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 1172 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1172 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.250.10.0/24, 10.250.11.0/24, 10.250.12.0/24, 10.250.13.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.250.10.0/24 … 10.250.13.0/24
p =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A12 ms
- B8 ms
- C1 ms
- D0.125 ms
4.Computer networksHard
Why was the TCP window scale option introduced?
- AThe 32-bit window field was too large and wasted header space in every TCP segment sent
- BUDP needed a way to control its sending rate on congested paths
- CThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- DWithout it, TCP could not detect lost segments on long-distance links
5.Number systemsMedium
The 8-bit pattern 1110 0111 stores a signed integer in two's complement. What is its value N in decimal?
1110 01118 bit
N =
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: 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 < 1172
211 − 2 = 2046 ≥ 1172 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
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
10 = 00001010
13 = 00001101
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.250.8.0/21 (it also covers 4 network(s) that are not in the list).
3.Computer networks
Answer: C · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
C · 1 ms
4.Computer networks
Answer: C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
Why this is the answer
The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.
Answer
C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: N = −25
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 + 64 + 32 + 4 + 2 + 1 = −25
Answer
N = −25
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
ExamProblems · examproblems.com/courses/computer-networks