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 192.168.111.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 3?
192.168.111.0/24/26subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.140.232/28. What is the fourth octet x of its network address?
10.20.140.232/28
x =
3.Computer networksMedium
What is the role of a default gateway on a host?
- AIt assigns MAC addresses
- BIt is the router used for traffic to other networks
- CIt is the switch port the host is connected to
- DIt is the DNS server
4.Computer networksHard
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
- A8 Mbit/s
- B255 Mbit/s
- C2 Mbit/s
- D1 Mbit/s
5.Number systemsMedium
Set the bits so they store −48 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −488 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: x = 191
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
26 = 64
Subnet number 3
network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191
Answer
x = 191
2.IPv4 subnetting
Answer: x = 224
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: ⌊232/16⌋·16 = 224
broadcast: 224 + 16 − 1 = 239
usable hosts: 225 … 238
Answer
x = 224
3.Computer networks
Answer: B · 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
B · It is the router used for traffic to other networks
4.Computer networks
Answer: A · 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
A · 8 Mbit/s
5.Number systems
Answer: U = 208
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
48 = 0011 0000
inverted: 1100 1111
+1 → 1101 0000
Read it as unsigned
1101 0000₂ = 256 − 48 = 208
Answer
U = 208
6.Number systems
Answer: M = 1536 KiB
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.
Bits and bytes
1024·768·16 = 12582912 bit = 1572864 B
Convert to KiB
1572864/1024 = 1536 KiB
Answer
M = 1536 KiB
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