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 subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?
255.255.255.248
H =
2.IPv4 subnettingHard
A host has the address 10.20.39.225/29. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.39.225/29
x =
3.Computer networksMedium
How does TLS typically use asymmetric and symmetric cryptography?
- ASymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
- BAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
- COnly asymmetric cryptography is used throughout
- DOnly symmetric cryptography is used, with a fixed shared password
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIts 128-bit address space is large enough to give every device a global address
- BIPv6 only works on local networks, so addresses never need to be shared globally
- CIPv6 encrypts all addresses
- DIPv6 does not use addresses
5.Number systemsMedium
Set the bits so they store −71 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −718 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 0101 << 1. What is the result's value R in decimal?
1101 0101<< 1
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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).
Prefix length
255.255.255.248 = /29
Usable hosts
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 231
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 of this host
network: ⌊225/8⌋·8 = 224
broadcast: 224 + 8 − 1 = 231
usable hosts: 225 … 230
Answer
x = 231
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: A · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
A · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 185
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.
Invert and add one
71 = 0100 0111
inverted: 1011 1000
+1 → 1011 1001
Read it as unsigned
1011 1001₂ = 256 − 71 = 185
Answer
U = 185
6.Number systems
Answer: R = 170
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
Shifting left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1101 0101 << 1
= 1010 1010
Value of the result
1010 1010₂ = 170
Answer
R = 170
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