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2013 6 2 1 Solution Chapter 1 14 The propagation delay is the time that is required for the energy of a signal to propagate from one point to another a Find the propagation delay for a signal traversing the following networks at the speed of light in cable 2 3 x 108meters second iibd 10 a circuit board 10 cm a room 10 m a building 100 m a metropolitan area 100 km a continent 5000 km up and down to a geostationary satellite 2 x 36000 km up and down to a geostationary satellite 2 x 36000 km 2013 6 2 2 Solution To find the propagation delay divide distance by the speed of light in cable Thus we have a circuit board tprop 4 347 x 10 10seconds 8 a room tprop 4 3478 x 10 8seconds a building tprop 4 3478 x 10 7seconds a metropolitan area tprop 4 3478 x 10 4seconds a continent tprop 0 02174 seconds up and down to a geostationary satellite tprop 0 31304 seconds b How many bits are in transit during the propagation delay in the above cases if bits are entering the above networks at the following transmission speeds 10 000 bits second 1 megabit second 100 megabits second 10 gigabits second Solution f The number of bits in transit is obtained by multiplying the transmission rate R by the propagation delay Distance m 10 Kbps1 Mbps100 Mbps10 Gbps 0 14 347 x 10 64 347 x 10 40 043474 3478 104 3478 x 10 40 0434784 3478434 780 1004 3478 x 10 30 4347843 4784347 800 100 0004 3478434 78434784 3478x106 5 000 000217 42174021740002 174x108 72 000 0003130 4313040313040003 1304x109 2013 6 2 3 15 In problem 14 how long does it take to send an L byte file and to receive a 1 byte acknowledgment back Let L 109 bytes S liSolution The total time required to send a file and receive an acknowledgment of its receipt is given by ttotal Lmessage R Lack R 2 tprop Lmessage R Lack R 2 d c where Lis the message length in bits Lis the where Lmessageis the message length in bits Lackis the acknowledgment length in bits R is the transmission bit rate d is the distance traversed and c is the speed of light The above equation shows that there are two main factors that determine total delay 1 Message and ACK transmission time which depends on the message length and the transmission bit rate 2 Propagation delay which depends solely on distance When the propagation delay is small message and ACK transmission times determine the total delay On the othertransmission times determine the total delay On the other hand when the bit rate becomes very large the propagation delay provides a delay component that cannot be reduced no matter how fast the transmission rate becomes The tables below show the two main components of the total delay in microseconds The message transmission time isdelay in microseconds The message transmission time is shown in red and the propagation delay is shown in blue The entries in the total delay tables are colored according to which delay component is dominant 2013 6 2 4 Message ACK delay 10 kbps message ACK delay 1 Mbps message ACK delay 100 Mbps message ACK delay 10 Gbps 8 00E 118 00E 098 00E 078 00E 05 Table 1 Message length 109bytes Distance meters 2 prop delay microseconds Total delay 10 kbps total delay 1 Mbps total delay 100 Mbps total delay 10 Gbps 0 10 000878 00E 118 00E 098 00E 078 00E 05 100 0869578 00E 118 00E 098 00E 078 00E 05 1000 8695658 00E 118 00E 098 00E 078 00E 05 100000869 56528 00E 118 00E 098 00E 078 01E 05 500000043478 268 00E 118 00E 098 00E 078 43E 05 720000003130448 00E 118 00E 098 06E 071 13E 06 Message ACK delay 10 kbps message ACK delay 1 Mbps message ACK delay 100 Mbps message ACK delay 10 Gbps 8 00E 118 00E 098 00E 078 00E 05 Table 1 Message length 1000 bytes Distance meters 2 prop delay microseconds Total delay 10 kbps total delay 1 Mbps total delay 100 Mbps total delay 10 Gbps 0 10 00087800800 0018008 00087 80 08087 0 80167 100 086957800800 087 8008 08696 80 16696 0 887757 1000 869565800800 8698008 86957 80 94957 1 670365 100000869 5652801669 565 8877 56522 949 6452 870 366 500000043478 26844278 260 51486 2609 43558 34 43479 06 720000003130441113044 321044313124 313044 8 2013 6 2 1 Solution Chapter 2 19 Suppose an application layer entity wants to send an L byte message to its peer process using an existing TCP connection The TCP segment consists of the message plus 20 bytes of header The segment is encapsulated into an IP packet that has an additional 20 bytes of header The IP packet in turn goesan additional 20 bytes of header The IP packet in turn goes inside an Ethernet frame that has 18 bytes of header and trailer What percentage of the transmitted bits in the physical layer correspond to message information if L 100 bytes 500 bytes 1000 bytes 2013 6 2 2 Solution TCP IP over Ethernet allows data frames with a payload size up to 1460 bytes Therefore L 100 500 and 1000 bytes are within this limit Thhd il d The message overhead includes TCP 20 bytes of header IP 20 bytes of header Ethernet total 18 bytes of header and trailer Therefore L 100 bytes100 158 63 3 efficiency L 100 bytes 100 158 63 3 ef。