hw1_fall09_solutions

hw1_fall09_solutions - CS 421 HW#1, due Nov. 12, 2009 1) I...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CS 421 HW#1, due Nov. 12, 2009 1) I wrote down a UDP based ping program, which can send ping request packets of variable size. I made some measurements using this tool by pinging the first hop router from my computer: the router has a ping time of 90 psec for packets of 500 Bytes long and 250 usec for packets of 1500 Bytes long (1 usec = 10‘6 sec). Assume that the router simply echoes back the ping request message and transmission rates at both directions are same. What is the propagation delay and transmission rate of the link connecting my computer to the router? 2) Consider a 1 Mbps channel with a 20 msec one—way propagation delay, i.e., 40 msec roundtrip propagation delay. We want to transfer a file of size 13500 Bytes. Each segment has a total size of 1625 Bytes including the 125 Bytes header, i.e., each segment contains 1500 Bytes of data. When there is data to be transmitted, each segment contains the maximum number of bytes. Assume that ACK segments are of 125 Bytes long and there is a processing delay of l msec after a segment is fully received at the receiver until the transmission of the corresponding ACK is started. Go-Back-N protocol is used with a window size of N = 4 segments. Assume that every 6th segment crossing the forward channel is lost while ACKs are not lost or corrupted. Assume that the processing delay at the sender after an ACK is received is negligible. How much time is required to complete the transfer of the whole file and receive the final ACK at the sender? Assume that the timeout for each window is set to 50 msec starting from the beginning of the window. 3) Consider the use of Go-Back—6 protocol for communication from Node A to Node B. Assume that when the sender reaches at the end of the window, i.e., all packets in the window are sent but not ACK’d, the sender goes into timeout, i.e., it goes to the beginning of the window and retransmits all packets in the window, as if the timeout occurred. In the following figure, indicate the sequence number (SN) for packets sent from A to B, the ACK number (ACKN) for packets sent from B to A, the times and SN of the packets at B delivered to the application layer, and the window kept at A. Note that packets received during the transmission of another packet will be immediately processed, but the corresponding action, e.g., update of SN/ACKN, will take effect with the start of transmission of the next packet. Sender’s window [base,base+N-1] [0,5] U451 117;“ E?! n} {'35) i illilillliii {T i l ‘ l l , Delivered ‘7 y % packets 0 l 7‘ . I Node B X: Transmissron error SN Q", Km, v . }‘$“}lv's1 12/ 1’) V a 14 4) 5) 6) 7) Assume that the bandwidth of a connection is 100 Mbps (lOOXlO6 bits/sec) and the round-trip propagation delay for the connection is 20 msec. Each data packet is 2,500 Bytes long and the ACK packets have negligible lengths. Assume that you use the Selective Repeat protocol. What should be the minimum window size (in segments) in order to achieve a bandwidth utilization of at least 80%, i.e., Umder 2 0.8? What is the minimum number of bits necessary to represent the sequence number for proper operation of the Selective Repeat protocol? Consider the TCP round-trip time and timeout estimation algorithm as we discussed in class. Suppose that currently EstimatedRTT = 20 ms and DeVRTT = 10 ms. Assume that the next measured SampleRTT = 30 ms. Immediately after calculating the updated values, TCP transmits a new segment, and this segment experiences an RTT of 50 ms. a) Assuming that neither this new data segment nor its ACK are lost, will this new segment timeout? b) An earlier version of TCP did not use DeVRTT, and simply set Timeoutlnterval = 2 X EstimatedRTT. Now answer the above question in a. Assume that a TCP connection uses window scaling that allows receive window of up to l MBytes. Suppose that the TCP connection runs over a l Gbits/sec (1x109 bits/sec) link with a round-trip propagation delay of 100 msec and there is no other traffic on the link. We want to transfer a 255 KByte file over this TCP connection. Assume that TCP receive buffer is l MBytes, and the application process removes the data as soon as it is placed in the receive buffer. Assume further that TCP connection uses 1 KByte segments and no loss event occurs during the entire file transfer (for simplicity, assume that the whole file is transferred using 255 segments). Further assume that the slow start threshold (ssthresh) at the beginning of the connection is infinitely large. Ignore all processing and queueing delays. What is the total time from the beginning of the transmission of the first data segment until the sender receives the ACK for the last data segment? A TCP sender is transmitting over a 1 Gbps (1x109 bits/sec) channel which has a 10 msec one—way delay. Assume further that a very large amount of data is transferred and no packers are lost or errored. The sender utilization (Uganda) is defined as the percentage of time the sender is busy transmitting bits (same as defined in your textbook). Remember that the receive window field in the TCP header is 2 Bytes long. a) Assume that window scaling is not used for this connection. What is the maximum value of Usender possible for this connection? b) Now suppose that the TCP connection uses a Window scaling factor of 64. What is the maximum value of Usender possible for this connection? 8) Assume a congestion feedback model for a system composed of two flows sharing a bottleneck link with bandwidth R bits/sec where both connections have the same RTT. Each flow gets binary synchronous feedback in discrete time steps of one RTT. If the aggregate consumption of the two flows is above the bottleneck bandwidth R, both senders receive a congestion notification signal (CN), otherwise they receive no CN. The flows use a simple congestion control scheme: When no CN is received in a time step, each sender increases its window to BW, where W is the current window size and B > 1. On the other hand, for each congested time step, i.e., when senders receive CN, each sender decreases its window to OLW, where 0 < 0L < 1 (typical values can be: OL = 0.5, B = 1.01). We can call this algorithm as Multiplicative Increase-Multiplicative Decrease (MIMD). Prove or disprove that MIMD achieves a fair allocation of bandwidth between the flows, i.e., each flow getting R/2, by using graphical arguments similar to the one we made in class in showing that TCP’s AIMD algorithm is fair. Use the following figure for showing the evolution of the throughputs. Assume that the initial throughputs achieved by the two connections correspond to “A”. 4,1an b/w Shaw’é Eating Connection 2 throughput Connection 1 throughput M I M D m wai’ min are 4.2 “€2.17 allot a ii? a ~ 3 w behvedm {Leak W‘s. awvvi Wt mo WE levied; Cum/pi field at b away; [tjym v 4: V J j v 46' tinlccaw {A MM; TV/fbir‘i'. :2 f + P ‘3: WM 9 , s i . R 2 3a ’0 g‘ 7 Xm;5§:‘vn (“Je [v1 bids ‘ ‘2 , ,- . 3: Wm: M w 14: .2 i D \3': 23393106 ' ‘_ i f k I? 3‘109/ Kiddie/f W‘ M . r ‘ ,-, ~,, : \3 1933:; {131? 106: 80. we“ :7 R= 5'“ .= 40g= "00 “M”! Q 52 8.7m"S D .__ ,6 11060 .1_ , "6. "g «I»! : 1/5310 _. . :: (‘15—‘70)40 = {:10 :7 D3 "(900 M4 310 403’ 12: 11069!“ mss=m§cm life; at am“ + 418’ 1.17M W LTT: 1‘10 MS I Am = «12: ladle; 0:43,Soo 1231'“ TM“: ’1 “$9 = 41.. m; 9’; ngwh : [10 1:0] .{06 TV“; A m 03' ,§oo 7MB 11qu :4“; L mam (5. Illa L—“f ’10"; o L $9M:wa Chan-Ull— 11'5"" Eve/j 42ft“ in HM). Tonufago ms M meow , "MP-"‘3‘. “Gav hfvluJa 51> 11:11“. fifiuw-af; 2:; £5 WS‘ ‘ ‘ .-, 4,. » P~ ‘ mauugtumrariflnnnn Lipid :AY \ofofaimgg Jismfii ACK ‘) Lamas Ludo A‘— 263 m; 1 Lu.” Sffmw” '9 1“; . reWsm-H-eo‘ owl Acts tam-n bow/L air 3’13 ms.r s’ 052 0,2“;0” N N Lg”? ‘\ o 9 a > 0-8 => N >/ 20.3 “5 ::_ x ~— 20,2:4om3 29 {a 34. 23790.? => 4: WM 8 my HT} E" 7+" r :3, .mlrrld . ~ r ~— 11 t ~«.)tlzl1,l£ + 5X. TT‘fifi BET—[2 :40 Ari .W, ~— 8 9 E'ZVQMZ % 20+ 4-30: saw “:30 g g ‘1') I r—fi’i —‘ “ _ '- ‘ I; 4+1- 5O >T5mecuf’gfl \ wae/ r" w)” 'fimfiflvd'a i§§ = 4+l+‘11&/‘+46‘r§2+6¥(+(28 HQ/Lqf/ 7(6) WI” 'Mlfie E flung in Jaw gin“; V r. ;" {oh‘l “me: 8 5" «amowh 145;; 4000‘ 3/ g . . ‘\ «0‘7 1 «a "l @ a) 2M" 190433“ rage/We Damion; $Vv4£ AREN‘th Sinus, we do MI- Jul .s'ejwwml size 45kqu (.5 aer)a;mexm/qu:{lj we Incwe M t A? 48 z 2 kg X40 M = .2. = 0.02 62%“? :2.é4‘1‘1% 20.404 40" 5) 220 bales (ace/{v6 chL-vw w?“ r1ML 14%" HM “BURN” \llaue'fi n: we M12 M=4:100%, f ...
View Full Document

This note was uploaded on 01/09/2012 for the course CS cs464 at Bilkent University.

Page1 / 5

hw1_fall09_solutions - CS 421 HW#1, due Nov. 12, 2009 1) I...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online