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
A subnet uses the mask 255.255.254.0. How many usable host addresses H does it have?
255.255.254.0
H =
2.IPv4 subnettingHard
A host has the address 10.20.5.177/26. What is the fourth octet x of the last usable host address in its subnet?
10.20.5.177/26
x =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A10.0.2.3
- B10.1.200.5
- C11.1.2.3
- D10.2.2.3
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AIs used only by UDP
- BHalves the congestion window every round trip until a packet loss is detected
- CKeeps the window fixed at one segment
- DIncreases the congestion window exponentially until a threshold or loss
5.Number systemsMedium
Write −26 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −268 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 0001 AND 1010 0100. What is the result's value R in decimal?
1010 0001AND1010 0100
R =
3Solutions
1.IPv4 subnetting
Answer: H = 510
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).
Prefix length
255.255.254.0 = /23
Usable hosts
29 − 2 = 510
Answer
H = 510
2.IPv4 subnetting
Answer: x = 190
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 of this host
network: ⌊177/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190
Answer
x = 190
3.Computer networks
Answer: B · 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
B · 10.1.200.5
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 = 230
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
26 = 0001 1010
inverted: 1110 0101
+1 → 1110 0110
Read it as unsigned
1110 0110₂ = 256 − 26 = 230
Answer
U = 230
6.Number systems
Answer: R = 160
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
AND gives 1 only where both bits are 1.
1010 0001
AND 1010 0100
= 1010 0000
Value of the result
1010 0000₂ = 160
Answer
R = 160
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