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 192.168.179.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 8?
192.168.179.0/24/27subnet 8 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.70.226.0/24, 10.70.227.0/24, 10.70.228.0/24, 10.70.229.0/24, 10.70.230.0/24, 10.70.231.0/24, 10.70.232.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.70.226.0/24 … 10.70.232.0/24
p =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- ASYN
- BFIN
- CURG
- DPSH
4.Computer networksHard
What does TCP fast retransmit do?
- ADoubles the window after every loss
- BResends every segment twice to be safe
- CSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
- DResends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systemsMedium
Write −52 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −528 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0001 >> 1. What is the result's value R in decimal?
1001 0001>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: x = 224
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
25 = 32
Subnet number 8
network: (8 − 1)·32 = 224
broadcast: 224 + 32 − 1 = 255
Answer
x = 224
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
226 = 11100010
232 = 11101000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.70.224.0/20 (it also covers 9 network(s) that are not in the list).
3.Computer networks
Answer: B · FIN
Why this is the answer
Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.
Answer
B · FIN
4.Computer networks
Answer: D · 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
D · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: U = 204
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
52 = 0011 0100
inverted: 1100 1011
+1 → 1100 1100
Read it as unsigned
1100 1100₂ = 256 − 52 = 204
Answer
U = 204
6.Number systems
Answer: R = 72
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
Shifting right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1001 0001 >> 1
= 0100 1000
Value of the result
0100 1000₂ = 72
Answer
R = 72
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