Name (Print) ______________________________ Email Address ECE4110 Quiz 2 PRACTICE answers p. 1 ECE4110, Internetwork Programming, QUIZ 2 - PRACTICE Spring 2006 Prof. John A. Copeland Practice for April 11, 2006 Tel.: 404-894-5177 E-Mail: [email protected] RULES. i This quiz is not open book. One original sheet of hand-written notes may be used. Calculators are ok. ii Answer all questions and show all work to receive full credit. Use back of sheets only if necessary. iii All sub-questions have the same weight. iv Please do not ask the proctors any questions during the exam about exam questions. Part of the test is understanding the question, as written, without supplemental information. If you feel additional data is needed to solve the problem, make (and state) an assumption and then work the problem. v. This is a time-limited test. All papers must be turned in 50 minutes after the start. If you find you are taking more than 10 minutes on a particular problem, move on and come back to that problem after finishing the others. The Georgia Honor Code applies (see last page). The actual test will be about 60% as long as this. Question 1 – TCP with no Congestion. A 100 Mbps network connection has a 100 ms round trip time (RTT). A server begins sending maximum-size packets (after the initial SYN-ACK) using Slow Start. The client ACK's every packet (col. 2) or every-other (col.3), and has a receiver-window which holds 16 maximum-size packets. Show how many packets are sent in each RTT period below: Time Period Start (ms) No. Segments, when all ACKed No. Segments, when 1/2 ACKed 0 1 1 100 2 2 200 4 3 300 8 3 400 16 6 500 16 9 600 16 12 700 16 16 ___ 400 or 700 ms _ How long does it take the server to ramp up to maximum transfer rate (ms). ____16,000 / 0.1s _____ If the maximum packet size is 1000 bytes, what is the maximum transfer rate (bytes/second). How and when is the receiver's maximum segment size (MSS) sent to the sent to the other host? How? ___ TCP option ____ When _______ in SYN or SYN-ACK ________________ How and when is the receiver's TCP window (WIN) sent to the sent to the other host? ? How? _____Field in TCP header______ When ______ with every segment ___________ The last WIN received was 8000, the last ACK number was 12000, and the last TCP segment sent had a SEQ number of 14000. How many more bytes can be sent? ___8000 - (14000 - 12000) = 6000 ____ Once those bytes are sent, name three conditions that will allow additional bytes to be sent: a. ____Larger WIN received ____ b.___ larger ACK number received_____ c. __"Persist" timer expires _______ A TCP host on an Ethernet LAN receives an MSS of 1460 bytes. It sends IP headers with no options (20 bytes) and TCP headers with 32 bytes (12 bytes for options). What is the maximum number of data bytes (SMSS) it can send per TCP segment? _____________ (Ethernet max, 1500 ) -20 -32 = 1448 bytes _______________Name (print) ______________________________ ECE4110 Quiz 2 PRACTICE - answers p. 2 of 4 Question 3 – Bandwidth-Delay Product a. If the round-trip time (RTT) for a dedicated 16 Mbit/s connection to California is 50 ms, what is the bandwidth-delay product of the link? ___16,000,000 bits * 0.050 s = 800,000 _(bits). [= 100,000 bytes] b. What is the minimum TCP widow size that will allow this channel to be fully utilized? __ 100 kbytes __ (bytes) c. What will the maximum throughput be if the window is half the value in b. above. ___ 8 Mbit/s __ (bits/s) d. What will the maximum throughput be if the window is twice the value in b. above. ____ 16 Mbit/s ___ (bits/s) Question 4 – TCP Response to Congestion - Lost and Out-of-Order Packets Once a TCP segment times out, a variable collision window-size is cwnd is used. For consistency, we consider cwnd to govern the number of segments sent during the initial Slow Start mode. SMSS (sender maximum segment size) was called segsize in the class slides, MSS in some of the books. Conditions at the sender may require SMSS to be smaller than the MSS value received for the other TCP host. What places an upper limit on the maximum number of outstanding (sent, but not ACKed) bytes besides cwnd? _________ win [value received in last packet from receiver] __________ After a time-out, the value of cwnd drops to ___one___ times the maximum segment size (SMSS or segsize). After a time-out, the value of ssthresh drops to ____1/2____ times the previous value of cwnd (but at least 2 SMSS). cwnd then grows in _ slow-start ____ mode until it reaches ssthresh. which is reduced to _1/2__ times the value of __ min(cwnd, win) but at least 2 SMSS __ before the time-out. After cwnd reaches ssthresh, cwnd grows linearly by adding SMSS bytes for approximately every Y number of maximum segments sent (ignore the incorrect SMSS/8 term in the slides and book). What is this linear mode (of cwnd increase) called? ___ congestion avoidance ______ For the case where the receiver ACKs every packet, Y = __ cwnd/SMSS _____ (in terms of cwnd, SMSS) For the case where the receiver ACKs every other packet, Y = _______ 2 cwnd/SMSS _______ (ditto) What is the maximum value of cwnd? _ moot, since win used if smaller __(or is it moot, see TCP Illustrated, p. 310, rule 1). Does cwnd or win (from the receiver) govern the bytes (by sequence number) that can be sent? _ the smaller value ___ Question 5 – TCP Fast Retransmission and Fast Recovery After a TCP sender receives __ 3 ___ duplicate ACKs it retransmits the first missing segment. How many additional bytes (at a minimum) can be sent after each additional duplicate ACK is received? _SMSS______ [because cwnd increases by SMSS, and the ACK number does not change] How much does cwnd increase after each additional duplicate ACK is received? ___ SMSS ______ When does Fast Recovery operation start? ______ when 1st non-duplicate ACK value received _______ When FR starts, cwnd is set equal to _1/2___ x cwnd-before-dup-ACK + ____ 1 _____ SMSS. In Fast Recovery, how many SMSS bytes are added each RTT is all segments are ACKed (do not use SMSS/8 term): if all segments are ACKed_____ 1 _________ if every other segment is ACKed? _________ ½ _____________Name (print) ______________________________ ECE4110 Quiz 2 PRACTICE - answers p. 3 of 4 Question 6– Window Size versus Time SMSS =1420 bytes, MSS=1420 bytes, RTO = 50 ms,