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 172.31.0.0/16 is divided into equal subnets with prefix /22. How many subnets N are there?
172.31.0.0/16/22
N =
2.IPv4 subnettingHard
A host has the address 10.20.225.56/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.225.56/28
x =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A0.125 ms
- B12 ms
- C1 ms
- D8 ms
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe server's disk space
- BThe DNS cache of the client
- CThe switch's MAC address table
- DThe server's table of half-open connections
5.Number systemsMedium
The 8-bit pattern 1100 0100 stores a signed integer in two's complement. What is its value N in decimal?
1100 01008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0111 1100 << 3. What is the result's value R in decimal?
0111 1100<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: N = 64
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
22 − 16 = 6
Number of subnets
N = 26 = 64
Answer
N = 64
Each /22 subnet has 210 − 2 = 1022 usable hosts.
2.IPv4 subnetting
Answer: x = 63
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
24 = 16
Subnet of this host
network: ⌊56/16⌋·16 = 48
broadcast: 48 + 16 − 1 = 63
usable hosts: 49 … 62
Answer
x = 63
3.Computer networks
Answer: C · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
C · 1 ms
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: N = −60
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 4 = −60
Answer
N = −60
6.Number systems
Answer: R = 224
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 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0111 1100 << 3
= 1110 0000
Value of the result
1110 0000₂ = 224
Answer
R = 224
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