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 department needs 205 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
205 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.188.171/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.188.171/25
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe frame is too large for one port
- BThe switch is acting as a router
- CThe frame has a wrong IP checksum and must be checked by every host
- DThe destination MAC address is not yet in its address table
4.Computer networksHard
By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?
- A6 kbit/s
- B3 kbit/s
- C12 kbit/s
- D24 kbit/s
5.Number systemsMedium
Set the bits so they store −90 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −908 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0011 >> 2. What is the result's value R in decimal?
1001 0011>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: p = 24
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).
Host bits needed
27 − 2 = 126 < 205
28 − 2 = 254 ≥ 205 → h = 8
Prefix length
p = 32 − 8 = 24
Answer
p = /24
2.IPv4 subnetting
Answer: x = 255
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
27 = 128
Subnet of this host
network: ⌊171/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 255
3.Computer networks
Answer: D · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
D · The destination MAC address is not yet in its address table
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: U = 166
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
90 = 0101 1010
inverted: 1010 0101
+1 → 1010 0110
Read it as unsigned
1010 0110₂ = 256 − 90 = 166
Answer
U = 166
6.Number systems
Answer: R = 36
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 right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1001 0011 >> 2
= 0010 0100
Value of the result
0010 0100₂ = 36
Answer
R = 36
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