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 172.25.0.0/16 is divided into equal subnets with prefix /19. How many subnets N are there?
172.25.0.0/16/19
N =
2.IPv4 subnettingHard
A host has the address 172.20.36.122/21. What is the third octet x of its network address?
172.20.36.122/21
x =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A10.2.2.3
- B10.0.2.3
- C10.1.200.5
- D11.1.2.3
4.Computer networksHard
What does TCP fast retransmit do?
- AResends a segment after three duplicate ACKs, without waiting for the timeout
- BResends every segment twice to be safe
- CSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
- DDoubles the window after every loss
5.Number systemsMedium
Set the bits so they store −8 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −88 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 16-bit two's complement?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: N = 8
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).
Borrowed bits
19 − 16 = 3
Number of subnets
N = 23 = 8
Answer
N = 8
Each /19 subnet has 213 − 2 = 8190 usable hosts.
2.IPv4 subnetting
Answer: x = 32
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
23 = 8
Subnet of this host
network: 172.20.32.0
broadcast: 172.20.39.255
Answer
x = 32
3.Computer networks
Answer: C · 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
C · 10.1.200.5
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 = 248
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
8 = 0000 1000
inverted: 1111 0111
+1 → 1111 1000
Read it as unsigned
1111 1000₂ = 256 − 8 = 248
Answer
U = 248
6.Number systems
Answer: N = −32768
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.
Most negative value
−215 = −32768
Answer
N = −32768
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