d t Thus, L R ECS 152A Computer Networks Fall 2002 HW 7 Solutions

Transcription

1 ECS 52A Computer Network Fall 2002 HW 7 Solution. (5 pt) Conider four tation that are attached to two different bu cable. The tation exchange fixed-ize packet of length ec. Time i divided into lot of ec. When a tation ha a packet to tranmit, the tation chooe either bu with equal probability and tranmit at the beginning of the next lot with probability p. Find the value of p that imize the rate at which packet are uccefully tranmitted. To imize the ucceful tranmiion rate i to imize the probability of ucceful tranmiion. Pr{Succe} (# tation) x Pr{one tation tranmit on one bu and at the next lot} x Pr{no other tation tranmit on the ame bu and at the next lot} 4 (½ p) (- ½ p) 3 2p ( - ½ p) 3 Take the derivative with repect to p and et it to 0: d/dp 2 ( - ½ p) 3 - (3p)( - ½ p) 2 0 p ½ 2. (20 pt.) M terminal are attached by a dedicated pair of line to a hub in a tar topology. The ditance from each terminal to the hub i d meter, the peed of the tranmiion line i bit/econd, all packet are of length 2500 byte, and the ignal propagate on the line at a peed of 2.5x0 meter/econd. For the four combination of the following parameter (d 25 meter or d 2500 meter; 0 Mbit/ or 0 Gbit/), compare the imum network throughput achievable when the hub i implementing: Slotted AOHA; CSMA/CD. 2500byte *2500bit 00,000bit d t prop 2.5x0 meter / Thu, x] time to tranfer packet 00,000bit (Continued on next page )

2 2 cont.) Slotted Aloha: We know that the imum throughput i imply /e (0.37) percent of the idealized throughput (ee p.35), the actual value can be found by: bit / / e * z packet / bit / packet Note that thee value below are percentage of the ideal throughput. Ditance ate 25 meter 2500 meter 0 Mbit/ Gbit/ CSMA/CD: From the derivation (p. 34) we know that the imum throughput for CSMA/CD i: t prop d / v d D ; where a + + ) a ]/ v ] v ] + + ) d / v ] ) d /(2.5x0 m / *0 bit / ) Note that thee value below are percentage of the ideal throughput. Ditance ate 25 meter 2500 meter 0 Mbit/ Gbit/

3 3. (20 pt.) Conider the tar-topology network in #2 when the -ring protocol i ued for medium acce control. Aume ingle-packet operation, eight-bit latency at each tation, M 25 tation. Aume a free i three byte long. (a) Find the effective packet tranmiion time for the four combination of d and. Given: M b v packet 25 bit 2. 5x0 3 byte 24bit 5 2, 500 byte 0 bit m / The ditance from each terminal to the hub i d meter, aume the ditance traveled inide the hub i 0. The total ditance around the ring i then M*2*d. X eff tranmiion time + packet tranmiion time + ring latency packet M 2d Mb packet 2d b M + v v (b)aume that each tation can tranmit up to a imum of k packet per. Find the imum network throughput for the four cae of d and. The imum throughput occur when all tation tranmit k packet per. After completing the tranmiion of k packet, each tation then tranmit a free into the ring. On pg 375 we re given for ingle packet per ecall that X] ]/. M * M ( + τ ') + τ ' MB ; τ ' τ + ; + a' ( + / M ) a' τ ' However, in thi cenario, we re trying to end k packet per. Thi can be thought of a ending a ingle packet of ize: k*]. Thu: for k packet per M * k * ; M ( k * + X + τ ') + τ ' 2d τ ' M + v b

4 3 cont.) i) d 25m; 0Mbit/ 0x0 bit/ M 25; X] * 2500bit/ 0x0 bit/ 0.0 X *3bit /0x0 bit / 2.4x0 2* 25m bit τ ' 25* µ 2.5x0 m / 0x0 bit / 25* k * * ( k * µ + 25µ ) + 25µ µ % Similarly, we can find for the other value of d and : ii) d 25m, 0Gbit/ 0x0 9 bit/ 2% iii) d 2500m, 0Mbit/ 0x0 bit/ 79% iv) d 2500m, 0Gbit/ 0x0 9 bit/ 0.40% 4. (5 pt.) Suppoe that a group of 32 tation i erviced by a -ring AN. Suppoe packet i 000 bit long. The data rate of the link i 0 Mbp. There i a 2.5-bit latency per adapter. Each tation i 50 meter apart. Calculate the time it take to tranfer a packet uing the three reinertion trategie: after completion of tranmiion, after return of, and after return of packet. M 32 v 2 x 0 m/ ize of packet 000 bit b latency per adapter 2.5 bit-time einertion after completion of tranmiion: 2.4µ Aume the tranmiion time i mall and can be ignored. The one-hop walk time i for paing the to the next tation. X eff packet tranmiion time + one hop walk time / + (d/v +b/) 00.5 µec

5 einertion after return of : X eff {packet tranmiion time, ring latency} + one hop walk time {/, τ } + (d/v + b/) X eff 0 µec τ M(d/v + b/) einertion after return of packet: X eff packet tranmiion time + ring latency + one hop walk time (/) + M(d/v + b/) + (d/v + b/) 7 µec 5. (30 pt.) A meage of ize m i to be tranmitted over an -hop path in a tore-and-forward packet network a a erie of N conecutive packet, each containing k data bit and h header bit. Aume that m >> k + h. The bit rate of each link i bit/econd. Propagation delay and queuing delay are negligible. (a) (5) What i the total number of bit that mut be tranmitted? Total number of bit ent all data bit + all header bit N (k + h) (b) (5) What i the total delay experienced by the meage (i.e. the time between the firt tranmitted bit at the ender and the lat received bit at the receiver)? Uing pipelining, when the t packet arrive, there are N- packet left to be ent. The 2 nd packet will be at the next to lat router. The 3 rd packet will be at the router before where the 2 nd packet i at, and o on. The 2 nd packet take more hop to arrive at the detination, and everything after the 2 nd packet advance more hop, and o on. Total time (time to end packet via hop) + (time to end N- packet through hop) [()(k+h)/] + [(N-)(k+h)/] (c) (0) What value of k minimize the total delay? From part (b), we have delay D [()(k+h)/] + [(N-)(k+h)/] We know N m/k, o D [()(k+h)/] + [(m/k - )(k+h)/] Take the derivative of D with repect to k, then et it to 0. And then olve for k. The k that minimize the delay would be: Mh k