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 770 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
770 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.232.11/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.232.11/30
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A100 ms
- B10 ms
- C0.1 ms
- D1 ms
4.Computer networksHard
Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?
- ABy routing A's frames through C
- BOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
- CBy increasing the TTL of the frames
- DBy making A and C use collision detection so they stop sending when frames overlap
5.Number systemsMedium
Set the bits so they store −3 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −38 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 20 bits?
20 bit
N =
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 < 770
210 − 2 = 1022 ≥ 770 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 11
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
22 = 4
Subnet of this host
network: ⌊11/4⌋·4 = 8
broadcast: 8 + 4 − 1 = 11
usable hosts: 9 … 10
Answer
x = 11
3.Computer networks
Answer: B · 10 ms
Why this is the answer
2 × 10⁶ m / 2 × 10⁸ m/s = 0.01 s = 10 ms. Propagation delay depends on distance and medium, not packet size.
Answer
B · 10 ms
4.Computer networks
Answer: B · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
Why this is the answer
A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.
Answer
B · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems
Answer: U = 253
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
3 = 0000 0011
inverted: 1111 1100
+1 → 1111 1101
Read it as unsigned
1111 1101₂ = 256 − 3 = 253
Answer
U = 253
6.Number systems
Answer: N = 1.049 × 10⁶
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.
Largest value
220 − 1 = 1048575
Answer
N = 1048575
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