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 192.168.10.0/24 must be divided into at least 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2410 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.92.189/25. What is the fourth octet x of its network address?
10.20.92.189/25
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ATDMA
- BCSMA/CA
- CCSMA/CD
- DToken passing
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AThe retransmission time-out, calculated from the measured round-trip time of each segment
- BHow much unacknowledged data the sender may send, based on the receiver's buffer
- CThe size of each IP packet header
- DThe number of open ports
5.Number systemsMedium
Write −36 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −368 bit
U =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 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).
Bits to borrow
24 = 16 ≥ 10 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.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 of this host
network: ⌊189/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 128
3.Computer networks
Answer: C · 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
C · CSMA/CD
4.Computer networks
Answer: B · 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
B · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 220
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
36 = 0010 0100
inverted: 1101 1011
+1 → 1101 1100
Read it as unsigned
1101 1100₂ = 256 − 36 = 220
Answer
U = 220
6.Number systems
Answer: M = 192 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
256·256·24 = 1572864 bit = 196608 B
Convert to KiB
196608/1024 = 192 KiB
Answer
M = 192 KiB
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