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 898 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
898 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.179.20/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.179.20/28
x =
3.Computer networksMedium
Which protocol does the "ping" command use?
- ATCP
- BHTTP
- CICMP
- DUDP
4.Computer networksHard
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
- AThe server can push updates over UDP
- BSeveral requests reuse one TCP connection, avoiding repeated handshakes
- CThe connection stays encrypted without TLS
- DResponses can arrive in any order on one stream, so slow replies do not block others
5.Number systemsMedium
Set the bits so they store −23 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −238 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 0110 AND 1101 0110. What is the result's value R in decimal?
1000 0110AND1101 0110
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 < 898
210 − 2 = 1022 ≥ 898 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 31
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: ⌊20/16⌋·16 = 16
broadcast: 16 + 16 − 1 = 31
usable hosts: 17 … 30
Answer
x = 31
3.Computer networks
Answer: C · ICMP
Why this is the answer
Ping sends ICMP echo requests and measures the time until echo replies return.
Answer
C · ICMP
4.Computer networks
Answer: B · Several requests reuse one TCP connection, avoiding repeated handshakes
Why this is the answer
Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.
Answer
B · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems
Answer: U = 233
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
23 = 0001 0111
inverted: 1110 1000
+1 → 1110 1001
Read it as unsigned
1110 1001₂ = 256 − 23 = 233
Answer
U = 233
6.Number systems
Answer: R = 134
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
AND gives 1 only where both bits are 1.
1000 0110
AND 1101 0110
= 1000 0110
Value of the result
1000 0110₂ = 134
Answer
R = 134
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