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 1888 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1888 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.45.135/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.45.135/30
x =
3.Computer networksMedium
Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?
- A192.168.0.0/22
- B192.168.0.0/25
- C192.168.0.0/23
- D192.168.1.0/23
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo encrypt the payload of each segment
- BTo compress the data more efficiently
- CTo select the best route to the destination for the segments of that connection
- DTo avoid confusion with old segments and make spoofed segments harder to inject
5.Number systemsMedium
The 8-bit pattern 0011 1101 stores a signed integer in two's complement. What is its value N in decimal?
0011 11018 bit
N =
6.Number systemsHard
A system must give every one of 2809 items its own binary code of the same length. What is the smallest number of bits b needed?
2809
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1888
211 − 2 = 2046 ≥ 1888 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 135
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: ⌊135/4⌋·4 = 132
broadcast: 132 + 4 − 1 = 135
usable hosts: 133 … 134
Answer
x = 135
3.Computer networks
Answer: C · 192.168.0.0/23
Why this is the answer
0 and 1 share all bits of the third octet except the last, so one bit less of prefix (/23) starting at 192.168.0.0 covers both exactly.
Answer
C · 192.168.0.0/23
4.Computer networks
Answer: D · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
D · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: N = 61
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.
32 + 16 + 8 + 4 + 1 = 61
Answer
N = 61
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 2809
212 = 4096 ≥ 2809
Answer
b = 12
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