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 525 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
525 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.75.157/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.75.157/30
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A126
- B64
- C62
- D30
4.Computer networksHard
With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?
- AGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
- BNeither ever resends lost frames
- CGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
- DBoth resend the whole window from the start
5.Number systemsMedium
Set the bits so they store −15 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −158 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0010 0111 OR 1011 1010. What is the result's value R in decimal?
0010 0111OR1011 1010
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 < 525
210 − 2 = 1022 ≥ 525 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 158
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: ⌊157/4⌋·4 = 156
broadcast: 156 + 4 − 1 = 159
usable hosts: 157 … 158
Answer
x = 158
3.Computer networks
Answer: C · 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
C · 62
4.Computer networks
Answer: A · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
Why this is the answer
A Go-Back-N receiver discards out-of-order frames, so the sender must repeat everything after the loss. Selective Repeat buffers them and asks only for the missing one.
Answer
A · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systems
Answer: U = 241
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
15 = 0000 1111
inverted: 1111 0000
+1 → 1111 0001
Read it as unsigned
1111 0001₂ = 256 − 15 = 241
Answer
U = 241
6.Number systems
Answer: R = 191
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.
0010 0111
OR 1011 1010
= 1011 1111
Value of the result
1011 1111₂ = 191
Answer
R = 191
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