Data Communications and Networking Solution Manual

c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line fails, only the corresponding device cannot operate.......Page 1
25. The telephone network was originally designed for voice communication; the Internet was originally designed for data communi.........Page 2
15. The International Standards Organization, or the International Organization of Standards, (ISO) is a multinational body dedi.........Page 3
25. The errors between the nodes can be detected by the data link layer control, but the error at the node (between input port and output port) of the node cannot be detected by the data link layer.......Page 4
b. (8 / 1000) s = 0. 008 s = 8 ms......Page 5
SNR= (signal power)/(noise power).......Page 6
c. Number of bits = bandwidth ¥ delay = 100 Mbps ¥ 2 ms = 200,000 bits......Page 7
17. See Figure 4.2. Bandwidth is proportional to (12.5 / 8) N which is within the range in Table 4.1 (B = N to B = 2N) for the Manchester scheme.......Page 9
d. f /N = 150/100 = 1.5 Æ P = 0.0......Page 10
N = 4 ¥ B = 4 ¥ 1 MHz = 4 Mbps......Page 11
11. We use the formula S = (1/r) ¥ N, but first we need to calculate the value of r for each case.......Page 13
17. We use the formula B = (1 + d) ¥ (1/r) ¥ N, but first we need to calculate the value of r for each case.......Page 14
21.......Page 15
b. Supergroup level: overhead = 240 KHz - (5 ¥ 48 KHz) = 0 Hz.......Page 17
25. See Figure 6.2.......Page 18
29. Random numbers are 11, 13, 10, 6, 12, 3, 8, 9 as calculated below:......Page 19
13. We can use Table 7.1 to find the power for different frequencies:......Page 21
b. The incident angle (60 degrees) is the same as the critical angle (60 degrees). We have refraction. The light ray travels along the interface.......Page 22
c. The incident angle (80 degrees) is greater than the critical angle (60 degrees). We have reflection. The light ray returns back to the more dense medium.......Page 23
b. 78 + 25 + 100 = 203 ms......Page 25
b. In a virtual-circuit network, the VCIs are local. A VCI is unique only in relationship to a port. In other words, the (port, .........Page 26
b. Total crosspoints ³ 4N[(2N)1/2 -1] ³ 174,886. With less than 200,000 crosspoints we can design a three-stage switch. We can use n = (N/2)1/2 =23 and choose k = 45. The total number of crosspoints is 178,200.......Page 27
11. Packet-switched networks are well suited for carrying data in packets. The end-to- end addressing or local addressing (VCI) .........Page 29
21. The cable modem technology is based on the bus (or rather tree) topology. The cable is distributed in the area and customers.........Page 30
9. At least three types of error cannot be detected by the current checksum calculation. First, if two data items are swapped du.........Page 31
c. Dataword: 1111 Æ Codeword: 1111111 Æ Corrupted: 0111110 Æ s2s1s0 = 111 Change b1 (Table 10.5) Æ Corrected codeword: 0101110Æ dataword: 0101 The dataword is found, but it is incorrect. C(7,4) cannot correct two errors.......Page 32
31. Figure 10.1 shows the generation of the codeword at the sender and the checking of the received codeword at the receiver using polynomial division.......Page 33
Figure 10.2 Solution to Exercise 33......Page 34
11. Piggybacking is used to improve the efficiency of bidirectional transmission. When a frame is carrying data from A to B, it .........Page 35
17. A five-bit sequence number can create sequence numbers from 0 to 31. The sequence number in the Nth packet is (N mod 32). This means that the 101th packet has the sequence number (101 mod 32) or 5.......Page 36
Algorithm 11.4 A bidirectional algorithm for Stop-And-Wait ARQ......Page 37
Algorithm 11.5 A bidirectional algorithm for Selective-Repeat ARQ......Page 38
25. State Rn = 0 means the receiver is waiting for Frame 0. State Rn = 1 means the receiver is waiting for Frame 1. We can then say......Page 39
31. In the worst case, we send the a full window of size 7 and then wait for the acknowledgment of the whole window. We need to send 1000/7 ª 143 windows. We ignore the overhead due to the header and trailer.......Page 40
Tfr = (1000 bits) / 1 Mbps = 1 ms G = ns ¥ nfs ¥ Tfr = 100 ¥ 10 ¥ 1 ms = 1 For pure ALOHA Æ S = G ¥ e-2G ª 13.53 percent......Page 43
Figure 12.1 Solution to Exercise 19......Page 44
25. We can say:......Page 45
¨ 01011000 11010100 00111100 11010010 01111010 11110110......Page 47
19. We can calculate the propagation time as t = (2500 m) / (200,000.000) = 12.5 ms. To get the total delay, we need to add propagation delay in the equipment (10 ms). This results in T = 22.5 ms.......Page 48
11. In CSMA/CD, the protocol allows collisions to happen. If there is a collision, it will be detected, destroyed, and the frame will be resent. CSMA/CA uses a technique that prevents collision.......Page 49
15. Figure 15.2 shows one possible solution.......Page 51
Figure 15.3 Solution to Exercise 19......Page 52
21. A bridge has more overhead than a repeater. A bridge processes the packet at two layers; a repeater processes a frame at onl.........Page 53
19. In GSM, separate bands are assigned for each direction in communication. This means 124 analog channels are available in eac.........Page 55
Period = (1/100) (distance) 1.5 = (1/100) (7778)1.5 = 6860 s = 1.9 hours......Page 56
11. Each STS-n frame carries (9 ¥ n ¥ 86) bytes of bytes. SONET sends 8000 frames in each second. We can then calculate the user data rate as follows:......Page 57
h. The only J byte is at the path layer to show the continuous stream of data at the path layer (end-to-end). The user uses a pa.........Page 58
j. Z bytes are unused bytes. All of the bytes in SOH are assigned, but in LOH and POH some bytes are still unused.......Page 59
13. We first need to look at the EA bits. In each byte, the EA bit is the last bit (the eight bit from the left). If EA bit is 0, the address ends at the current byte; if it 1, the address continues to the next byte.......Page 61
a. The minimum number of cells is 1. This happens when the data size £ 36 bytes. Padding is added to make it exactly 36 bytes. Then 8 bytes of header creates a data unit of 44 bytes at the SAR layer.......Page 62
c. When user (user data + 43 + 8) mod 48 = 0.......Page 63
b. Class B (first bit is 1 and second bit is 0)......Page 65
b. 232- 29 = 8 Addresses per subnet......Page 66
255......Page 67
b. 0:AA::119A:A231......Page 68
37. The node identifier is 0000:0000:1211. Assuming a 32-bit subnet identifier, the subnet address is 581E:1456:2314:ABCD:0000 where ABCD:0000 is the subnet identifier.......Page 69
9. The checksum is eliminated in IPv6 because it is provided by upper-layer protocols; it is therefore not needed at this level.......Page 71
23. Let us first find the value of header fields before answering the questions: VER = 0x4 = 4 HLEN =0x5 = 5 Æ 5 ¥ 4 = 20 Servic.........Page 72
g. The type of service is normal.......Page 73
17. It could happen that host B is unreachable, for some reason. The error message generated by an intermediate router could the.........Page 75
27. Ethernet: Supported number of groups using 23 bits = 223 = 8,388,608 groups IP: Supported number of groups using 28 bits =228 = 268,435,456 groups Address space lost: 268,435,456 - 8,388,608 = 260,046,848 groups......Page 76
13. A host that is totally isolated needs no routing information. The routing table has no entries.......Page 77
m1......Page 78
31. No, RPF does not create a shortest path tree because a network can receive more than one copy of the same multicast packet. RPF creates a graph instead of a tree.......Page 79
33. Yes, RPM creates a shortest path tree because it is actually RPB (see previous answer) with pruning and grafting features. The leaves of the tree are the networks.......Page 80
Explanation......Page 81
29. The largest number in the sequence number field is 232 -1. If we start at 7000, it takes [(232 - 1) -7000] / 1,000,000 = 4295 seconds.......Page 82
33. See Figure 23.5. Note that the value of cumTSN must be updated to 8.......Page 83
Figure 23.5 Solution to Exercise 33......Page 84
19. Input: (100/60) ¥ 12 + 0 ¥ 48 = 20 gallons Output: 5 gallons Left in the bucket: 20 - 5 = 15......Page 85
23. CTD is the average cell transfer delay. If each cell takes 10 ms to reach the destination, we can say that CTD = [(10 ms ¥ n) / n] in which n is the total number of cells transmitted in a period of time. This means that CTD = 10 ms......Page 86
21. The number of question sections and answer sections must be the same. The relationship is one-to-one.......Page 87
17. Three transmissions, each with a minimum size of 72 bytes, mean a total of 216 bytes or 1728 bits.......Page 89
23. There should be limitations on anonymous FTP because it is unwise to grant the public complete access to a system. If the co.........Page 90
17. HTTP/1.1 400 Bad Request......Page 91
29. PUT /bin/letter HTTP /1.1......Page 92
13. OCTET STRING tag: 04......Page 93
counter, length, value (2313)......Page 94
15. The web server and media server can be two distinct machines since it is the metafile-data file combination that is important.......Page 95
19. H.323 can also be used for video, but it requires the use of videophones. Currently most people don’t have videophones.......Page 96
a. We can show the encryption character by character. We encode characters A to Z as 0 to 25. To wrap, we subtract 26. The encrypted message is NBCM CM UH YRYLWCMY.......Page 97
13. We can, but it is not safe at all. The best we can do is to change a 0 sometimes to 0 and sometimes to 1 and to change a 1 sometimes to 0 and sometimes to 1. It can be easily broken using trial and error.......Page 98
29. Nothing happens in particular. Assume both Alice and Bob choose x = y = 9. We have the following situation with g = 7 and p .........Page 99
b. The algorithm does not meet the second criteria (weak collision). If the digest is given, we can create 10 numbers that hash .........Page 101
21. Figure 31.1. shows one scheme. Note that the scheme forces Bob to use the timestamp which is related to the timestamp used by Alice (T+1), this ensures that the two messages belongs to the same session.......Page 102
31. See Figure 31.3. The shaded area shows the encryption/decryption layer.......Page 103
Figure 31.3 Solution to Exercise 31......Page 104
17. A VPN is a technology that allows an organization to use the global Internet yet safely maintain private internal communication.......Page 105
25. IPSec uses the services of IKE to create a security association that includes session keys. However, this does not start fro.........Page 106
31. IPSec uses IKE to create security parameters. IKE has defined several methods to do so. Each method uses a different set of .........Page 107