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 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.217.25/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.217.25/30
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AOnly faster loading
- BConfidentiality, integrity and server authentication
- CProtection against all malware and phishing on the visited site
- DAnonymity of the user
4.Computer networksHard
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
- A2 Mbit/s
- B1 Mbit/s
- C8 Mbit/s
- D255 Mbit/s
5.Number systemsMedium
Set the bits so they store −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −138 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 1101 << 3. What is the result's value R in decimal?
0101 1101<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 26
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
22 = 4
Subnet of this host
network: ⌊25/4⌋·4 = 24
broadcast: 24 + 4 − 1 = 27
usable hosts: 25 … 26
Answer
x = 26
3.Computer networks
Answer: B · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
B · Confidentiality, integrity and server authentication
4.Computer networks
Answer: C · 8 Mbit/s
Why this is the answer
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer
C · 8 Mbit/s
5.Number systems
Answer: U = 243
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
13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011
Read it as unsigned
1111 0011₂ = 256 − 13 = 243
Answer
U = 243
6.Number systems
Answer: R = 232
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 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0101 1101 << 3
= 1110 1000
Value of the result
1110 1000₂ = 232
Answer
R = 232
ExamProblems · examproblems.com/courses/computer-networks