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 router has routes to these networks: 10.10.35.0/24, 10.10.36.0/24, 10.10.37.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.10.35.0/24 … 10.10.37.0/24
p =
3.Computer networksMedium
A browser receives HTTP status code 404. What does it mean?
- AThe server had an internal error
- BThe request succeeded
- CThe client must log in first
- DThe requested resource was not found
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
Set the bits so they store −92 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −928 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 = 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: p = 21
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
35 = 00100011
37 = 00100101
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.10.32.0/21 (it also covers 5 network(s) that are not in the list).
3.Computer networks
Answer: D · The requested resource was not found
Why this is the answer
404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.
Answer
D · The requested resource was not found
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 = 164
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
92 = 0101 1100
inverted: 1010 0011
+1 → 1010 0100
Read it as unsigned
1010 0100₂ = 256 − 92 = 164
Answer
U = 164
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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