Solutions to Select Exercises

Transcription

1 Solutions to Select Exercises CHAPTER. We will count the transfer as completed when the last data bit arrives at its destination (a). MB = 89 bits. initial RTTs (6 ms) +,8,9/,, bps (transmit) + RTT/ (propagation).8 seconds. (b) Number of packets required =. MB/KB = 6. To the above we add the time for RTTs (the number of RTTs between when packet arrives and packet 6 arrives), for a total of =.8 seconds. (c) Dividing the 6 packets by gives This will take 76. RTTs (half an RTT for the first batch to arrive, plus 76 RTTs between the first batch and the 77th partial batch), plus the initial RTTs, for 6.8 seconds. (d) Right after the handshaking is done we send one packet. One RTT after the handshaking we send two packets. At n RTTs past the initial handshaking we have sent n = n+ packets. At n = we have thus been able to send all 6 packets; the last batch arrives. RTT later. Total time is +. RTTs, or second. 6. Propagation delay is m/( 8 m/s) = µs. 8 bits / µs is. Mbps. For -byte packets, this rises to 6. Mbps.. (a) Propagation delay on the link is ( 9 )/( 8 ) = 8 seconds. Thus, the RTT is 68 seconds. (b) The delay bandwidth product for the link is 8 8 =.8 MB. (c) After a picture is taken, it must be transmitted on the link and be completely propagated before Mission Control can interpret it. Transmit delay for MB of data is,9, 8

2 8 Solutions to Select Exercises bits/8 = 8 seconds. Thus, the total time required is transmit delay + propagation delay = = seconds. 7. (a) For each link, it takes Gbps / kb = µs to transmit the packet on the link, after which it takes an additional µs for the last bit to propagate across the link. Thus, for a LAN with only one switch that starts forwarding only after receiving the whole packet, the total transfer delay is two transmit delays + two propagation delays = µs. (b) For three switched and thus four links, the total delay is four transmit delays + four propagation delays = 6 µs. (c) For cut-through, a switch need only decode the first 8 bits before beginning to forward. This takes 8 ns. This delay replaces the switch transmit delays in the previous answer for a total delay of one transmit delay + three cut-through decoding delays + four propagation delays =.8 µs. 7. (a) 9 8 =,9,99,. Gbps. (b) 8 8 = 6 Kbps. (c) 6 = Kbps. (d) 88, = 6,8. Mbps. CHAPTER. The B/B encoding of the given bit sequence is the following: Bits NRZ 7. Let mark each position where a stuffed bit was removed. There was one error where the sever consecutive s are detected (err) At the end of the bit sequence, the end of frame was detected (eof). err eof 9. (a) We take the message, append 8 zeros and divide by (x 8 + x + x + ). The remainder

3 Solutions to Select Exercises 8 is. We transmit the original message with this remainder appended, resulting in. (b) Inverting the first bit gives. Dividing by (x 8 + x + x + ) gives a a remainder of.. One-way latency of the link is ms. (Bandwidth) (roundtrip delay) is about pps. sec, or packets. SWS should be this large. (a) If RWS=, the necessary sequence number space is 6. Therefore, bits are needed. (b) If RWS=SWS, the sequence number space must cover twice the SWS, or up to. Therefore, 6 bits are needed.. The figure that follows gives the timeline for the first case. The second case reduces the total transaction time by roughly RTT. Frame[] Frame[] Frame[] Frame[] Frame[] Frame[] ACK[] Frame[] Frame[] ACK[] RTT Frame[] DUPACK[] DUPACK[] DUPACK[] RTT Frame[] Frame[] DUPACK[] DUPACK[] DUPACK[] RTT Timeout Frame[] RTT Frame[6] ACK[] RTT Frame[6] ACK[] RTT ACK[6] ACK[6] RTT RTT

4 8 Solutions to Select Exercises CHAPTER. The following table is cumulative; at each part the VCI tables consist of the entries at that part and also all previous entries. Note that at stage (d) we assume that VCI on port of switch cannot be reused (it was used for a connection to H in part (a)). This would correspond to the case where VCIs are bidirectional, as they commonly are. Exercise Part Switch Input Port VCI Output Port VCI (a) (b) (c) (d) (e) (f). The following list shows the mapping between LANs and their designated bridges. B dead B A,B,D

5 Solutions to Select Exercises 8 B E,F,G,H B I B idle B6 J B7 C 6. All bridges see the packet from D to C. Only B, B, and B see the packet from C to D. Only B, B, and B see the packet from A to C. B A-interface : A B-interface : D (not C) B B-interface : A B-interface : C B-interface : D B C-interface : C B-interface : A,D B D-interface : D B-interface : C (not A) 7. Since the I/O bus speed is less than the memory bandwidth, it is the bottleneck. Effective bandwidth that the I/O bus can provide is / Mbps because each packet crosses the I/O bus twice. Therefore, the number of interfaces is (/) =. 7. By definition, path MTU is 76 bytes. Maximum IP payload size is 76 = 6 bytes. We need to transfer + = bytes in the IP payload. This would be fragmented into fragments, the first of size bytes (because the fragment needs to be a multiple of 8 bytes, so it can t be exactly 6) and the second of size = 9 bytes. There are packets in total if we use path MTU. In the previous setting we needed packets. 7. (a) Information Stored at Node A Distance to Reach Node B C D E F A B C D E F

6 A Distance to Reach Node 6 B A 6 F E B 86 Solutions to Select Exercises (b) Information Distance to Reach Node Stored at Node C D A B C D E F (c) Information Stored at Node C D E F A B C D E F. The following is an example network topology. C A D E B F 6. Apply each subnet mask and, if the corresponding subnet number matches the SubnetNumber column, then use the entry in Next-Hop. (a) Applying the subnet mask..., we get Use interface as the next hop. (b) Applying subnet mask..., we get (Next hop is Router.) Applying subnet mask..., we get (Next hop is Router.) However,... is a longer prefix, so use Router as the next hop.

7 Solutions to Select Exercises 87 (c) None of the subnet number entries match, so use default Router R. (d) Applying subnet mask..., we get Use interface as the next hop. (e) Applying subnet mask..., we get Use Router as the next hop. 6. Step Confirmed Tentative 6 (A,,-) (A,,-) (A,,-) (B,,B) (A,,-) (B,,B) (D,,B) (A,,-) (B,,B) (D,,B) (C,,B) (A,,-) (B,,B) (D,,B) (C,,B) (E,6,B) (B,,B) (D,,D) (D,,B) (C,7,B) (C,,B) (E,7,B) (E,6,B) 7. (a) F (b) B (c) E (d) A (e) D (f) C CHAPTER. The following figures illustrate the multicast trees for sources D and E. CHAPTER. The advertised window should be large enough to keep the pipe full; delay (RTT) bandwidth here is ms Gbps = Mb = 7. MB of data. This requires bits ( =,,) for the

8 88 Solutions to Select Exercises AdvertisedWindow field. The sequence number field must not wrap around in the maximum segment lifetime. In 6 seconds, 7. GB can be transmitted. bits allows a sequence space of 8.6 GB, and so will not wrap in 6 seconds.. (a) B / ( GB) = 89 ms. (b) ticks in 89 ms is once each 89 µs indicating wrap around in.7 Ms or approximately days. 7. Using initial Deviation = it took iterations for TimeOut to fall below.. Iteration SampleRTT EstRTT Dev diff TimeOut

9 Solutions to Select Exercises 89 CHAPTER 6. (a) First we calculate the finishing times F i. We don t need to worry about clock speed here since we may take A i = for all the packets. F i thus becomes just the cumulative per-flow size: F i = F i + P i. Packet Size Flow F i We now send in increasing order of F i : Packet, Packet, Packet 6, Packet, Packet 7, Packet, Packet, Packet 8. (b) To give flow a weight of we divide each of its F i by : F i = F i + P i /. To give flow a weight of we divide each of its F i by : F i = F i + P i /. To give flow a weight of we divide each of its F i by : F i = F i + P i /. Again, we are using the fact that there is no waiting. Packet Size Flow Weighted F i

10 8 Solutions to Select Exercises Transmitting in increasing order of the weighted F i we send as follows: Packet, Packet, Packet 6, Packet, Packet, Packet 7, Packet 8, Packet.. (a) For the ith arriving packet on a given flow we calculate its estimated finishing time F i by the formula F i = max{a i,f i } +, where the clock used to measure the arrival times A i runs slow by a factor equal to the number of active queues. The A i clock is global; the sequence of F i values calculated as above is local to each flow. The following table lists all events by wall clock time. We identify packets by their flow and arrival time; thus, packet A is the packet that arrives on flow A at wall clock time (ie., the third packet). The last three columns are the queues for each flow for the subsequent time interval, including the packet currently being transmitted. The number of such active queues determines the amount by which A i is incremented on the subsequent line. Multiple packets appear on the same line if their F i values are all the same; the F i values are in italic when F i = F i + (versus F i = A i + ). Wall Clock A i Arrivals F i Sent A s Queue B s Queue C s Queue. A,B,C. A A B. C. B B.8 A. C A. B. A A B C..666 A. C A B 6. A6. B A,A6 B C6. 7. B7 C7. 6. A A,A6 B A8 B8 6.. C A6,A8 B7,B8 9 A9 7. B7 A6,A8,A9 B7,B8,B9 B9 6. C C,C C,C C,C C,C C,C6 C,C6,C7 C,C6,C7 C6,C7 (Continued)

11 Solutions to Select Exercises 8 Wall Clock A i Arrivals F i Sent A s Queue B s Queue C s Queue A C B B A6 C6 B8 A8 C7 B9 A9 C B A B A6,A8,A9 A8,A9,A A8,A9,A A8,A9,A A9,A A9,A A9,A A A A B8,B9 B8,B9 B8,B9 B9,B B9,B B9,B,B B,B B,B B,B B B C6,C7 C7 C7,C C7,C C7,C C C C (b) For weighted fair queuing we have, for flow B,. A,C. B A B C B C A B C A A6 C6 B7 C7 A8 B8 F i = max{a i,f i } +. For flows A and C, F i is as before. Here is the table corresponding to the one above: Wall Clock A i Arrivals F i Sent A s Queue B s Queue C s Queue A C B A C A C A A A,A A,A6 A,A6 A6,A8 B B7 B7,B8 C,C C,C C,C C,C C,C,C6 C,C6,C7 C6,C7 (Continued)

12 8 Solutions to Select Exercises Wall Clock A i Arrivals F i Sent A s Queue B s Queue C s Queue 9. A9 7. B7 A6,A8,A9 B7,B8,B9 C6,C7 B B8 A6,A8,A9 B8,B9 C6,C7. A 8. A6 A6,A8,A9,A B9 C6,C7. C 7. C6 A8,A9,A B9 C6,C7,C.666 B 6.66 B9 A8,A9,A B9,B C7,C 6. A8 A9,A B C7,C 6. B 6.8 C7 A9,A B,B C B A9,A B,B C 7 7. B A B C 8 7. A9 A C C A C 8. A A. (a) We have TempP = MaxP AvgLen MinThreshold MaxThreshold MinThreshold. AvgLen is halfway between MinThreshold and MaxThreshold, which implies that the fraction here is / and so TempP = MaxP/ = p/. We now have P count = TempP/( count TempP) = /(x count), where x = /p. Therefore, P count = Evaluating the product gives where x = /p. x x x x x (count + ). x count ( P ) ( P n ) x (n + ) x n = x (n + ), x

13 Solutions to Select Exercises 8 (b) From the result of previous question, Therefore, Accordingly, x = α = x (n + ). x (n + ) α α p = ( α) (n + ) α = /p. 8. At every second, the bucket volume must not be negative. For a given bucket depth D and token rate r, we can calculate the bucket volume v(t) at time t seconds and enforce v(t) being non-negative: v() = D + r = D ( r) v() = D + r = D ( r) v() = D + r = D ( r) v() = D + r = D ( r) v() = D 6 + r = D (7 r) v() = D 6 + 6r = D 6( r) We define the functions f (r),f (r),...,f 6 (r) as follows: f (r) = r f (r) = ( r) = f (r) f (r) (for r ) f (r) = r f (r) (for r ) f (r) = r < f (r) (for r ) f (r) = 7 r f 6 (r) = 6( r) f (r) (for r ) First of all, for r, f i (r) for all i. This means if the token rate is faster than packets per second any positive bucket depth will suffice (i.e., D ). For r, we only need to consider f (r) and f (r), since other functions are less than these functions.

14 8 Solutions to Select Exercises One can easily find f (r) f (r) = r 7. Therefore, the bucket depth D is enforced by the following formula: f (r) = 7 r (r =,) D f (r) = ( r) (r =,,) (r ) CHAPTER 7. Each string is preceded by a count of its length; the array of salaries is preceded by a count of the number of elements. That leads to the following sequence of integers and ASCII characters being sent: M A R Y 77 7 J A N U A R Y INT INT INT be le be le 8998 be 8998 le