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
A host has the address 10.20.120.99/27. What is the fourth octet x of its network address?
10.20.120.99/27
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
The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?
- ADistance-vector; split horizon and poison reverse
- BLink-state; flooding link updates
- CStatic routing; adding default routes
- DPath-vector; longer AS paths
5.Number systemsMedium
Write −78 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −788 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
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 = 96
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: ⌊99/32⌋·32 = 96
broadcast: 96 + 32 − 1 = 127
usable hosts: 97 … 126
Answer
x = 96
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: A · Distance-vector; split horizon and poison reverse
Why this is the answer
After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.
Answer
A · Distance-vector; split horizon and poison reverse
5.Number systems
Answer: U = 178
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
78 = 0100 1110
inverted: 1011 0001
+1 → 1011 0010
Read it as unsigned
1011 0010₂ = 256 − 78 = 178
Answer
U = 178
6.Number systems
Answer: M = 1024 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
512·512·32 = 8388608 bit = 1048576 B
Convert to KiB
1048576/1024 = 1024 KiB
Answer
M = 1024 KiB
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