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 192.168.6.0/24 is divided into /28 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 9?
192.168.6.0/24/28subnet 9 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.38.4.0/24, 10.38.5.0/24, 10.38.6.0/24, 10.38.7.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.38.4.0/24 … 10.38.7.0/24
p =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- AToken passing
- BCSMA/CD
- CCSMA/CA
- DTDMA
4.Computer networksHard
A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?
- AA duplicate MAC address on the internet
- BThe TCP window is too small
- CA missing or wrong default gateway
- DA faulty DNS server setting
5.Number systemsMedium
Set the bits so they store −118 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1188 bit
U =
6.Number systemsHard
A system must give every one of 2243 items its own binary code of the same length. What is the smallest number of bits b needed?
2243
b = bit
3Solutions
1.IPv4 subnetting
Answer: x = 143
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
24 = 16
Subnet number 9
network: (9 − 1)·16 = 128
broadcast: 128 + 16 − 1 = 143
Answer
x = 143
2.IPv4 subnetting
Answer: p = 22
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
4 = 00000100
7 = 00000111
common leading bits: 6
Prefix length
p = 16 + 6 = 22
Answer
p = /22
Summary route: 10.38.4.0/22
3.Computer networks
Answer: B · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
B · CSMA/CD
4.Computer networks
Answer: C · A missing or wrong default gateway
Why this is the answer
Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.
Answer
C · A missing or wrong default gateway
5.Number systems
Answer: U = 138
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
118 = 0111 0110
inverted: 1000 1001
+1 → 1000 1010
Read it as unsigned
1000 1010₂ = 256 − 118 = 138
Answer
U = 138
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 2243
212 = 4096 ≥ 2243
Answer
b = 12
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