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 192.168.10.0/24 must be divided into at least 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2430 subnets
H =
2.IPv4 subnettingHard
A host has the address 172.20.196.177/20. What is the third octet x of its network address?
172.20.196.177/20
x =
3.Computer networksMedium
How does TLS typically use asymmetric and symmetric cryptography?
- AOnly symmetric cryptography is used, with a fixed shared password
- BOnly asymmetric cryptography is used throughout
- CAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
- DSymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
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 0.5 Mbit/s
- BAbout 50 Mbit/s
- CAbout 640 Mbit/s
- DAbout 5 Mbit/s
5.Number systemsMedium
Set the bits so they store −64 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −648 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 1010 OR 0111 0001. What is the result's value R in decimal?
1111 1010OR0111 0001
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).
Bits to borrow
25 = 32 ≥ 30 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
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
24 = 16
Subnet of this host
network: 172.20.192.0
broadcast: 172.20.207.255
Answer
x = 192
3.Computer networks
Answer: C · 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
C · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
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: U = 192
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
64 = 0100 0000
inverted: 1011 1111
+1 → 1100 0000
Read it as unsigned
1100 0000₂ = 256 − 64 = 192
Answer
U = 192
6.Number systems
Answer: R = 251
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 0111 0001
= 1111 1011
Value of the result
1111 1011₂ = 251
Answer
R = 251
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