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 255.255.240.0. How many usable host addresses H does it have?
255.255.240.0
H =
2.IPv4 subnettingHard
A host has the address 172.20.214.45/20. What is the third octet x of its subnet's broadcast address?
172.20.214.45/20
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/8
- B/32
- C/24
- D/16
4.Computer networksHard
What does TCP fast retransmit do?
- AResends a segment after three duplicate ACKs, without waiting for the timeout
- BDoubles the window after every loss
- CResends every segment twice to be safe
- DSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
5.Number systemsMedium
Set the bits so they store −37 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −378 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 4094
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
255.255.240.0 = /20
Usable hosts
212 − 2 = 4094
Answer
H = 4094
2.IPv4 subnetting
Answer: x = 223
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 = 223
3.Computer networks
Answer: C · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
C · /24
4.Computer networks
Answer: A · 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
A · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: U = 219
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
37 = 0010 0101
inverted: 1101 1010
+1 → 1101 1011
Read it as unsigned
1101 1011₂ = 256 − 37 = 219
Answer
U = 219
6.Number systems
Answer: M = 468.8 KiB
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.
Bits and bytes
800·600·8 = 3840000 bit = 480000 B
Convert to KiB
480000/1024 = 468.75 KiB
Answer
M = 468.75 KiB
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