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FTP: is application layer protocol, uses more than one port at server side. Out-of-Band Control: FTP sends its control information, which includes user identification, password, and put/get commands, on one connection, and sends data files on a separate parallel connection. Because it uses a separate connection for the control information, FTP uses out-of-band control. Components of web searching: Crawler/Index/Rank. Propagation delay is defined as how long it takes for a certain amount of bytes to transferred over a medium. Propagation delay is the distance between the two routers divided by the propagation speed. d/s where d is the distance and s is the speed. Transmission delay: R=link bandwidth (bps) L=packet length (bits) time to send bits into link = L/R Propagation delay: d = length of physical link s = propagation speed in medium (2x108m/sec) propagation delay = d/s. Consider two hosts, A and B, are separated by 5,000 kilometers and are directly connected by a link with transmission rate R=20Mbps. The propagation speed over the link is 2.5*108meters/sec. What is the propagation delay of the link? Propagation delay = distance/speed, = (5000x1000)/2.5 x 10^8, (50 x 100000)/(25 x 10^7), 2/10^2, = .02 secs Consider only one file of 50,000 bits from Host A to Host B. What is the transmission delay? D trans = L/R, t= 50000/(20x10^6) + 0.02, = .0025 + .02, = .0225 secs. Suppose the file is sent continuously as one big message. What is the maximum number of bits that will be in the link at any given time? D (end-to-end) = ( m/s + L/R), At 0.02 seconds, the first bit reaches host B, we have to find out how many bits were transmitted into the link in 0.02 seconds, R=20mbps, In 1 second 0mb are sent, In 0.02 seconds (20x0.02) mbs are sent, In 0.02 seconds, 0.4mb or 4000,000 bits are sent, Now 400,000 > 50,000(file size), Hence a maximum of 50,000 bits can be in the link during the file transfer, A max of 50,000 bits will be in the link during the file transfer. Let x denote the size of the file, what is the minimum value of x for the link to be continuously transmitting? P2P architecture: no always-on server, arbitrary end systems directly communicate, peers are intermittently connected and change IP addresses. Email System: [email protected]  [user agent]  [mail server]  smtp[mail server][user agent] [email protected] 1) Ann uses UA to compose message and “to bob. 2) The UA sends the message packet to the corresponding MS where the message is placed in the queue. This link works utilizing SMTP. 3) The mail server looks up the address of the recipients MS using DNS, sends the message to the MS again using SMTP. 4) Now the message stays in the queue of the recipients MS. 5) [email protected] can now invoke the UA at his end to procure the message from the MS queue. The protocols used to do this are POP3 or IMAP4, however bob is using web-based mail, and he will most probably invoke the UA and obtain the message using HTTP. 6) Also if for some reason, Ann’s MS realizes that bob’s MS is inaccessible or down, it will retain the message in its queue and try again at a rescheduled time. All Mailing protocols are application layer protocols and work on top of the TCP, which is transport layer protocol. This is because reliability is more important than speed as compared to UDP. UDP and TCP uses 1’s complement for their checksums. Suppose UDP and TCP use 8-bit words in computing the checksum write down the checksum of this message 010101010111000001001100 11101101. Write down the information used in UDP for de-multiplexing Destination IP add, Dest port no.; write down the information used in TCP for de-multiplexing source IP, Source port no, dest IP, dest port no. Suppose within your web browser, you click on a link to obtain webpage. The IP address for associated URL is not cached in your local host. Suppose that DNS servers should be visited before your host receives the IP address. The successive visits incur an RTT of RTT1,RTT2,…,RTTn. Suppose that the base html file associated with the link references two very small objects on the same server A and another three small objects in another server B. Let RTT_A denote the RTT between the local host and server A and RTT_B denote the RTT between the local host and server B. Suppose the IP address of server B is not cached in the local host. Neglecting transmission times, a) Nonpersistent http with no parallel TCP connections? Time = DNS for A + time for base + Time for 2 objects A + DNS for B + Time for 3 objects B, Time = (i=1)E^n RTTi +(j=1)E^n RTTj + 2RTT_A + 2x2 RTT_A + 3x2 RTT_B. b) nonpersistent http with parallel connections? Time= DNS for A + Time for Base + 2 RTT_A + DNS for B + 2 RTTB c) Persistent http without pipelining or parallel connection? Time = DNS A + Time for Base + Time for 2 object A + DNs B + Time for 3 Object B, =(i=1)E^n RTTi + 2RTT_A + 2 RTT_A + (j=1)E^n RTTj + 4 RTT_B, = (i=1)E^n RTTi + (j=1)E^n RTTj + 2 RTT_A + 2RTT_A + 4RTT_B d) Persistent http with pipelining and parallel connections? Time = DNS A + time for base + 2 objects A + DNS B + Time for 3 Objects, = (i=1)E^n RTTi + (j=1)E^n RTTj + 2 RTT_A + mRTTA + 2 RTT_B. Host A requests a HTML file of size O = 100kbytes from web server B. suppose RTT = 100 msec and S = 100bytes. Transmission rate R = 1mbps. The Transmission goes through a slow start process. No error and no congestion in the network. Calculate the total time for Host A to get to file. O =100kb, RTT = 100ms, S =100B, R=1Mbps, Latency = 2 RTT + O/R + P[RTT +(S/R)]-((2^p)-1)S/R. P = min(Q, k -1), K = [log2 ((O/S)+1), = [log2((100x1000/100) +1)], [log2(1001)] =10. Q =[log2 [(1+ (RTT/(S/R)))] +1], =[log2 [( 1 + (100 x 1000 x 1000/1000x 100 x 4))] +1 ], =[log2(251) +1], =8. log2(z)=log(z)/log(2). Latency = 2 x (100/1000) + (100x1000)/(10^6)/4 + 8[(100/1000) + (100/(10^6/4))] – ((2^8-1)-1)x 4(10^4). = 1.3520 seconds. Window size vs. Sequence #, suppose that the lowest sequence number that the receiver is waiting for is packet m. In this case, its window is [m, m+w-1] and it has received and ACKed packet m-1 and the w-1 packets before that, where w is the size of the window. If none of those w ACKs have been yet received by the sender, then ACK messages with values of [m-w,m-1] may still be propagating back. If no ACKs with these ACK numbers have been received by the sender, then the sender’s window would be [m-w, m-1]. Thus, the lower edge of the senders’ window is m-w, leading edge of the receivers window is m+w-1. In order for the leading edge of receiver’s window to not overlap with the training edge of the sender’s window, the sequence number space must thus be big enough to accommodate 2w sequence numbers. That is, the sequence number space must be at least twice as large as the window size, k > w. Q = [log2 (1 + (RTT/(S/R))] +1 , K= [log2((o/s) +1)]. Propagation delay d/s = length of phys link/pop.speed (2x108m/sec Transmission delay: R=link bandwidth (bps) L=packet length (bits) time to send bits into link = L/R Maximum no of bits in the link: max {no of bits sent in dprop, file size} Nodal delay: dproc + dqueue + dtransm + dprop Queue delay: L * a/R or [NL+(L-X)]/R withN:packets already in queue, x:bits of current transmitted pkt P2P architecture: no always-on server, arbitrary end systems directly communicate, peers are intermittently connected and change IP addresses, highly scalable. Response time: 2*RTT (one to initiate TCP con. + one for HTTP request and first few bytes of response) + transmit time RTT: EstRTT = (1­α)EstRTT + α SampRTT. Packet switching: data sent through net in discrete chunks, each end­to­end data stream divided into packets, packets share network  resources, each packet uses full link BW Congestion: packets queue and wait  asynchr. Stat. multiplexing: sequence of packets does not have to have fixed pattern, shared on demand Store&Forward: entire packets must  arrive at router before it can be transmitted to next link  Out-of-Band Control: FTP sends its control information, which includes user identification, password, and put/get commands, on one connection, and sends data files on a separate parallel connection. Because it uses a separate connection for the control information, FTP uses out-of-band control. FTP: is application layer protocol, uses more than one port at server, Messages “in­band” control & data messages may be interleaved (HTTP, DNS, SMTP) “out­of­bound”  control and data messages carried in separate connections (FTP) Multiplexing (at sender): gathering data from multiple sockets, enveloping with header De-multiplexing (at receiver): delivering received segments to correct socket, each datagram has source IP +dest IP, each segment has source port + dest port. Delay modeling: R: rate/BW S: MSS(bits) O: Object size(bits) W: window size/segments P: number of idle times Q: object infinite size K: no.of windows cover the object P=min{Q,K-1} K=[log2 (O/S +1)] Latency =2RTT + O/R + P[RTT+ S/R] – (2P-1)S/R Q=[log2(1+RTT/S/R)]+1 Slow start: fixed window WS/R>RTT+S/R  delay=2RTT+O/R , WS/R
A TCP connection is used to send 120kb of data. The sender has an MSS(maximum segment size) of 4kbytes and a congestion window threshold of 16kbytes. Assuming that the sender detects a triple dup ACK at the end of the 3th RTT and a timeout at end of 8th RTT. The receiver’s buffer size is 22kb. Show the values of congestion window size, congestion window threshold, and amount of data sent over time for each RTT. RTT Threshold Congestion window Size Data Sent 1 16k 4k 4k 2 16k 8k 12k 3 16k 16k 28k 4 8k 8k 36k 5 8k 12k 48k 6 8k 16k 64k 7 8K 20k 84k 8 8k 22k (rcv buffer) 9 11k 4k 100 11k 8k