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 192.168.186.0/24 is divided into /29 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 9?
192.168.186.0/24/29subnet 9 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.194.67/19. What is the third octet x of its network address?
172.20.194.67/19
x =
3.Computer networksMedium
How does TLS typically use asymmetric and symmetric cryptography?
- AOnly symmetric cryptography is used, with a fixed shared password
- BAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
- CSymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
- DOnly asymmetric cryptography is used throughout
4.Computer networksHard
By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?
- A24 kbit/s
- B6 kbit/s
- C12 kbit/s
- D3 kbit/s
5.Number systemsMedium
The 8-bit pattern 0101 1111 stores a signed integer in two's complement. What is its value N in decimal?
0101 11118 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 1010 OR 1011 1000. What is the result's value R in decimal?
1111 1010OR1011 1000
R =
3Solutions
1.IPv4 subnetting
Answer: x = 64
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
23 = 8
Subnet number 9
network: (9 − 1)·8 = 64
broadcast: 64 + 8 − 1 = 71
Answer
x = 64
2.IPv4 subnetting
Answer: x = 192
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 in the third octet
25 = 32
Subnet of this host
network: 172.20.192.0
broadcast: 172.20.223.255
Answer
x = 192
3.Computer networks
Answer: B · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
Why this is the answer
Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.
Answer
B · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
4.Computer networks
Answer: C · 12 kbit/s
Why this is the answer
2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.
Answer
C · 12 kbit/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.
64 + 16 + 8 + 4 + 2 + 1 = 95
Answer
N = 95
6.Number systems
Answer: R = 250
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
OR gives 1 where at least one bit is 1.
1111 1010
OR 1011 1000
= 1111 1010
Value of the result
1111 1010₂ = 250
Answer
R = 250
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