IPv4 subnetting
- Hosts per subnet
- 2(32 − prefix) − 2 (/31: 2, /32: 1)
- Block size
- 256 − (mask octet)
- Network address
- IP AND mask
- Broadcast address
- IP OR (NOT mask): all host bits set to 1
- Subnets from n borrowed bits
- 2n
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
2Practice problems
1.IPv4 subnettingMedium
The network 192.168.193.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 3?
192.168.193.0/24/26subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.81.101.0/24, 10.81.102.0/24, 10.81.103.0/24, 10.81.104.0/24, 10.81.105.0/24, 10.81.106.0/24, 10.81.107.0/24, 10.81.108.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.81.101.0/24 … 10.81.108.0/24
p =
3.Computer networksMedium
HTTP is stateless. How do websites usually keep a user logged in across requests?
- ABy keeping the TCP connection open for days
- BWith cookies holding a session identifier or token
- CBy giving each user a fixed IP address
- DBy storing the password in the URL
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 does not allow packets larger than 576 bytes
- BIPv6 fragments at the transport layer inside TCP
- CRouters never fragment IPv6 packets; only the source can, using path MTU discovery
- DIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
5.Number systemsMedium
Write −67 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −678 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 20-bit two's complement?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 191
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Block size
26 = 64
Subnet number 3
network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191
Answer
x = 191
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
101 = 01100101
108 = 01101100
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.81.96.0/20 (it also covers 8 network(s) that are not in the list).
3.Computer networks
Answer: B · With cookies holding a session identifier or token
Why this is the answer
The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.
Answer
B · With cookies holding a session identifier or token
4.Computer networks
Answer: C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
Why this is the answer
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer
C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 189
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Invert and add one
67 = 0100 0011
inverted: 1011 1100
+1 → 1011 1101
Read it as unsigned
1011 1101₂ = 256 − 67 = 189
Answer
U = 189
6.Number systems
Answer: N = −524288
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Most negative value
−219 = −524288
Answer
N = −524288
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