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 192.168.10.0/24 must be divided into at least 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.92.110/28. What is the fourth octet x of its network address?
10.20.92.110/28
x =
3.Computer networksMedium
Which of these is a private IPv4 address range?
- A200.1.1.0/24
- B100.0.0.0/8
- C8.8.8.0/24
- D192.168.0.0/16
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AIs used only by UDP
- BKeeps the window fixed at one segment
- CHalves the congestion window every round trip until a packet loss is detected
- DIncreases the congestion window exponentially until a threshold or loss
5.Number systemsMedium
Write −22 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −228 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 16 bits?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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).
Bits to borrow
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 96
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: ⌊110/16⌋·16 = 96
broadcast: 96 + 16 − 1 = 111
usable hosts: 97 … 110
Answer
x = 96
3.Computer networks
Answer: D · 192.168.0.0/16
Why this is the answer
Private ranges are 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16; they are not routed on the public internet.
Answer
D · 192.168.0.0/16
4.Computer networks
Answer: D · Increases the congestion window exponentially until a threshold or loss
Why this is the answer
Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.
Answer
D · Increases the congestion window exponentially until a threshold or loss
5.Number systems
Answer: U = 234
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.
Invert and add one
22 = 0001 0110
inverted: 1110 1001
+1 → 1110 1010
Read it as unsigned
1110 1010₂ = 256 − 22 = 234
Answer
U = 234
6.Number systems
Answer: N = 65535
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
216 − 1 = 65535
Answer
N = 65535
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