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 1611 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1611 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.110.143.0/24, 10.110.144.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.110.143.0/24 … 10.110.144.0/24
p =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- APSH
- BFIN
- CSYN
- DURG
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A1 collision domain, 1 broadcast domain
- B8 collision domains, 8 broadcast domains
- C1 collision domain, 8 broadcast domains
- D8 collision domains, 1 broadcast domain
5.Number systemsMedium
Set the bits so they store −74 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −748 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 1000 >> 2. What is the result's value R in decimal?
0100 1000>> 2
R =
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 < 1611
211 − 2 = 2046 ≥ 1611 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
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
143 = 10001111
144 = 10010000
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.110.128.0/19 (it also covers 30 network(s) that are not in the list).
3.Computer networks
Answer: B · FIN
Why this is the answer
Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.
Answer
B · FIN
4.Computer networks
Answer: D · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
D · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 182
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
74 = 0100 1010
inverted: 1011 0101
+1 → 1011 0110
Read it as unsigned
1011 0110₂ = 256 − 74 = 182
Answer
U = 182
6.Number systems
Answer: R = 18
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 right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
0100 1000 >> 2
= 0001 0010
Value of the result
0001 0010₂ = 18
Answer
R = 18
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