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EE555_Homework_01_Solutions

# EE555_Homework_01_Solutions - EE555 HW#1 SOLUTIONS FALL...

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Unformatted text preview: EE555 HW#1 SOLUTIONS, FALL 2007 Instructor: Prof. Ali Zahid Stallin ’8 Problem Solutions 8.10= 8.,10Ts = (1)(0.?) + (3}(02) + 10(0.1) = 2.3 ms Deﬁne S as the random variable that represents service time. Then (1%: = mm = E{(s— m2]: {1— 2.3)2(0.7) +(3— 2.3)2(0.2)+ (10— 2.332(01) = 7.21ms Using the equations in Figure 8.63: A = 0.5 (1+ (7.21/5.29)) = 1.18 a. p = XTS = 0.33 x 2.3 = 0.767 Tr = Ts + (stA)/(1 — p) = 2.3 + (0.767 x 2.3 x1.18)/(0.233)= 11.23 ms 1‘ = p + (92A)/(1 — p) = 3.73 messages I). p = 751‘s = 0.25 x 2.3 = 0.575 Tr = 2.3 + (0.575 x 2.3 x 1.18)/(0.425) = 5.971115 r: 0.576 + (0.675 x 0.573 x 1.18)/(0.425) = 1.49 messages c. p = 7515 = 0.2 x 2.3 = 0.46 Tr = 2.3 + (0.46 x 2.3 x 1.18)/(0.54) =- 4.61 ms 1' = 0.46 + (0.46 x 0.46 x 1.18)/(0.54) = 0.92 messages 8.11 k = 0.05 msg/sec; T5 = (14.400 x 8)/9600 =12 sec; p = 0.6 21. TW = st/(l — p) = 18 sec 1). w = pz/(l — p) = 0.9 8.17 a. System throughput: A = 7-. + PA; therefore A = U (1 — P) Server utilization: p = ATS = ATS/(l — P) T = T5 = (1—P)TS 1— p l — P — ATS b. A single customer may cycle around multiple times before leaving the system. The mean number of passes] is given by: I=1(1—P}+2(l—P)P+3{l—P)P2+...=(l—P)§iPH'=ﬁ i=l — Because the number of passes is independent of the queuing time, the total time in the system is given by the product of the two means, II}. Solution for problem 2 (2093, part gal=ISQts, Qart gbl=52ts ): a) This is an WN Queuing Model Packet delay = queuing delay = Tq N = number of servers = 10 Ts = time to service for one packet = 1000/10Mbps = 100 its 1) = MTs) 9.. = mean arrival rate into each queue = 50000110 = 5000 packets/sec per interface p = 5000 x 100u = 0.5 Tq = CTS / (N(1-p)) + Ts, where C is the Erlang C function = probability that all servers are busy. C = (1-K) / (l-pK)‘ where K is the Poisson ratio function K = K] / K2 'Where K1 = 2:;‘(Aavw is. K2 = Zprfns For N = 10. p = 0.5, we get: M (IX 1562 7832 39062 195312 976562 ' 1.70139 15 50099 9.688 ' 5.382289 2.691144 0'1 K3 26.0415 — ﬂ || 143 .6895 146 3806 5 H K 0.981615 0 =(1-K)/('i-pK) 0.036105 Ts 00001 Tq = C(Ts)/{N(§-p)) + Ts 0.000101 Therefore, Tq = 101 usec b) One fast interface M = 100 Mbps Ts = 1000/100Mbps = 10 as p = MTS), 9's. = mean arrival rate into each queue = 50000 packetsl'sec per interface p = 50000x1011= 0.5 Tq = Ts I (1—p) = 20 u ms Solution for problem 4 (EMS, 5 for each part): a) Let X denote the number of hops per transmission. X can take the values of 1 (with probability p), 2 (with probability p(1-p)) or 3 (with probability (1-p)2). Hence, 1300 = 1(p) + 2p<1-p) + 30-102 = p2-3p+3 b) Let Y denote the number of transmissions a packet makes. The probability of successful transmission is (1-p)2. Y is Geometric R.V. with parameter (1-p)2. Hence, E(Y) = 1/(1-p)2. c) The average # of hops per received packet is hence = (p2-3p+3)/(1-p)2. Solution for problem 5 1209ts, IOQts for each gart): 0 Average Arrival Rate 7» =1/2 calls/minute 0 Average Service Time Up = 4 minutes/call 0 Number of Servers c=4 a) This system can be modeled as M/M/4 Server utilization p = K/cu = 1/2 Probability of having to wait for a line = P (N24) = C (4,1/2,1/4) C (4,1/2,1/4) = Erlang C-formula C(4,l/2,1/4) = ((Mu)° /c!)/[((7»/u)° /c!)+ (1-p)2((7»/u)n /n!)] Where the summation is from n=0 to n=c-1 Substituting we get P (N24) = 4/23 E 0.17 E(Nw) = Average # of calls waiting in Queue = (p /(1-p)) C (4,1/2,1/4) = 4/23 calls E(W) = Average waiting time in Queue = E(Nw)/ 7» E 0.34 minutes E(N) = Average # of calls in system = E(Nw)+ Mu E 2.17 calls b) This system can be modeled as M/M/4/4 Probability that a call is redirected to PSTN = P (N=4) = B(4,1/2,1/4) B(4,1/2,1/4) = Erlang B-formula B(4,1/2,1/4) = (Qt/pf /c!)/Z((7»/u)n /n!) where the summation is from n=0 to n=c Substituting, we get P (N=4) E 9.5% E(N) = Average # of calls in the private system = ((Mu))(1- B(4,1/2,1/4)) E 1.81 BOMB 5a In ‘5’ w A printer is attached to the LAN of the department. The printing jobs are assumed to ar— rive with a Poissonian intensity /\ and the actual printing times are assumed to obey the distribution Exp(,u). The capacity of the printer has become insufﬁcient with regard to the increased load. In order to improve the printing service, there are three alternatives: 1. Replace the old printer by a new one twice as fast, i.e. G ——» nu with service rate 2M. ““2” 2. Add another similar printer (service rate ,LL) and divide ® ——» m: the users in two groups of equal size directing the works p =A/2u in each group to their own printer. The arrival rate of jobs to each printer is A/2. __. nu p=kl2u 3. The same as alternative 2, but now there is a common printer queue where all jobs are taken and the job at the Q _, I“, head of the queue is sent to whichever printer becomes free ﬁrst. p=M2u formance we use the mean sojourn time of a job T ( time in system, from the arrival of the printing job to the full completion of the job) _ /\ T _ 1 -_ 1 1 [0—2/1 1—2/1—A—1—p 2/1 2 Now we have two separate M /M / 1 queues With parameters /\ / 2 and ,u _ 1 1 u 2M u—A/2 1—10 u The load per server is the same as before. Now ' (both arrivals and the service). Take case 1 as the reference: calculate the sojourn times in cases 2 and 3 in relation to that in case 1. 0 Alternative 1, i.e. one fast printer is the best one. o In alternative 2, the sojourn time is twice as long as in case 1. o In case 3, the second printer does not help at all at low loads: each job is taken directly into the service (without waiting) but the actual printing takes twice the time as with the fast printer. 0 At heavy loads, the mean sojourn time of case 3 is the same as in case 1 (in both cases it consists mainly of the waiting). Two slow printers fed by a common queue discharge the work in the queue as efﬁciently as one fast printer. 0 This is not the case for the alternative 2. When the queues are separate, it is possible that one printer stays idle while there are jobs waiting in the queue for the other printer. This deteriorates the overall performance in such a way that also at high loads alternative 2 is on the average two times slower than alternative 1. ...
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EE555_Homework_01_Solutions - EE555 HW#1 SOLUTIONS FALL...

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