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 489 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
489 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.140.206.0/24, 10.140.207.0/24, 10.140.208.0/24, 10.140.209.0/24, 10.140.210.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.140.206.0/24 … 10.140.210.0/24
p =
3.Computer networksMedium
A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?
- A10.1.0.0/16, because it is the longest matching prefix
- B10.0.0.0/8, because it covers more addresses than the /16
- CBoth routes, splitting the traffic equally
- D10.0.0.0/8, because it was added first
4.Computer networksHard
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
- A1 Mbit/s
- B2 Mbit/s
- C255 Mbit/s
- D8 Mbit/s
5.Number systemsMedium
Write −7 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −78 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 1110 XOR 1110 0111. What is the result's value R in decimal?
1010 1110XOR1110 0111
R =
3Solutions
1.IPv4 subnetting
Answer: p = 23
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
28 − 2 = 254 < 489
29 − 2 = 510 ≥ 489 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
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
206 = 11001110
210 = 11010010
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.140.192.0/19 (it also covers 27 network(s) that are not in the list).
3.Computer networks
Answer: A · 10.1.0.0/16, because it is the longest matching prefix
Why this is the answer
IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.
Answer
A · 10.1.0.0/16, because it is the longest matching prefix
4.Computer networks
Answer: D · 8 Mbit/s
Why this is the answer
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer
D · 8 Mbit/s
5.Number systems
Answer: U = 249
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
7 = 0000 0111
inverted: 1111 1000
+1 → 1111 1001
Read it as unsigned
1111 1001₂ = 256 − 7 = 249
Answer
U = 249
6.Number systems
Answer: R = 73
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.
Bit by bit
XOR gives 1 where the two bits differ.
1010 1110
XOR 1110 0111
= 0100 1001
Value of the result
0100 1001₂ = 73
Answer
R = 73
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