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
The network 10.34.16.0/20 is divided into equal subnets with prefix /25. How many subnets N are there?
10.34.16.0/20/25
N =
2.IPv4 subnettingHard
A host has the address 10.20.102.33/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.102.33/29
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BCSMA/CD
- CToken passing
- DTDMA
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe DNS cache of the client
- BThe switch's MAC address table
- CThe server's disk space
- DThe server's table of half-open connections
5.Number systemsMedium
Set the bits so they store −47 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −478 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 1101 >> 2. What is the result's value R in decimal?
0101 1101>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: N = 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).
Borrowed bits
25 − 20 = 5
Number of subnets
N = 25 = 32
Answer
N = 32
Each /25 subnet has 27 − 2 = 126 usable hosts.
2.IPv4 subnetting
Answer: x = 38
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
23 = 8
Subnet of this host
network: ⌊33/8⌋·8 = 32
broadcast: 32 + 8 − 1 = 39
usable hosts: 33 … 38
Answer
x = 38
3.Computer networks
Answer: B · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
B · CSMA/CD
4.Computer networks
Answer: D · 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
D · The server's table of half-open connections
5.Number systems
Answer: U = 209
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
47 = 0010 1111
inverted: 1101 0000
+1 → 1101 0001
Read it as unsigned
1101 0001₂ = 256 − 47 = 209
Answer
U = 209
6.Number systems
Answer: R = 23
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 right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
0101 1101 >> 2
= 0001 0111
Value of the result
0001 0111₂ = 23
Answer
R = 23
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