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 1741 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1741 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.35.12/22. What is the third octet x of its network address?
172.20.35.12/22
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BAAAA
- CCNAME
- DMX
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo encrypt the payload of each segment
- BTo avoid confusion with old segments and make spoofed segments harder to inject
- CTo select the best route to the destination for the segments of that connection
- DTo compress the data more efficiently
5.Number systemsMedium
Set the bits so they store −41 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −418 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 0101 << 1. What is the result's value R in decimal?
0110 0101<< 1
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 < 1741
211 − 2 = 2046 ≥ 1741 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 32
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
22 = 4
Subnet of this host
network: 172.20.32.0
broadcast: 172.20.35.255
Answer
x = 32
3.Computer networks
Answer: B · 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
B · AAAA
4.Computer networks
Answer: B · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
B · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 215
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
41 = 0010 1001
inverted: 1101 0110
+1 → 1101 0111
Read it as unsigned
1101 0111₂ = 256 − 41 = 215
Answer
U = 215
6.Number systems
Answer: R = 202
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 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0110 0101 << 1
= 1100 1010
Value of the result
1100 1010₂ = 202
Answer
R = 202
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