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 192.168.67.0/24 is divided into /25 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 2?
192.168.67.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.176.0/24 is split with VLSM for these LANs: A: 13 hosts, B: 4 hosts, C: 5 hosts, D: 20 hosts, E: 15 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN A?
A: 13B: 4C: 5D: 20E: 15
x =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- ACollision detection only works on fibre cables, not on shared wireless radio channels
- BRadio signals never collide
- CA wireless station cannot reliably detect collisions while it is transmitting
- DWi-Fi only allows one device per network
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AFor responses too large for UDP and for zone transfers
- BFor every query sent from a web browser
- COnly when the domain name is longer than ten characters
- DOnly for reverse lookups of IP addresses
5.Number systemsMedium
The 8-bit pattern 1000 1111 stores a signed integer in two's complement. What is its value N in decimal?
1000 11118 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1000 AND 0110 0111. What is the result's value R in decimal?
1011 1000AND0110 0111
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 128
2.IPv4 subnetting
Answer: x = 64
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
D: 20 hosts → 25 = 32 (/27)
E: 15 hosts → 25 = 32 (/27)
A: 13 hosts → 24 = 16 (/28)
C: 5 hosts → 23 = 8 (/29)
B: 4 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .31
E: .32 – .63
A: .64 – .79
C: .80 – .87
B: .88 – .95
Answer
x = 64
LAN A: 192.168.176.64/28.
3.Computer networks
Answer: C · A wireless station cannot reliably detect collisions while it is transmitting
Why this is the answer
A radio transmitter cannot hear other weak signals while sending, and some stations are hidden from each other. So 802.11 waits random backoff times and uses acknowledgements.
Answer
C · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: A · For responses too large for UDP and for zone transfers
Why this is the answer
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer
A · For responses too large for UDP and for zone transfers
5.Number systems
Answer: N = −113
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.
Add the place values
The other bits keep their normal positive values.
−128 + 8 + 4 + 2 + 1 = −113
Answer
N = −113
6.Number systems
Answer: R = 32
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
AND gives 1 only where both bits are 1.
1011 1000
AND 0110 0111
= 0010 0000
Value of the result
0010 0000₂ = 32
Answer
R = 32
ExamProblems · examproblems.com/courses/computer-networks