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 subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?
255.255.255.248
H =
2.IPv4 subnettingHard
A host has the address 10.20.5.113/27. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.5.113/27
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- Aff02::1
- B::
- Cfe80::1
- D::1
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo compress the data more efficiently
- BTo avoid confusion with old segments and make spoofed segments harder to inject
- CTo encrypt the payload of each segment
- DTo select the best route to the destination for the segments of that connection
5.Number systemsMedium
Write −23 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −238 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 0100 AND 1101 1110. What is the result's value R in decimal?
0100 0100AND1101 1110
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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).
Prefix length
255.255.255.248 = /29
Usable hosts
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 127
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: ⌊113/32⌋·32 = 96
broadcast: 96 + 32 − 1 = 127
usable hosts: 97 … 126
Answer
x = 127
3.Computer networks
Answer: D · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
D · ::1
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 = 233
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.
Invert and add one
23 = 0001 0111
inverted: 1110 1000
+1 → 1110 1001
Read it as unsigned
1110 1001₂ = 256 − 23 = 233
Answer
U = 233
6.Number systems
Answer: R = 68
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
AND gives 1 only where both bits are 1.
0100 0100
AND 1101 1110
= 0100 0100
Value of the result
0100 0100₂ = 68
Answer
R = 68
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