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 220 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
220 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.44.125.0/24, 10.44.126.0/24, 10.44.127.0/24, 10.44.128.0/24, 10.44.129.0/24, 10.44.130.0/24, 10.44.131.0/24, 10.44.132.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.44.125.0/24 … 10.44.132.0/24
p =
3.Computer networksMedium
Which protocol does the "ping" command use?
- AHTTP
- BICMP
- CTCP
- DUDP
4.Computer networksHard
Why was the TCP window scale option introduced?
- AUDP needed a way to control its sending rate on congested paths
- BWithout it, TCP could not detect lost segments on long-distance links
- CThe 32-bit window field was too large and wasted header space in every TCP segment sent
- DThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systemsMedium
The 8-bit pattern 0100 0000 stores a signed integer in two's complement. What is its value N in decimal?
0100 00008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 1000 XOR 1011 0001. What is the result's value R in decimal?
1111 1000XOR1011 0001
R =
3Solutions
1.IPv4 subnetting
Answer: p = 24
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
27 − 2 = 126 < 220
28 − 2 = 254 ≥ 220 → h = 8
Prefix length
p = 32 − 8 = 24
Answer
p = /24
2.IPv4 subnetting
Answer: p = 16
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
125 = 01111101
132 = 10000100
common leading bits: 0
Prefix length
p = 16 + 0 = 16
Answer
p = /16
Summary route: 10.44.0.0/16 (it also covers 248 network(s) that are not in the list).
3.Computer networks
Answer: B · ICMP
Why this is the answer
Ping sends ICMP echo requests and measures the time until echo replies return.
Answer
B · ICMP
4.Computer networks
Answer: D · 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
D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: N = 64
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.
64 = 64
Answer
N = 64
6.Number systems
Answer: R = 73
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.
Bit by bit
XOR gives 1 where the two bits differ.
1111 1000
XOR 1011 0001
= 0100 1001
Value of the result
0100 1001₂ = 73
Answer
R = 73
ExamProblems · examproblems.com/courses/computer-networks