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 192.168.137.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 8?
192.168.137.0/24/29subnet 8 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.82.0/24 is split with VLSM for these LANs: A: 13 hosts, B: 24 hosts, C: 17 hosts, D: 6 hosts, E: 5 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 13B: 24C: 17D: 6E: 5
x =
3.Computer networksMedium
Which protocol does the "ping" command use?
- ATCP
- BICMP
- CUDP
- DHTTP
4.Computer networksHard
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
- A255 Mbit/s
- B8 Mbit/s
- C2 Mbit/s
- D1 Mbit/s
5.Number systemsMedium
Set the bits so they store −45 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −458 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 10 bits?
10 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 56
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 number 8
network: (8 − 1)·8 = 56
broadcast: 56 + 8 − 1 = 63
Answer
x = 56
2.IPv4 subnetting
Answer: x = 80
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
B: 24 hosts → 25 = 32 (/27)
C: 17 hosts → 25 = 32 (/27)
A: 13 hosts → 24 = 16 (/28)
D: 6 hosts → 23 = 8 (/29)
E: 5 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .31
C: .32 – .63
A: .64 – .79
D: .80 – .87
E: .88 – .95
Answer
x = 80
LAN D: 192.168.82.80/29.
3.Computer networks
Answer: B · ICMP
Why this is the answer
Ping sends ICMP echo requests and measures the time until echo replies return.
Answer
B · ICMP
4.Computer networks
Answer: B · 8 Mbit/s
Why this is the answer
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer
B · 8 Mbit/s
5.Number systems
Answer: U = 211
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
45 = 0010 1101
inverted: 1101 0010
+1 → 1101 0011
Read it as unsigned
1101 0011₂ = 256 − 45 = 211
Answer
U = 211
6.Number systems
Answer: N = 1023
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.
Largest value
210 − 1 = 1023
Answer
N = 1023
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