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 192.168.187.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 7?
192.168.187.0/24/27subnet 7 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.49.175.0/24, 10.49.176.0/24, 10.49.177.0/24, 10.49.178.0/24, 10.49.179.0/24, 10.49.180.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.49.175.0/24 … 10.49.180.0/24
p =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A4
- B1
- C2
- D0
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo avoid confusion with old segments and make spoofed segments harder to inject
- BTo encrypt the payload of each segment
- CTo select the best route to the destination for the segments of that connection
- DTo compress the data more efficiently
5.Number systemsMedium
Write −17 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −178 bit
U =
6.Number systemsHard
A system must give every one of 836 items its own binary code of the same length. What is the smallest number of bits b needed?
836
b = bit
3Solutions
1.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 number 7
network: (7 − 1)·32 = 192
broadcast: 192 + 32 − 1 = 223
Answer
x = 192
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
175 = 10101111
180 = 10110100
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.49.160.0/19 (it also covers 26 network(s) that are not in the list).
3.Computer networks
Answer: D · 0
Why this is the answer
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer
D · 0
4.Computer networks
Answer: A · 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
A · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 239
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.
Invert and add one
17 = 0001 0001
inverted: 1110 1110
+1 → 1110 1111
Read it as unsigned
1110 1111₂ = 256 − 17 = 239
Answer
U = 239
6.Number systems
Answer: b = 10 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
29 = 512 < 836
210 = 1024 ≥ 836
Answer
b = 10
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