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 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2416 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.65.45.0/24, 10.65.46.0/24, 10.65.47.0/24, 10.65.48.0/24, 10.65.49.0/24, 10.65.50.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.65.45.0/24 … 10.65.50.0/24
p =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A1 ms
- B8 ms
- C12 ms
- D0.125 ms
4.Computer networksHard
Head-of-line blocking in HTTP/2 over TCP occurs because:
- AHTTP/2 uses UDP
- BServers cannot handle parallel requests
- CA lost TCP segment delays all multiplexed streams until it is retransmitted
- DEach request needs its own new TCP connection, so later requests wait in a queue
5.Number systemsMedium
The 8-bit pattern 1100 0111 stores a signed integer in two's complement. What is its value N in decimal?
1100 01118 bit
N =
6.Number systemsHard
What is the largest unsigned integer N that fits in 12 bits?
12 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 16 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: p = 19
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
45 = 00101101
50 = 00110010
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.65.32.0/19 (it also covers 26 network(s) that are not in the list).
3.Computer networks
Answer: A · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
A · 1 ms
4.Computer networks
Answer: C · 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
C · A lost TCP segment delays all multiplexed streams until it is retransmitted
5.Number systems
Answer: N = −57
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 4 + 2 + 1 = −57
Answer
N = −57
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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