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 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.204.130.0/24, 10.204.131.0/24, 10.204.132.0/24, 10.204.133.0/24, 10.204.134.0/24, 10.204.135.0/24, 10.204.136.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.204.130.0/24 … 10.204.136.0/24
p =
3.Computer networksMedium
What is the network address of the host 192.168.10.77/26?
- A192.168.10.128
- B192.168.10.0
- C192.168.10.64
- D192.168.10.76
4.Computer networksHard
In the TCP/IP model, which layer adds port numbers?
- ALink layer
- BTransport layer
- CApplication layer
- DInternet layer
5.Number systemsMedium
Set the bits so they store −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1128 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 1110 OR 1101 0101. What is the result's value R in decimal?
0100 1110OR1101 0101
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: p = 20
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
130 = 10000010
136 = 10001000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.204.128.0/20 (it also covers 9 network(s) that are not in the list).
3.Computer networks
Answer: C · 192.168.10.64
Why this is the answer
/26 subnets start at 0, 64, 128 and 192. 77 lies in the block 64–127, so the network address is 192.168.10.64.
Answer
C · 192.168.10.64
4.Computer networks
Answer: B · Transport layer
Why this is the answer
TCP and UDP headers contain source and destination ports; the internet layer (IP) adds addresses.
Answer
B · Transport layer
5.Number systems
Answer: U = 144
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
112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000
Read it as unsigned
1001 0000₂ = 256 − 112 = 144
Answer
U = 144
6.Number systems
Answer: R = 223
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
OR gives 1 where at least one bit is 1.
0100 1110
OR 1101 0101
= 1101 1111
Value of the result
1101 1111₂ = 223
Answer
R = 223
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