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 172.30.0.0/16 is divided into equal subnets with prefix /22. How many subnets N are there?
172.30.0.0/16/22
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.248.11.0/24, 10.248.12.0/24, 10.248.13.0/24, 10.248.14.0/24, 10.248.15.0/24, 10.248.16.0/24, 10.248.17.0/24, 10.248.18.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.248.11.0/24 … 10.248.18.0/24
p =
3.Computer networksMedium
How many usable host addresses does a /24 IPv4 network have?
- A256
- B252
- C255
- D254
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 only works on local networks, so addresses never need to be shared globally
- BIPv6 does not use addresses
- CIts 128-bit address space is large enough to give every device a global address
- DIPv6 encrypts all addresses
5.Number systemsMedium
Set the bits so they store −78 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −788 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 1100 OR 1001 0110. What is the result's value R in decimal?
0100 1100OR1001 0110
R =
3Solutions
1.IPv4 subnetting
Answer: N = 64
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).
Borrowed bits
22 − 16 = 6
Number of subnets
N = 26 = 64
Answer
N = 64
Each /22 subnet has 210 − 2 = 1022 usable hosts.
2.IPv4 subnetting
Answer: p = 19
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
11 = 00001011
18 = 00010010
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.248.0.0/19 (it also covers 24 network(s) that are not in the list).
3.Computer networks
Answer: D · 254
Why this is the answer
A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.
Answer
D · 254
4.Computer networks
Answer: C · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
C · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 178
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Invert and add one
78 = 0100 1110
inverted: 1011 0001
+1 → 1011 0010
Read it as unsigned
1011 0010₂ = 256 − 78 = 178
Answer
U = 178
6.Number systems
Answer: R = 222
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Bit by bit
OR gives 1 where at least one bit is 1.
0100 1100
OR 1001 0110
= 1101 1110
Value of the result
1101 1110₂ = 222
Answer
R = 222
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