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 10.196.68.0/22 is divided into equal subnets with prefix /24. How many subnets N are there?
10.196.68.0/22/24
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.197.84.0/24, 10.197.85.0/24, 10.197.86.0/24, 10.197.87.0/24, 10.197.88.0/24, 10.197.89.0/24, 10.197.90.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.197.84.0/24 … 10.197.90.0/24
p =
3.Computer networksMedium
Which metric does RIP use to choose routes?
- AMeasured delay
- BBusiness policy
- CHop count
- DLink bandwidth
4.Computer networksHard
What does TCP fast retransmit do?
- ADoubles the window after every loss
- BResends a segment after three duplicate ACKs, without waiting for the timeout
- 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 −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1128 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 4
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
24 − 22 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /24 subnet has 28 − 2 = 254 usable hosts.
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
84 = 01010100
90 = 01011010
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.197.80.0/20 (it also covers 9 network(s) that are not in the list).
3.Computer networks
Answer: C · Hop count
Why this is the answer
RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.
Answer
C · Hop count
4.Computer networks
Answer: B · 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
B · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: U = 144
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
112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000
Read it as unsigned
1001 0000₂ = 256 − 112 = 144
Answer
U = 144
6.Number systems
Answer: M = 3072 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
1024·768·32 = 25165824 bit = 3145728 B
Convert to KiB
3145728/1024 = 3072 KiB
Answer
M = 3072 KiB
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