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 192.168.116.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 7?
192.168.116.0/24/29subnet 7 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.52.19.0/24, 10.52.20.0/24, 10.52.21.0/24, 10.52.22.0/24, 10.52.23.0/24, 10.52.24.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.52.19.0/24 … 10.52.24.0/24
p =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A1 ms
- B8 ms
- C0.125 ms
- D12 ms
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ATLS certificates are sent over ICMP
- BTCP cannot open connections without ICMP
- CDNS needs ICMP to resolve names
- DPath MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systemsMedium
Set the bits so they store −59 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −598 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 32-bit two's complement?
32 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 55
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
23 = 8
Subnet number 7
network: (7 − 1)·8 = 48
broadcast: 48 + 8 − 1 = 55
Answer
x = 55
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
19 = 00010011
24 = 00011000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.52.16.0/20 (it also covers 10 network(s) that are not in the list).
3.Computer networks
Answer: A · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
A · 1 ms
4.Computer networks
Answer: D · 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
D · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 197
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
59 = 0011 1011
inverted: 1100 0100
+1 → 1100 0101
Read it as unsigned
1100 0101₂ = 256 − 59 = 197
Answer
U = 197
6.Number systems
Answer: N = −2.147 × 10⁹
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
−231 = −2147483648
Answer
N = −2147483648
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