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 563 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
563 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.91.87/27. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.91.87/27
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AAAAA
- BCNAME
- CMX
- DA
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ATCP cannot open connections without ICMP
- BPath MTU discovery fails because "fragmentation needed" messages are blocked
- CTLS certificates are sent over ICMP
- DDNS needs ICMP to resolve names
5.Number systemsMedium
Set the bits so they store −7 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −78 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1000 << 2. What is the result's value R in decimal?
1001 1000<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: p = 22
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
29 − 2 = 510 < 563
210 − 2 = 1022 ≥ 563 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 95
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
25 = 32
Subnet of this host
network: ⌊87/32⌋·32 = 64
broadcast: 64 + 32 − 1 = 95
usable hosts: 65 … 94
Answer
x = 95
3.Computer networks
Answer: A · AAAA
Why this is the answer
An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.
Answer
A · AAAA
4.Computer networks
Answer: B · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
B · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 249
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
7 = 0000 0111
inverted: 1111 1000
+1 → 1111 1001
Read it as unsigned
1111 1001₂ = 256 − 7 = 249
Answer
U = 249
6.Number systems
Answer: R = 96
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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1001 1000 << 2
= 0110 0000
Value of the result
0110 0000₂ = 96
Answer
R = 96
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