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 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.41.201/27. What is the fourth octet x of its network address?
10.20.41.201/27
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe switch is acting as a router
- BThe frame has a wrong IP checksum and must be checked by every host
- CThe destination MAC address is not yet in its address table
- DThe frame is too large for one port
4.Computer networksHard
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
- AResponses can arrive in any order on one stream, so slow replies do not block others
- BThe connection stays encrypted without TLS
- CThe server can push updates over UDP
- DSeveral requests reuse one TCP connection, avoiding repeated handshakes
5.Number systemsMedium
Set the bits so they store −62 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −628 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 12 bits?
12 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.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
25 = 32
Subnet of this host
network: ⌊201/32⌋·32 = 192
broadcast: 192 + 32 − 1 = 223
usable hosts: 193 … 222
Answer
x = 192
3.Computer networks
Answer: C · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
C · The destination MAC address is not yet in its address table
4.Computer networks
Answer: D · 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
D · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems
Answer: U = 194
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
62 = 0011 1110
inverted: 1100 0001
+1 → 1100 0010
Read it as unsigned
1100 0010₂ = 256 − 62 = 194
Answer
U = 194
6.Number systems
Answer: N = 4095
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
212 − 1 = 4095
Answer
N = 4095
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