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 department needs 495 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
495 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.135.62/27. What is the fourth octet x of its network address?
10.20.135.62/27
x =
3.Computer networksMedium
What does ARP do?
- AResolves domain names
- BRoutes packets on the internet
- CFinds the MAC address that belongs to an IP address on the local network
- DAssigns IP addresses automatically to hosts when they join the local network
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
- BThe size of each IP packet header
- CThe number of open ports
- DHow much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systemsMedium
Write −13 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −138 bit
U =
6.Number systemsHard
A system must give every one of 1298 items its own binary code of the same length. What is the smallest number of bits b needed?
1298
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 23
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).
Host bits needed
28 − 2 = 254 < 495
29 − 2 = 510 ≥ 495 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 32
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
25 = 32
Subnet of this host
network: ⌊62/32⌋·32 = 32
broadcast: 32 + 32 − 1 = 63
usable hosts: 33 … 62
Answer
x = 32
3.Computer networks
Answer: C · Finds the MAC address that belongs to an IP address on the local network
Why this is the answer
ARP broadcasts "who has this IP?" and the owner replies with its MAC address, so frames can be delivered locally.
Answer
C · Finds the MAC address that belongs to an IP address on the local network
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 = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: b = 11 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
210 = 1024 < 1298
211 = 2048 ≥ 1298
Answer
b = 11
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