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
A subnet uses the mask 255.255.254.0. How many usable host addresses H does it have?
255.255.254.0
H =
2.IPv4 subnettingHard
A host has the address 172.20.129.223/20. What is the third octet x of its subnet's broadcast address?
172.20.129.223/20
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A62
- B30
- C126
- D64
4.Computer networksHard
A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?
- AChange the PC's subnet mask to /32
- BTurn off the PC's DNS client
- CGive the PC a different MAC address so the router can find it from outside
- DConfigure port forwarding on the router to the PC's private address
5.Number systemsMedium
Set the bits so they store −14 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −148 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 0100 OR 1011 1101. What is the result's value R in decimal?
1011 0100OR1011 1101
R =
3Solutions
1.IPv4 subnetting
Answer: H = 510
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).
Prefix length
255.255.254.0 = /23
Usable hosts
29 − 2 = 510
Answer
H = 510
2.IPv4 subnetting
Answer: x = 143
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 in the third octet
24 = 16
Subnet of this host
network: 172.20.128.0
broadcast: 172.20.143.255
Answer
x = 143
3.Computer networks
Answer: A · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
A · 62
4.Computer networks
Answer: D · Configure port forwarding on the router to the PC's private address
Why this is the answer
NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.
Answer
D · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: U = 242
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
14 = 0000 1110
inverted: 1111 0001
+1 → 1111 0010
Read it as unsigned
1111 0010₂ = 256 − 14 = 242
Answer
U = 242
6.Number systems
Answer: R = 189
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.
1011 0100
OR 1011 1101
= 1011 1101
Value of the result
1011 1101₂ = 189
Answer
R = 189
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