Solutions to End of Chapter Problems
Problem 1:
MAC address: No; IP address: Yes.
Problem 2:
(a)
Time to live and Header checksum.
(b)
HLEN = 5; TOTAL LENGTH = 1024.
(c)
The time to live field is a number decremented by one on each hop. When the time to
live (hop-counter) reaches zero, the packet is discarded. The purpose is to ensure the
packet does not wander around the Internet forever; it can only touch the number of
routers equal to the initial value in the time to live field.
(d)
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(e)
It uses a source address of 0.0.0.0 and a destination address of 255.255.255.255.
(f)
It is discarded (not routed any further).
(a)
10.64.128.192 / 28
(b)
14 (also accept 16)
(c)
10.64.128.193
(d)
10.64.128.206
(e)
10.64.128.207
(f)
No; this address is in the private address domain
(a)
(i) /24
(ii) /12
(b)
(i) 255.255.0.0
(ii) 255.128.0.0
Problem 3:
Problem 4:
Problem 5:
Answer is (a)
Problem 6:
(a) Version 4
(b) 83 78 a8 1f = 131.120.168.31
Problem 7:
(a)
255.255.224.0
(b)
136.52.96.0
(c)
136.52.127.255
(d)
8192 – 2 = 8190.
Problem 8:
Every packet is an independent entity, possible traveling over different paths from source to
destination
Problem 9:
False
Problem 10:
146.25.128.0
Problem 11:
Answer is (b)
7
Problem 12:
Answer is (a)
Problem 13:
32 ; 8 ; 0 ; 255
Problem 14:
To extract the network ID from an IP address (or to extract the network address).
Problem 15:
(a)
128.32.14.0
(b)
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(c)
510
(d)
128.32.15.254
Problem 16:
software / 32
network / host
host / network
host / broadcast
Problem 17:
(a)
address:
10011100 10001111 00001010 00110111
mask: AND 11111111 11111111 11111000 00000000
first address:
10011100 10001111 00001000 00000000
156.143.8.0
(b)
232-21-2 = 211-2 = 2046 hosts
(c)
First address assigned to a host:
10011100 10001111 00001000 00000001
156.143.8.1
last address assigned to a host:
10011100 10001111 00001111 11111110
156.143.15.254
(d)
Problem 18:
No. These private IPv4 addresses are provided to users to be used internally within private
networks. This allows users the ability to have an allotment of IP addresses to use within a
private network whose addresses can be distributed by the network administrator rather than the
Internet Service Provider. These addresses must be unique within a private network, but do not
need to be unique globally. All IP routers know that these addresses are for private networks
only.
Problem 19:
(a) True
Problem 20:
The valid ones are (b), (d) and (e)
Problem 21:
(a)
2^5 – 2 = 30
(b)
137.18.129.10000001 = 137.18.129.129
(c)
137.18.129.10011110 = 137.18.129.158
(d)
137.18.129. 10011111 = 137.18.129.159
(b) True
(c) False
(d) False
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Problem 22:
Compare network address and broadcast address (i.e. 230.8.16.0 and 230.8.31.255). View them
in bits.
230.8.16.0 11100110.00001000.00010000.00000000
230.8.31.255 11100110.00001000.00011111.11111111
Count the number of bits that are the same, and this is your prefix. In this example, the prefix is
20. Let the prefix in general be denoted as n. Then the number of addresses available for
assignment to hosts is 232n  2 .
Problem 23:
85.133.51.31
Problem 24:
N = 232-n = 232-14 = 218 = 262,144 total addresses. Of these 262,144 addresses, 262,142 are
assignable to hosts.
Problem 25:
11111111
11111100
00000000
00000000 =>
255.252.0.0
Problem 26: Change the dotted-decimal to binary: 11111111 11110000
We see there are 12 1's, so the prefix is 12. In slash notation: /12.
00000000
00000000
Problem 27: 140.150.16.17/18
First address:
address:
10001100
mask: AND 11111111
first address: 10001100
10010110
11111111
10010110
00010000
11000000
00000000
00010001
00000000
00000000
140.150.0.0
10001100
10010110
00010000
00010001
00111111
11111111
address:
Set the 32 – n = 14 rightmost bits to 1:
10001100
last address:
Problem 28:
Problem 29:
10010110
140.150.63.255
(a) No. When TTL is decremented to 0, the packet is discarded.
(b) Yes. The TTL field is 8 bits which means the maximum value of TTL is 255.
(c) No. The TTL field is 8 bits which means the maximum value of TTL is 255.
4
5
VER HLEN
00
Service
Type
004E 0010
Total
ID
Length
0000
Flag &
Frag Offset
12
TTL
06...
Protocol
(a) There are 5 x 4 = 20 bytes in the header.
(b) Size of the data and the 20-bytes header totals to 0x004E which is 78. Thus the size of the
data (without the header) is 58 bytes.
(c) The TTL field is 0x12 which is 18, so this packet can travel to 18 more routers. The 18th
would discard it.
Problem 30:
The TTL field and the Header checksum.
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