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 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2430 subnets
H =
2.IPv4 subnettingHard
A host has the address 172.20.72.22/17. What is the third octet x of its subnet's broadcast address?
172.20.72.22/17
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- ADetecting transmission errors
- BCorrecting every bit error automatically
- CEncrypting the frame
- DChoosing the best route
4.Computer networksHard
Head-of-line blocking in HTTP/2 over TCP occurs because:
- AA lost TCP segment delays all multiplexed streams until it is retransmitted
- BServers cannot handle parallel requests
- CHTTP/2 uses UDP
- DEach request needs its own new TCP connection, so later requests wait in a queue
5.Number systemsMedium
Set the bits so they store −97 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −978 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 24 bits?
24 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 30 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
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 in the third octet
27 = 128
Subnet of this host
network: 172.20.0.0
broadcast: 172.20.127.255
Answer
x = 127
3.Computer networks
Answer: A · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
A · Detecting transmission errors
4.Computer networks
Answer: A · A lost TCP segment delays all multiplexed streams until it is retransmitted
Why this is the answer
TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.
Answer
A · A lost TCP segment delays all multiplexed streams until it is retransmitted
5.Number systems
Answer: U = 159
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
97 = 0110 0001
inverted: 1001 1110
+1 → 1001 1111
Read it as unsigned
1001 1111₂ = 256 − 97 = 159
Answer
U = 159
6.Number systems
Answer: N = 1.678 × 10⁷
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
224 − 1 = 16777215
Answer
N = 16777215
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