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 342 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
342 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.189.205/18. What is the third octet x of its subnet's broadcast address?
172.20.189.205/18
x =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- AFIN
- BPSH
- CURG
- DSYN
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 does not use addresses
- CIPv6 encrypts all addresses
- DIPv6 only works on local networks, so addresses never need to be shared globally
5.Number systemsMedium
Set the bits so they store −81 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −818 bit
U =
6.Number systemsHard
A system must give every one of 344 items its own binary code of the same length. What is the smallest number of bits b needed?
344
b = bit
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 < 342
29 − 2 = 510 ≥ 342 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 191
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).
Block size in the third octet
26 = 64
Subnet of this host
network: 172.20.128.0
broadcast: 172.20.191.255
Answer
x = 191
3.Computer networks
Answer: A · 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
A · FIN
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: U = 175
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
81 = 0101 0001
inverted: 1010 1110
+1 → 1010 1111
Read it as unsigned
1010 1111₂ = 256 − 81 = 175
Answer
U = 175
6.Number systems
Answer: b = 9 bit
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.
Powers of two
28 = 256 < 344
29 = 512 ≥ 344
Answer
b = 9
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