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 1606 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1606 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.57.193/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.57.193/26
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- AEncrypting the frame
- BDetecting transmission errors
- CChoosing the best route
- DCorrecting every bit error automatically
4.Computer networksHard
DNS queries normally use UDP. When does DNS typically use TCP instead?
- AFor every query sent from a web browser
- BOnly when the domain name is longer than ten characters
- COnly for reverse lookups of IP addresses
- DFor responses too large for UDP and for zone transfers
5.Number systemsMedium
The 8-bit pattern 1000 0111 stores a signed integer in two's complement. What is its value N in decimal?
1000 01118 bit
N =
6.Number systemsHard
What is the most negative integer N that can be stored in 16-bit two's complement?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1606
211 − 2 = 2046 ≥ 1606 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 255
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
26 = 64
Subnet of this host
network: ⌊193/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254
Answer
x = 255
3.Computer networks
Answer: B · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
B · Detecting transmission errors
4.Computer networks
Answer: D · For responses too large for UDP and for zone transfers
Why this is the answer
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer
D · For responses too large for UDP and for zone transfers
5.Number systems
Answer: N = −121
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.
Add the place values
The other bits keep their normal positive values.
−128 + 4 + 2 + 1 = −121
Answer
N = −121
6.Number systems
Answer: N = −32768
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.
Most negative value
−215 = −32768
Answer
N = −32768
ExamProblems · examproblems.com/courses/computer-networks