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 255.255.248.0. How many usable host addresses H does it have?
255.255.248.0
H =
2.IPv4 subnettingHard
A host has the address 10.20.233.128/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.233.128/29
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CA
- BToken passing
- CTDMA
- DCSMA/CD
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A8 collision domains, 1 broadcast domain
- B8 collision domains, 8 broadcast domains
- C1 collision domain, 8 broadcast domains
- D1 collision domain, 1 broadcast domain
5.Number systemsMedium
Set the bits so they store −28 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −288 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0110 << 3. What is the result's value R in decimal?
1100 0110<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 2046
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
255.255.248.0 = /21
Usable hosts
211 − 2 = 2046
Answer
H = 2046
2.IPv4 subnetting
Answer: x = 134
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: ⌊128/8⌋·8 = 128
broadcast: 128 + 8 − 1 = 135
usable hosts: 129 … 134
Answer
x = 134
3.Computer networks
Answer: D · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
D · CSMA/CD
4.Computer networks
Answer: A · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
A · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 228
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
28 = 0001 1100
inverted: 1110 0011
+1 → 1110 0100
Read it as unsigned
1110 0100₂ = 256 − 28 = 228
Answer
U = 228
6.Number systems
Answer: R = 48
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.
1100 0110 << 3
= 0011 0000
Value of the result
0011 0000₂ = 48
Answer
R = 48
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