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 255.255.255.224. How many usable host addresses H does it have?
255.255.255.224
H =
2.IPv4 subnettingHard
A host has the address 10.20.131.196/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.131.196/26
x =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AOne attacker guesses a user's password
- BMany compromised machines flood a target with traffic
- CAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- DA virus deletes files on a single PC
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
- CIncreases the congestion window exponentially until a threshold or loss
- DKeeps the window fixed at one segment
5.Number systemsMedium
Write −95 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −958 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 20 bits?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
255.255.255.224 = /27
Usable hosts
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 255
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: ⌊196/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254
Answer
x = 255
3.Computer networks
Answer: B · Many compromised machines flood a target with traffic
Why this is the answer
In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.
Answer
B · Many compromised machines flood a target with traffic
4.Computer networks
Answer: C · 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
C · Increases the congestion window exponentially until a threshold or loss
5.Number systems
Answer: U = 161
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
95 = 0101 1111
inverted: 1010 0000
+1 → 1010 0001
Read it as unsigned
1010 0001₂ = 256 − 95 = 161
Answer
U = 161
6.Number systems
Answer: N = 1.049 × 10⁶
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
220 − 1 = 1048575
Answer
N = 1048575
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