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
The network 10.237.51.0/24 is divided into equal subnets with prefix /29. How many subnets N are there?
10.237.51.0/24/29
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.140.95.0/24, 10.140.96.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.140.95.0/24 … 10.140.96.0/24
p =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo give each device a public IP address
- BTo encrypt all traffic between switches so other departments cannot read it
- CTo split one physical switched network into separate broadcast domains
- DTo increase the speed of every switch port
4.Computer networksHard
A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?
- AAbout 50 Mbit/s
- BAbout 0.5 Mbit/s
- CAbout 640 Mbit/s
- DAbout 5 Mbit/s
5.Number systemsMedium
The 8-bit pattern 1010 0001 stores a signed integer in two's complement. What is its value N in decimal?
1010 00018 bit
N =
6.Number systemsHard
A system must give every one of 4002 items its own binary code of the same length. What is the smallest number of bits b needed?
4002
b = bit
3Solutions
1.IPv4 subnetting
Answer: N = 32
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).
Borrowed bits
29 − 24 = 5
Number of subnets
N = 25 = 32
Answer
N = 32
Each /29 subnet has 23 − 2 = 6 usable hosts.
2.IPv4 subnetting
Answer: p = 18
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
95 = 01011111
96 = 01100000
common leading bits: 2
Prefix length
p = 16 + 2 = 18
Answer
p = /18
Summary route: 10.140.64.0/18 (it also covers 62 network(s) that are not in the list).
3.Computer networks
Answer: C · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
C · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: D · About 5 Mbit/s
Why this is the answer
Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.
Answer
D · About 5 Mbit/s
5.Number systems
Answer: N = −95
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 + 32 + 1 = −95
Answer
N = −95
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 4002
212 = 4096 ≥ 4002
Answer
b = 12
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