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 subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?
255.255.240.0
H =
2.IPv4 subnettingHard
The network 192.168.24.0/24 is split with VLSM for these LANs: A: 27 hosts, B: 5 hosts, C: 26 hosts, D: 11 hosts, E: 51 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 B?
A: 27B: 5C: 26D: 11E: 51
x =
3.Computer networksMedium
Why does Wi-Fi use collision avoidance (CSMA/CA) instead of collision detection?
- AA wireless station cannot reliably detect collisions while it is transmitting
- BWi-Fi only allows one device per network
- CRadio signals never collide
- DCollision detection only works on fibre cables, not on shared wireless radio channels
4.Computer networksHard
A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?
- AThe three-way handshake
- BNagle's algorithm
- CSlow start
- DFast retransmit
5.Number systemsMedium
Write −8 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −88 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 1101 >> 2. What is the result's value R in decimal?
1000 1101>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 4094
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).
Prefix length
255.255.240.0 = /20
Usable hosts
212 − 2 = 4094
Answer
H = 4094
2.IPv4 subnetting
Answer: x = 144
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
E: 51 hosts → 26 = 64 (/26)
A: 27 hosts → 25 = 32 (/27)
C: 26 hosts → 25 = 32 (/27)
D: 11 hosts → 24 = 16 (/28)
B: 5 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
A: .64 – .95
C: .96 – .127
D: .128 – .143
B: .144 – .151
Answer
x = 144
LAN B: 192.168.24.144/29.
3.Computer networks
Answer: A · 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
A · A wireless station cannot reliably detect collisions while it is transmitting
4.Computer networks
Answer: B · Nagle's algorithm
Why this is the answer
Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.
Answer
B · Nagle's algorithm
5.Number systems
Answer: U = 248
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
8 = 0000 1000
inverted: 1111 0111
+1 → 1111 1000
Read it as unsigned
1111 1000₂ = 256 − 8 = 248
Answer
U = 248
6.Number systems
Answer: R = 35
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
Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1000 1101 >> 2
= 0010 0011
Value of the result
0010 0011₂ = 35
Answer
R = 35
ExamProblems · examproblems.com/courses/computer-networks