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 1069 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1069 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.42.143/25. What is the fourth octet x of the last usable host address in its subnet?
10.20.42.143/25
x =
3.Computer networksMedium
What is the role of a default gateway on a host?
- AIt assigns MAC addresses
- BIt is the DNS server
- CIt is the router used for traffic to other networks
- DIt is the switch port the host is connected to
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's table of half-open connections
- BThe DNS cache of the client
- CThe switch's MAC address table
- DThe server's disk space
5.Number systemsMedium
Write −91 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −918 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 0101 OR 1011 1001. What is the result's value R in decimal?
1011 0101OR1011 1001
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 < 1069
211 − 2 = 2046 ≥ 1069 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 254
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
27 = 128
Subnet of this host
network: ⌊143/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 254
3.Computer networks
Answer: C · It is the router used for traffic to other networks
Why this is the answer
The host sends packets for non-local destinations to the default gateway, usually its local router.
Answer
C · It is the router used for traffic to other networks
4.Computer networks
Answer: A · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
A · The server's table of half-open connections
5.Number systems
Answer: U = 165
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
91 = 0101 1011
inverted: 1010 0100
+1 → 1010 0101
Read it as unsigned
1010 0101₂ = 256 − 91 = 165
Answer
U = 165
6.Number systems
Answer: R = 189
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
OR gives 1 where at least one bit is 1.
1011 0101
OR 1011 1001
= 1011 1101
Value of the result
1011 1101₂ = 189
Answer
R = 189
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