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 192.168.175.0/24 is divided into /28 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 9?
192.168.175.0/24/28subnet 9 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.216.240/20. What is the third octet x of its network address?
172.20.216.240/20
x =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A11.1.2.3
- B10.0.2.3
- C10.2.2.3
- D10.1.200.5
4.Computer networksHard
A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?
- AAbout 0.5 Mbit/s
- BAbout 640 Mbit/s
- CAbout 50 Mbit/s
- DAbout 5 Mbit/s
5.Number systemsMedium
Set the bits so they store −35 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −358 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 24 bits?
24 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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 number 9
network: (9 − 1)·16 = 128
broadcast: 128 + 16 − 1 = 143
Answer
x = 128
2.IPv4 subnetting
Answer: x = 208
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 in the third octet
24 = 16
Subnet of this host
network: 172.20.208.0
broadcast: 172.20.223.255
Answer
x = 208
3.Computer networks
Answer: D · 10.1.200.5
Why this is the answer
With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.
Answer
D · 10.1.200.5
4.Computer networks
Answer: D · About 5 Mbit/s
Why this is the answer
Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.
Answer
D · About 5 Mbit/s
5.Number systems
Answer: U = 221
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
35 = 0010 0011
inverted: 1101 1100
+1 → 1101 1101
Read it as unsigned
1101 1101₂ = 256 − 35 = 221
Answer
U = 221
6.Number systems
Answer: N = 1.678 × 10⁷
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
224 − 1 = 16777215
Answer
N = 16777215
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