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 10.127.176.0/20 is divided into equal subnets with prefix /27. How many subnets N are there?
10.127.176.0/20/27
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.92.73.0/24, 10.92.74.0/24, 10.92.75.0/24, 10.92.76.0/24, 10.92.77.0/24, 10.92.78.0/24, 10.92.79.0/24, 10.92.80.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.92.73.0/24 … 10.92.80.0/24
p =
3.Computer networksMedium
Which subnet mask corresponds to /20?
- A255.255.255.240
- B255.255.248.0
- C255.255.240.0
- D255.255.224.0
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIts 128-bit address space is large enough to give every device a global address
- BIPv6 encrypts all addresses
- CIPv6 only works on local networks, so addresses never need to be shared globally
- DIPv6 does not use addresses
5.Number systemsMedium
The 8-bit pattern 1110 1100 stores a signed integer in two's complement. What is its value N in decimal?
1110 11008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 1110 << 1. What is the result's value R in decimal?
1100 1110<< 1
R =
3Solutions
1.IPv4 subnetting
Answer: N = 128
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).
Borrowed bits
27 − 20 = 7
Number of subnets
N = 27 = 128
Answer
N = 128
Each /27 subnet has 25 − 2 = 30 usable hosts.
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
73 = 01001001
80 = 01010000
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.92.64.0/19 (it also covers 24 network(s) that are not in the list).
3.Computer networks
Answer: C · 255.255.240.0
Why this is the answer
/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.
Answer
C · 255.255.240.0
4.Computer networks
Answer: A · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
A · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: N = −20
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 + 32 + 8 + 4 = −20
Answer
N = −20
6.Number systems
Answer: R = 156
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
Shifting left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 1110 << 1
= 1001 1100
Value of the result
1001 1100₂ = 156
Answer
R = 156
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