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 255.255.255.192. How many usable host addresses H does it have?
255.255.255.192
H =
2.IPv4 subnettingHard
A host has the address 10.20.115.89/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.115.89/26
x =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- ARadio signals never collide
- BA wireless station cannot reliably detect collisions while it is transmitting
- CWi-Fi only allows one device per network
- DCollision detection only works on fibre cables, not on shared wireless radio channels
4.Computer networksHard
What does TCP fast retransmit do?
- ASkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
- BResends every segment twice to be safe
- CResends a segment after three duplicate ACKs, without waiting for the timeout
- DDoubles the window after every loss
5.Number systemsMedium
The 8-bit pattern 1001 1100 stores a signed integer in two's complement. What is its value N in decimal?
1001 11008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1010 XOR 0110 1011. What is the result's value R in decimal?
1001 1010XOR0110 1011
R =
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).
Prefix length
255.255.255.192 = /26
Usable hosts
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 127
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: ⌊89/64⌋·64 = 64
broadcast: 64 + 64 − 1 = 127
usable hosts: 65 … 126
Answer
x = 127
3.Computer networks
Answer: B · A wireless station cannot reliably detect collisions while it is transmitting
Why this is the answer
A radio transmitter cannot hear other weak signals while sending, and some stations are hidden from each other. So 802.11 waits random backoff times and uses acknowledgements.
Answer
B · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: C · Resends a segment after three duplicate ACKs, without waiting for the timeout
Why this is the answer
Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.
Answer
C · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: N = −100
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.
Add the place values
The other bits keep their normal positive values.
−128 + 16 + 8 + 4 = −100
Answer
N = −100
6.Number systems
Answer: R = 241
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
XOR gives 1 where the two bits differ.
1001 1010
XOR 0110 1011
= 1111 0001
Value of the result
1111 0001₂ = 241
Answer
R = 241
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