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 department needs 843 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
843 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.66.79/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.66.79/25
x =
3.Computer networksMedium
How many usable host addresses does a /24 IPv4 network have?
- A254
- B256
- C252
- D255
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?
- ATurn off the PC's DNS client
- BChange the PC's subnet mask to /32
- 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
Write −92 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −928 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 0011 << 2. What is the result's value R in decimal?
1111 0011<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 843
210 − 2 = 1022 ≥ 843 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 127
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
27 = 128
Subnet of this host
network: ⌊79/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126
Answer
x = 127
3.Computer networks
Answer: A · 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
A · 254
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 = 164
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
92 = 0101 1100
inverted: 1010 0011
+1 → 1010 0100
Read it as unsigned
1010 0100₂ = 256 − 92 = 164
Answer
U = 164
6.Number systems
Answer: R = 204
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.
1111 0011 << 2
= 1100 1100
Value of the result
1100 1100₂ = 204
Answer
R = 204
ExamProblems · examproblems.com/courses/computer-networks