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 255.255.255.128. How many usable host addresses H does it have?
255.255.255.128
H =
2.IPv4 subnettingHard
A host has the address 10.20.91.14/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.91.14/30
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BCSMA/CD
- CToken passing
- DTDMA
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AThe size of each IP packet header
- BThe retransmission time-out, calculated from the measured round-trip time of each segment
- CThe number of open ports
- DHow much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systemsMedium
Set the bits so they store −6 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −68 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 126
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
255.255.255.128 = /25
Usable hosts
27 − 2 = 126
Answer
H = 126
2.IPv4 subnetting
Answer: x = 14
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
22 = 4
Subnet of this host
network: ⌊14/4⌋·4 = 12
broadcast: 12 + 4 − 1 = 15
usable hosts: 13 … 14
Answer
x = 14
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: D · How much unacknowledged data the sender may send, based on the receiver's buffer
Why this is the answer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer
D · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 250
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
6 = 0000 0110
inverted: 1111 1001
+1 → 1111 1010
Read it as unsigned
1111 1010₂ = 256 − 6 = 250
Answer
U = 250
6.Number systems
Answer: M = 1875 KiB
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.
Bits and bytes
800·600·32 = 15360000 bit = 1920000 B
Convert to KiB
1920000/1024 = 1875 KiB
Answer
M = 1875 KiB
ExamProblems · examproblems.com/courses/computer-networks