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 192.168.226.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
192.168.226.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.202.34.0/24, 10.202.35.0/24, 10.202.36.0/24, 10.202.37.0/24, 10.202.38.0/24, 10.202.39.0/24, 10.202.40.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.202.34.0/24 … 10.202.40.0/24
p =
3.Computer networksMedium
What uniquely identifies one end of a TCP connection on the internet?
- AA domain name alone
- BA subnet mask and gateway
- CA MAC address alone
- DAn IP address together with a port number
4.Computer networksHard
Which of these addresses is a public (not RFC 1918 private) IPv4 address?
- A192.168.255.1
- B172.32.0.1
- C172.31.0.1
- D10.255.0.1
5.Number systemsMedium
The 8-bit pattern 1100 0000 stores a signed integer in two's complement. What is its value N in decimal?
1100 00008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1110 1111 XOR 1011 1001. What is the result's value R in decimal?
1110 1111XOR1011 1001
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 128
2.IPv4 subnetting
Answer: p = 20
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
34 = 00100010
40 = 00101000
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.202.32.0/20 (it also covers 9 network(s) that are not in the list).
3.Computer networks
Answer: D · An IP address together with a port number
Why this is the answer
A TCP connection is identified by source IP and port plus destination IP and port. Ports separate the many applications on one host.
Answer
D · An IP address together with a port number
4.Computer networks
Answer: B · 172.32.0.1
Why this is the answer
172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.
Answer
B · 172.32.0.1
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.
−128 + 64 = −64
Answer
N = −64
6.Number systems
Answer: R = 86
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.
1110 1111
XOR 1011 1001
= 0101 0110
Value of the result
0101 0110₂ = 86
Answer
R = 86
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