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 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2420 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.37.182/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.37.182/30
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A100 ms
- B10 ms
- C0.1 ms
- D1 ms
4.Computer networksHard
A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?
- A500 kB
- B4 kB
- C50 kB
- D2.5 MB
5.Number systemsMedium
Set the bits so they store −94 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −948 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
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 ≥ 20 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 182
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: ⌊182/4⌋·4 = 180
broadcast: 180 + 4 − 1 = 183
usable hosts: 181 … 182
Answer
x = 182
3.Computer networks
Answer: B · 10 ms
Why this is the answer
2 × 10⁶ m / 2 × 10⁸ m/s = 0.01 s = 10 ms. Propagation delay depends on distance and medium, not packet size.
Answer
B · 10 ms
4.Computer networks
Answer: A · 500 kB
Why this is the answer
The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.
Answer
A · 500 kB
5.Number systems
Answer: U = 162
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
94 = 0101 1110
inverted: 1010 0001
+1 → 1010 0010
Read it as unsigned
1010 0010₂ = 256 − 94 = 162
Answer
U = 162
6.Number systems
Answer: M = 512 KiB
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.
Bits and bytes
512·512·16 = 4194304 bit = 524288 B
Convert to KiB
524288/1024 = 512 KiB
Answer
M = 512 KiB
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