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1、Chapter 1P18Suppose two hosts, A and B, are separated by 10,000 kilometers and are connected by a direct link of R =2 Mbps. Suppose the propagation speed over the link is 2.5*108 meters/sec.a) = =0.04(s) R=8.0(b) 104b) N= R=8.0(b) 104c) The bandwidth-delay product of a link is the maximum number of
2、bits that can be in the linkd) w= =125(m/b)，1 bit is 125 meters long, which is longer than a football fielde) w= = P19Referring to problem P18, suppose we can modify R. For what value of R is the width of a bit as long as the length of the link?依题意，w=m=1.0(m)，又 w= 107所以，R= =25(bps)P20Consider proble
3、m P18 but now with a link of R=1Gbps.a) R=4.0(b) 107b) N=4.04.0 4.0 107 105 = 105c) w= =0.25(m/b) P21Refer again to problem P18.a) = ttrans + tprop =+ =0.2+0.04=0.24(s)4.0 105 b) =10(ttrans +2 tprop)=1(s)c) P24In modern packet-switched networks, the source host segments long, application- layer mess
4、ages into smaller packets and sends the packets into the network. The receiver then reassembles the packets back into original message.We refer to this process as message segmentation.a) Time to send message from source host to first packet switch =. With store-and-forward switching, the total time
5、to move 668.0 10sec4sec2 10message from source host to destination host =4sec 312sechopsb) T=0.001(s)=1ms 20002 106Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = sec2sec12mmc) NumberArrival1st32nd54000th4002time at which last (4
6、000th) packet is received=. It can be 3s 3999*1 s4.002smmseen that delay in using message segmentation is significantly less (almost 1/3rd). d) Drawbacks:i.Packets have to be put in sequence at the destination. ii.Message segmentation results in many smaller packets. Since header size is usually the
7、 same for all packets regardless of their size, with message segmentation the total amount of header bytes is more.P26Consider sending a large file of F bits for Host A to Host B. There are two links between A and B, and the links are uncongested. Host A segments the file into segments of S bits eac
8、h and adds 40 bits of header to each segment, forming packets of L=40+S bits. Each link has a transmission rate of R bps. Find the value of S that minimizes the delay of moving the file from Host A to Host B. Disregard propagation dealy.Time at which the 1st packet is received at the destination：T=(
9、s)，After + 40 2this, one packet is received at destination everysec + 40=+()()= ()()总 + 40 2 1 + 40+ 1 + 40To calculate the value of S which leads to the minimum delay，=0 总21401()040FSSFR SSRChapter 2P7The total amount of time to get the IP address is.nRTTRTTRTTL21Once the IP address is known, elaps
10、es to set up the TCP connection and ORTTanother elapses to request and receive the small object. The total response time ORTTisT=noRTTRTTRTTRTTL212P8a) Non-persistent HTTP with no parallel TCP connections:=T+3 20b) Non-persistent HTTP with parallel TCP connections:=T+20c) Persistent HTTP:=T+0P9a) =0
11、.06(s)， =10，=0.69 1051.5 107/(1-)=0.15(s)=2+0.15=2.15(s)b) the average access delay is /(1-0.6) =(0.06 sec)/1 (0.6)(0.6) = 0.09375 seconds.the average response time is0 .12 sec +2 sec =2.09375 sec for cache missesthe average response time is (0.4)(0 sec) + (0.6)(2.09375 sec) =1.25625 seconds.Thus th
12、e average response time is reduced from 2.6 sec to 1.25625sec.P16F = 5Gbits = 5 * 1024 Mbits us = 20 Mbps dmin = di = 1 MbpsClient-ServerDcs = max NF/us, F/dmin DcsN101001000100kps512025600256000250kps512025600256000u500kps512025600256000Peer to Peer)u , NF/(u, F/dmaxF/uDN1iisminsPP 2Dp2pN1010010001
13、00kps512017201.043516.6250kps512011527.919383.6u500kps512051205120P19There are N nodes in the overlay network. There are N(N-1)/2 edges.P21Alice sends her query to at most N neighbors. Each of these neighbors forwards the query to at most M = N-1 neighbors. Each of those neighbors forwards the query
14、 to at most M neighbors. Thus the maximum number of query messages is N + NM + NM2 + + NM(K-1) = N(1 + M + M2 + + M(K-1) ) = N(1-MK)/(1-M) = N(N-1)K- 1/(N-2) P23In this problem we explore designing a hierarchical overlay that has ordinary peers, super peers, and super-duper peers.a) 100400=4 =100 10
15、44 1064 104Therefore, we would need about 100 super-duper peers to support 4 million nodes. b) Each super peer might store the meta-data for all of the files its children are sharing. A super-duper peer might store all of the meta-data that its super-peer children store. An ordinary node would first send a query to its super peer. The super peer would respond with matches and then possibly forward the message to its super-duper peer. The super-duper peer would respon
