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 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2410 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.219.146.0/24, 10.219.147.0/24, 10.219.148.0/24, 10.219.149.0/24, 10.219.150.0/24, 10.219.151.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.219.146.0/24 … 10.219.151.0/24
p =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/16
- B/8
- C/24
- D/32
4.Computer networksHard
A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?
- AChange the PC's subnet mask to /32
- BConfigure port forwarding on the router to the PC's private address
- CGive the PC a different MAC address so the router can find it from outside
- DTurn off the PC's DNS client
5.Number systemsMedium
Set the bits so they store −77 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −778 bit
U =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 10 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: p = 21
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
146 = 10010010
151 = 10010111
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.219.144.0/21 (it also covers 2 network(s) that are not in the list).
3.Computer networks
Answer: C · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
C · /24
4.Computer networks
Answer: B · Configure port forwarding on the router to the PC's private address
Why this is the answer
NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.
Answer
B · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: U = 179
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
77 = 0100 1101
inverted: 1011 0010
+1 → 1011 0011
Read it as unsigned
1011 0011₂ = 256 − 77 = 179
Answer
U = 179
6.Number systems
Answer: M = 600 KiB
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.
Bits and bytes
640·480·16 = 4915200 bit = 614400 B
Convert to KiB
614400/1024 = 600 KiB
Answer
M = 600 KiB
ExamProblems · examproblems.com/courses/computer-networks