much reduction will occur.5 i.. in the delay of getting served? Solution: This is a M / G /1 : / FIFO Process Given: Utilization rate 75% i. the overhead crane is 75% utilized.75 2 i.0714 8.0714 1 0.. What is the average calling rate for the services of the crane and what is the average delay in getting service? If the average service time is cut to 8.815 min 2 1 2 0.0714 min 4.8 minutes.5712 1/ 8 6 min 6 .8 0. 0.0714 min 0.. Time study observations gave the average slinging time as 10.286 hour To find the average delay in getting service Here 8.5 minutes with a standard deviation of 8.0714 0.0 minutes.e.75 Wq 26.3 minutes then 1 1 min E T 8.5 3 1 0.5 min We know that 1 1 E T 10.e.8.0 minutes. average calling rate for the services of the crane 0. 42 hrs =1.326 min The Average waiting time has a reduction of 26.75 car / hr 2 2 1 1/ 2 4 Lq Ls 3 1 4 2 0.5712 2 2 2 8.9192 2 1 53.8-8.5712 Wq 2 1 2 0. Given that the service time for all cars is constant and equal to 6 minutes determine LS .5 minutes 2 0. Cars arrive for loading by the crane according to a Poisson distribution with a mean of 5 cars per hour. WS and Wq . we have Ls 2 2 2 2 1 1 0 1/ 4 3 0. Solution: The given problem is in M / G /1 : / FIFO model. 5cars/hr cars/hr.Non-Markovian Queues and Queue Networks Average delay in getting service 2 2 2 0. 0.1149 0.113 Weeks .0808 0 0.0714 6 0. Lq .25 car / hr Ls 3 / 4 3 hr 9 minutes 1/ 5 20 7 Ws . The service time T is constant with mean E T 1 6 min Since the service time T is constant then var[T]= 2 0 1 1 1 6 min 6 hr hr.125 Problem: 11 A car manufacturing plant uses one big crane for loading cars into a truck.32=18. 60 10 10 hr 5 1 hr 10 2 Using k formula.0714 1 0. The parking lot is large enough to accommodate any number of cars.E T 2 1 E T 2 1 4 10 1 15 3 . if the time for washing and cleaning a car follows.2.3minutes.0.10 15 1 4 3 100 2 225 2 3 2 1 3 1 304 .8 and 15 minutes and corresponding probabilities 0. Solution: 4 Mean = 4 / hour = per minute 60 1 per minute 15 Let T be a continuous random variable. 1/ 5 20 Problem: 12 A car wish facility operates with only one bay. 2 225 3 2 1 304 3 2 152 302 .10 1 15 2 1 . (c) A discrete distribution with values equal to 4. 1 1 1 Then. Find the average number of cars waiting in the packing lot.Non-Markovian Queues and Queue Networks Wq Lq 1/ 4 1 hr 3 minutes.D. E(T) = mean of the uniform distribution a b 8 12 20 10 2 2 2 1 2 Var (T) b a 12 1 1 1 4 2 2 12 8 4 16 12 12 12 3 By the Pollaczek-Kninchine formula.6 and 0. 2 2 var T E T Ls .2. (a) Uniform distribution between 8 and 12 minutes (b) A normal distribution with mean 12 minutes and S. Cars arrive according to a Poisson distribution with a mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy. . 3 3 225 3 2 3 225 225 1 2 3 2 8 . 2 TP T 4 0.8 1.7 Cars 2 Cars 1 15 12 (c) The service time T follows the discrete distribution given below. Lq Ls 1 12 2.5 Cars. E T 12 min.2 0. 2 2 var T E T Ls .5 2.2 8 0.342 Cars Car By Little’s formula.153 9 12 4 1 4 153 15 15 225 . 1 15 5 5 225 6 3 2 1 . and Var(T)=9 15 1 1 E T 12 By the Pollaczek-Kninchine formula.5 15 2.5 0.6 15 0.8 4.6 minutes 9 .6 0. 1 1.12 2 15 15 4 153 2. 5 90 By Little’s formula.2 0.675 cars 1 car Lq Ls (b) 1 .E T 2 1 E T 1 2 1 . T : 4 8 15 P(T) : 0.12 .Non-Markovian Queues and Queue Networks 1.342 15 1 10 0.8 3 2 8. 024 1Car 225 12.64 8. 2 2 var T E T Ls . Ws .64 73.45 Car 5 1 E T Problem: 13 An automatic car wash facility operates with only one bay.2 0. If the service times for all cars is constant and equal to 10 minutes.573 0. Lq .573 1.024 5 1 86 8.6 15 225 0.6 225 15 1 2 1 8.96 86.6 12. Wq . Cars arrive according to a Poisson process with mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy. Lq Ls 1 1.8 8.8 4.Non-Markovian Queues and Queue Networks T 2 P T 42 0.2 8 0.6 15 0.025 0.6 1.6 8.6 2 1 15 By Little formula.8 3 8.E T 2 1 E T 1 2 12.6 15 1 12. Determine Lq . ’T’ follows a distribution with E(T)=10 and Var(T)=0. Solution: 4 per hour 4 per minute 60 1 per minute 15 Given service time ‘T’ is a constant. 10 .6 1 8.64 BY Pollaczek-Khinchine formula.6minutes var E T 2 E T 2 2 86. Wq 1 10 10 15 15 Problem: 14 In a big factory. which do the service of the damaged machines exponentially with respective rates of 1/hour and 2/hour.10 15 100 10 10 100 15 20 4 225 15 15 10 15 225 10 15 3 2 15 By Little formula. Solution: Given 0. i. there are a large number of operating machines and two sequential repair shops.5 / hour 1 per hour 2 1 = 1 per hour 2 = 2 per hour 11 .. Lq . (ii) the average number of machines in the service section of the factory and (ii) The average repair time of a machine. a customer has to spend 20 minutes in the system and 10 minutes in the queue.5 / hour. L 4 WS s 15 20 minutes 3 1 Wq Ws 2 20 E T 1 E T 20 10 10 minutes.E T 2 1 E T 1 2 0 10 1 15 10 1 15 2 1 . find (i) the probability that both repair shops are idle.e. 2 2 var T E T Ls .Non-Markovian Queues and Queue Networks By Pollaczek .Khinchine formula. If the cumulative failure rate of all the machines in the factory is 0. The cumulative failure rate of all the machines is 1 per hour. Solution: STEP-1: Model Identification The current situation comes under the sequence queue model. (i) P(both the service stations are idle) P 0.Non-Markovian Queues and Queue Networks The situation in this problem is comparable with 2-stage Tandem queue with single server at each state. (iii) the bottleneck of the repair facility. 0 1 0 . Assuming that the system behavior can be approximated by the above 2-stage tendon queue. 1 . 1 1 2 2 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2 1 1 1 1 1 1 1 1 1 3 3 1 2 4 2 4 8 (ii) The average number of machines in service 1 2 1 1/ 2 1/ 2 1/ 2 1 4 2 1 1 1 1/ 2 3 / 2 3 3 1 2 2 2 (iii) The average repair time 1 1 1 2 1 1 8 1 1 1 3 3 1 2 2 2 2 2 Problem: 15 A TVS company in Chennai containing a repair section shared by a large number of machines has 2 sequential stations with respective service rates of 3 per hour and 4 per hour. find (i) the probability that booth the service stations are idle (free) (ii) the average repair time including the waiting time. Since any number of machines can be repaired. . each station comes under the model M / M /1 : / FCFS STEP 2: Given Data Cumulative failure rate Service rate of station I 1 1 1 1 3 12 . There are 5 reservation counters in front of which also there is enough space for the customers to wait. (2) The average repair time and waiting time (3) The bottleneck of the repair facility. Each reservation clerk takes 5 minutes on the average to complete the business of a customer in an exponential manner. Customers arrive at the reservation counter section at the rate of 40 per hour and takes one minute on the average to fill up the forms.(i) Find the probability that a customer has to wait to get the service in the reservation counter section (ii) Find the total waiting time for a customer in the entire reservation section.Non-Markovian Queues and Queue Networks Service rate of station II 2 4 STEP 3: To find the following (1) The probability that both the servive stations are idle. form a queue and fill up the reservation forms. Problem: 16 In the Airport reservation section of a city junction. STEP 4: Required Computations (i) P( m customers in the I station and n customers in the II station Pmn 1 1 1 2 1 2 1 1 2 2 1 1 1 1 2 3 1 P00 1 1 3 3 4 4 3 4 2 (ii) The average number of machines in service at the system(both the stations) Ls 1 2 0 0 m n 1 1 1 1 5 3 1 4 1 2 3 6 = 1 machine (approximately) Average repair time (including waiting time) 1 1 1 2 5 hours =50 minutes. Solution: 13 . Assume that only those who have the filled up reservation forms will be allowed into the counter section. there is enough space for the customers to assemble. 6 1 1 (iii) Since 4 1 3 2 We get the service station 1 is the bottleneck of the repair facility. 288 3 18 162 24 81 21.Non-Markovian Queues and Queue Networks The queuing system in the form-filling portion is only a M/M/1 model.. 40 / hour and 60 /hour.. 40/hour) becomes the input of the M/M/s system] (i) P(a customer has to wait in the counter section) E Ns .032 14 .. 40 40 2 60 40 20 1 1 hour or 3 min E Ws 20 The queuing system in the reservation counter section is an M/M/s model with 40 /hour.P0 . 12 /hour and s 5 [Since the output of the M/M/1 system in the same as the input of that system. r 0 r s / s 1 s 1 1 5 10 r 41 1 10 3 Where .494 0. . by Burke’s theorem and the output of this system (namely. one server) For this system.206 10.. 1 s 1 s 2 r s 1 1 P0 . since each customers is served by himself or herself (viz.288 1 1 1 1 31. r 0 r 3 5!1 2 / 3 100000 10 1 2 3 4 1 10 1 10 1 10 1 10 1 10 243 1 0! 3 1 3 2 3 3 3 4 3 120 3 1 10 100 1000 1 100000 100000 3 1 3 18 162 24 81 243 120 10 100 1000 1 10000 1 10. 5528 600 1 1 E Ws E N s 3.2195 3.69 3. One worker is available to install the engine and can service an average of 60 cars per hour. the car goes to the tire station and waits for its tires to be attached.3292 1 120 3 1 / E Ns P 2 0 ss ! 1 s s 1 5 10 1 3 0. An average of 54 cars per hour arrives.Non-Markovian Queues and Queue Networks 10 0. 1 60. s1 1. After the engine is installed. Both inter arrival times and service times are exponential.329=11. Each works on one car at a time and can put tires on a car in an average of 3 minutes.3292 minutes. s2 3. Solution: 54. requiring these two tasks.032 243 0.3333 0.032 10 5120 1 3 3 1 131.0888 hour 6 5.032 10 3 5 5! 1 2 3 1000000 729 1 0.329 minutes Total waiting time of a customer in the entire reservation room =6+5.032 1 Required probability 3 2 5!1 3 100000 0. (i) Determine the mean queue length at each work station. Three workers serve at the tire station. (ii) Determine the total expected time a car spends waiting for service.3333 3. 2 20 15 .5528 40 0. Problem: 17 The last two things that are done to a car before its manufacture is complete are installing the engine and putting on the tires. the total expected time a car spends waiting for engine installation an tires is 0.1 0.81 8.1) (1.138 hour 54 Thus.15+0.0) (1) P00 2 P01 (1.47 0.1 1 1 Wq Lq 8.288 hour.stage (service point) sequential queue model with blockage.90 7. if 1 .47 cars 2 P01 1P10 1Pb1 1 2 P11 P01 2 Pb1 1 P 11 P00 P P01 P Pb1 1 10 11 P00 P01 2P P00 P 10 11 2 P01 2 P Pb1 10 3P P01 11 Pb1 2 P 11 16 1 P00 P00 2 P 11 (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) .90 0. compute the average number of customers in system and the average time that a customer has to spend in the system.83 1 0.83 0. Problem: 18 For a 2.90 s2 2 3 20 60 2 j 3 0. 2 1 .90 0. For stage 1 (engine) 54 0.90 s1 1 60 0.Non-Markovian Queues and Queue Networks Since 1 and 32 .0) (0.90 1 1 Wq Lq 7. 1 2 and 2 1 Solution: Given 1. neither queue will “below up” and Jackson’s theorem is applicable. 1 2.1) (b.15 hour 54 For stage 2 (tires) 54 54 0.138=0.1cars 2 Lq 1 1 0. The balanced equations are State Rate that the process leaves (0.1) (1) (2) (3) (4) (5) Lq 0. P 10 2 11 2 (13) Pb1 P00 3 2 3 3 11 Pb1 2 11 L P01 P 2 P Pb1 10 11 3 2 1 2 2 11 11 11 11 5 3 5 6 2 1 11 11 11 11 17 .Non-Markovian Queues and Queue Networks (7) & (10) (12) & (1) P00 P01 3P 11 P00 P01 3P 11 3 Pb 2 1 (12) (13) 1 (8) 2 P P00 P00 10 3 4 2 P P00 10 3 2 P P00 10 3 (6) P00 P P01 P Pb1 1 10 11 2 1 2 P00 P00 P00 P00 P00 1 3 3 3 1 2 2 P00 1 1 1 3 3 3 5 P00 2 1 3 11 P00 1 3 P00 3 11 (14) by (13) & (14) (13) P01 3 11 1 1 3 1 P P00 11 3 3 11 11 P 11 1 11 i.e. 2 3 2 (14) P 10 3 11 11 i..e.. . 10 .1 1 P00 2 P01 P00 2 P01 1 P P00 2 P 10 11 ..... 1 1.. 9 . 0 0................ 0 1......... 6 .... 7 ... 12 2 P01 1 P10 2 Pb1 1 2 P11 P01 2 Pb1 1 P 11 P00 P P01 P Pb1 1 10 11 P P00 2 P 10 11 3P01 P 2 Pb1 10 3P P01 11 2 Pb1 P 11 7 & 10 P00 2 P01 6 P11 12 & 11 P00 2 P01 6 P11 12 Pb1 8 1 P P00 P00 10 3 4 P P00 10 3 . if 1....Non-Markovian Queues and Queue Networks W P01 P 2 P Pb1 L 10 11 P00 P01 P00 P01 1 1 11 6 /11 6 3 3 1 11 11 Problem: 19 For a 2-stage (service point) sequential queue model with blockage. 2 2 The balanced equations are 0. 13 ..1 1. 4 ......... 8 . 2 .. 11 . 5 .. compute Ls and WS . 3 .1 b. Solution: Given: 1. 14 6 P00 4 1 1 1 P00 P00 P P00 1 by 13 & 14 11 3 2 6 12 4 1 1 1 P00 1 1 3 2 6 12 37 P00 1 12 P00 12 37 18 ... 1 ... 1 1 and 2 2 ..... that too when the clerk is free. Due to limited availability of space. one in charge of receiving payment and the other in charge of delivering the items.Non-Markovian Queues and Queue Networks 13 P01 1 P00 2 1 12 2 37 13 6 37 1 P P00 11 6 1 12 6 37 P01 P 11 2 37 13 Pb1 1 P00 12 1 12 12 37 1 37 Pb1 14 P 10 4 P00 3 4 12 3 37 16 37 L P01 P 2 P Pb1 10 11 P 10 6 6 1 22 2 3 28 2 2 37 37 37 37 37 37 37 W 28 / 37 28 / 37 14 L P00 P01 12 6 18 / 37 9 1 37 37 Problem: 20 There are two salesmen in a shop. The customer who has finished his job 19 . only one customer is allowed to enter the shop. ....Non-Markovian Queues and Queue Networks has to wait there until the delivery section becomes free... 4 ....... 9 .......... 13 8 1 P P00 P00 10 4 5 P P00 10 4 ...... 14 6 P00 5 1 1 1 P00 P00 P00 P00 1 by 13 & 14 4 3 12 36 5 1 1 1 P00 1 1 4 3 12 36 97 P00 1 36 36 P00 97 20 .. 7 . find (i) the proportion of customers who enter the ration shop (ii) the average number of customers in the shop and (iii) the average amount of time that an entering customer spends in the shop.1 1 P00 3P01 1 P P00 2 P 10 11 2 P01 1 P10 2 Pb1 1 2 P11 P01 2 Pb1 1 P 11 P00 P P01 P Pb1 1 10 11 P P00 3P 10 11 4 P01 P 3Pb1 10 4 P P01 11 3Pb1 P 11 7 & 10 P00 3P01 12 P11 12 & 11 P00 3P01 12 P11 36 Pb1 ..... 11 .. 2 . 1 1. 8 .... 10 ... 5 ... 12 1....1 b.....1 1. 3 .. 0 . 1 .. Solution: Given : 1. 0 0... It customers arrive in accordance with a Poisson process at rate 1 and the service times to two clerks are independent and have exponential rates of 1 and 3..... 2 3 The balanced equations are P00 2 P01 0. 6 ... P21 4 21 .Can be obtained from equation 1 rj i Pij i 1 k That is. The service rates 1 and 2 are respectively 8 and 10. whereas a departure from server 2 will go 25 percent of the time to server 1 and will depart the system otherwise (i.e.. Determine the limiting probabilities.P12=1/2). A customer upon completion of service at server 1 is equally likely to go to server 2 or to leave the system (i.P22=0). we have 1 4 i Pi1 4 1 P 2 P21 11 i 1 2 1 4 0 2 4 1 11 P 0.Non-Markovian Queues and Queue Networks 13 1 1 36 12 P01 P00 3 3 97 97 1 1 36 3 P P00 11 12 12 97 97 Pb1 1 1 36 1 P00 36 36 97 97 5 5 36 45 14 P10 P00 4 4 97 97 The proportion of customers entering the shop P00 P01 36 12 48 97 97 97 (ii) L P01 P 2 P Pb1 10 11 12 45 1 57 8 3 2 97 97 97 97 97 97 65 97 L 65 / 67 65 97 65 P00 P01 36 12 97 48 48 1 97 97 Problem: 21 Consider a system of two servers where customers from outside the system arrive at server 1 at a Poisson rate 4 and at server 2 at a Poisson rate 5.e. (iii) W Solution: The total arrival rates to servers 1 and 2 call them 1 and 2 . L and W.P11=0..P21=1/4. m at server 2] 1 1 1 2 1 2 1 1 2 2 1 2 3 3 3 4 4 3 1 16 1 1 1 4 4 5 5 4 4 25 5 125 n1 n2 1 2 1 1 2 2 6 8 6 8 3 4 7 8 6 10 8 2 2 1 1 W Lq 7 4 5 9 7 9 Lq 22 . 2 1 1 1 1 4 5 1 4 2 5 1 4 1 4 8 1 21 1 1 8 4 7 21 1 8 4 21 8 1 3 2 6 4 7 1 2 5 6 5 3 8 2 6 3 8 4 Given 1 8..... 1 2 5 i Pi 2 i 1 2 5 1P 2 P22 12 1 5 1 0 4 1 2 5 1 2 P12 1/ 2.. 2 1 8 4 2 10 5 Hence P[n at server 1.Non-Markovian Queues and Queue Networks 1 1 4 2 4 ... 1 . 2 0.. P22 0 ... The service times at both the sections are exponentially distributed with parameters 15 and 12 respectively. 1 2 2 i Pi 2 2 1 P 2 P22 12 i 1 2 1 2 2 1 ... 2... grocery section and perishable (vegetables and fruits) section.. j 1. 1 total arrival rates of S1 2 total arrival rates of S 2 Jackson’s flow balance equations for this open model are j rj i Pij . Find also the average number of customers in the store and the average waiting time of a customer in the store..... Assuming that there is only one salesman in each section. 1 . a customer is equally likely to go the Psection or to leave the store. 2 12...k i 1 2 k 1 10 i Pi1 10 i P i P21 11 i 1 1 1 1 10 0 2 P21 4 4 1 1 10 2 P11 0 4 . find the probability that there are 3 customers in the G-section and 2 customers in the P-section. Solution: This system given is a Jackson’s open queuing system.Non-Markovian Queues and Queue Networks Problem: 22 In a departmental store there are 2 sections.25 and leave the store otherwise. namely.. whereas a customer on finishing his job in the P-section will go to the G-section with probability 0.. 2 P22 0 2 1 1 2 1 1 10 2 1 10 1 4 2 4 8 7 42 1 8 4 7 1 42 2 42 2 1 12 7 1 2 2 12 8 2 12 4 2 8 2 Given 1 15.... On finishing the job in the G-section.. 1 15 5 2 12 3 23 . Customers from outside arrive at the G-Section according to a Poisson process at a mean rate of 10/hour and they reach the P-section at a mean rate of 2/hour. An average of 8 customers per hour arrive from outside at server 1 and an average of 17 customers per hour arrive from outside at server 2. 2. of the customers complete service. j 1. (iv) How would the answers to parts (i) –(iii) change if server 2 could serve only an average of 20 customers per hour? Solution: Given r1 8 E W r2 17 P 1/ 2 P21 1/ 4 12 P P22 0 11 Jackson’s flow balance equations for this open model are 3 2 n1 n2 j rj j Pij . (iii) Find the average time a customer spends in the system. 2 2 Problem: 23 Consider two servers. 4 4 (i) What fraction of the time is server 1 idle? (ii) Find the expected number of customers at each server. k i 1 2 k 1 r1 i Pi1 i 1 8 1 P 2 P21 11 8 0 2 1 4 P11 0 24 .. Inter arrival times are exponential. Server 1 can serve at an exponential rate of 20 customers per hour and server 2 can serve at an exponential rate of 30 customers per hour. and half go to server 2.Non-Markovian Queues and Queue Networks P(n1 customers in S1 and S2 customers in S2) P n1 . half of the customers leave the system. 2 1 1 5 5 3 3 256 64 1 4 1 0..0152 125 5 9 3 16875 1 2 12 8 L 42 6 1 1 2 2 15 12 12 8 1 1 1 L 6 10 2 12 12 2 1 1 hours 60 minutes 30 minutes. and return to server1.. After completing service 3 1 at server 2. After completing service at server 1. n2 1 1 1 2 1 2 1 2 2 4 4 2 2 P 3. 25 . 2 2 1 1 1 1 8 17 1 4 2 17 1 8 1 4 8 1 49 1 1 8 4 7 49 1 8 4 49 8 1 7 2 14 4 7 1 2 2 17 14 17 7 24 2 (i) Server 1 may be treated as an M/M/1/GP/ / system with 14.3 Thus server 1 is idle 30% of the time...7 14 24 20 14 30 24 14 24 7 19 4 6 6 3 3 1 1 19 19 (iii) W L r1 r2 8 17 25 25 3 75 2 20 2 24 ..... So no steady state solution exists.Non-Markovian Queues and Queue Networks 1 1 8 2 4 .7 0... 1 2 r2 i Pi 2 i 1 2 17 1 P 2 P22 12 1 17 1 0 P22 0 2 1 2 17 1 . 1 2 (ii) L 1 1 2 2 0. 20 P0 1 1 0.