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 192.168.170.0/24 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 network address of subnet 4?
192.168.170.0/24/26subnet 4 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.160.144/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.160.144/28
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A64
- B30
- C126
- D62
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?
- AConfigure port forwarding on the router to the PC's private address
- BGive the PC a different MAC address so the router can find it from outside
- CTurn off the PC's DNS client
- DChange the PC's subnet mask to /32
5.Number systemsMedium
Write −13 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −138 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0010 1010 << 2. What is the result's value R in decimal?
0010 1010<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: x = 192
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 4
network: (4 − 1)·64 = 192
broadcast: 192 + 64 − 1 = 255
Answer
x = 192
2.IPv4 subnetting
Answer: x = 159
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
24 = 16
Subnet of this host
network: ⌊144/16⌋·16 = 144
broadcast: 144 + 16 − 1 = 159
usable hosts: 145 … 158
Answer
x = 159
3.Computer networks
Answer: D · 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
D · 62
4.Computer networks
Answer: A · 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
A · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: R = 168
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
Shifting left by 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0010 1010 << 2
= 1010 1000
Value of the result
1010 1000₂ = 168
Answer
R = 168
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