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 department needs 93 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
93 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.38.215.0/24, 10.38.216.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.38.215.0/24 … 10.38.216.0/24
p =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- ACorrecting every bit error automatically
- BDetecting transmission errors
- CEncrypting the frame
- DChoosing the best route
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- APath MTU discovery fails because "fragmentation needed" messages are blocked
- BDNS needs ICMP to resolve names
- CTLS certificates are sent over ICMP
- DTCP cannot open connections without ICMP
5.Number systemsMedium
Write −73 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −738 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 1100 OR 0001 1100. What is the result's value R in decimal?
0100 1100OR0001 1100
R =
3Solutions
1.IPv4 subnetting
Answer: p = 25
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).
Host bits needed
26 − 2 = 62 < 93
27 − 2 = 126 ≥ 93 → h = 7
Prefix length
p = 32 − 7 = 25
Answer
p = /25
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
215 = 11010111
216 = 11011000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.38.208.0/20 (it also covers 14 network(s) that are not in the list).
3.Computer networks
Answer: B · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
B · Detecting transmission errors
4.Computer networks
Answer: A · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
A · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 183
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
73 = 0100 1001
inverted: 1011 0110
+1 → 1011 0111
Read it as unsigned
1011 0111₂ = 256 − 73 = 183
Answer
U = 183
6.Number systems
Answer: R = 92
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
OR gives 1 where at least one bit is 1.
0100 1100
OR 0001 1100
= 0101 1100
Value of the result
0101 1100₂ = 92
Answer
R = 92
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