# a2-sol - Solution to CSE 489/589Homework Assignment 2...

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Solution to CSE 489/589Homework Assignment 2 Prepared by Hung Q. Ngo * October 18, 2010 Problem 1 (Birthday attack) . To poison the cache of a DNS server, in the old days the attacker only needed to match the Query ID, which is a 16-bit number. Let n = 2 16 be the size of this ”sample space”. The attacker sends m DNS requests (for the same domain), which will trigger m DNS queries with different (presumably randomized) IDs being sent, and then the attacker sends k faked replies whose IDs are chosen randomly in the set [ n ] = { 1 , · · · , n } . The probability that no (faked) reply has a matching ID with a query was shown in class to be ( 1 - m n ) k e - mk/n . So the attacker’s failure probability can be approximated to be e - mk/n . Generally, the attacker wants to minimize the total number of packets he has to send (that’s m + k ) and mimimize the chance of failure too. For n = 2 16 , to guarantee at most a 1 / 1000 chance of failure, what’s the minimum number of packets he has to send? (Use e - mk/n as the failure probability.) Repeat the quesiton for n = 2 32 , which is the case with randomized port numbers and Query IDs. Answer. We want to minimize m + k subject to e - mk/n 1 / 1000 , which is equivalent to mk n ln 1000 . For a fixed integer m , the number of packets sent is at least f ( m ) = m + n ln 1000 m . The function is minimized at m n ln 1000 for a total of roughly 2 n ln 1000 . (We will be off by ± 1 at most, due to integral rounding. So, it’s ok for your answer to be within ± 1 of the given answers below. The entire process is an approximation anyway because we have used e - x to approximate 1 - x .) When n = 2 16 , 2 n ln 1000 is approximately 1346 . When n = 2 32 , 2 n ln 1000 is approximately 344 , 491 . Problem 2. Suppose we run the Go-Back-N protocol for a channel with end-to-end delay 30 ms (typical for coast-to-coast). Suppose the window size is 2 16 bytes long. (This number is indeed the maximum TCP window size without window scale option). The channel is perfectly reliable with bit rate 10 Gbps. What is the maximum utilization theoretically achievable? (Assume ACK transmission time is negligi- ble.) Answer. The amount of time it takes to send a full window is only 65536 · 8 10 10 = 524 , 288 · 10 - 7 ms, much less than RTT = 60 ms. The maximum utilization is the maximum fraction of time the channel is busy within an RTT, which is 524 , 288 · 10 - 7 60 = 0 . 000873813333 . This is horrible, illustrating the need for the Window Scale Option. Problem 3. In this problem, we attempt to derive a highly simplified “macroscopic” model for the steady-state behavior of TCP. The major question we’d like to answer is: given a certain loss rate, what’s the maximum throughput TCP (Reno) can achieve. For simplicity, we will assume that all losses are signaled by the receipts of triple duplicate ACKs. (If there was a timeout event, slow-start with recover * Please let me know of any mistake/typo ASAP.

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