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 10.20.174.201/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.174.201/28
x =
3.Computer networksMedium
What is the role of a default gateway on a host?
- AIt assigns MAC addresses
- BIt is the router used for traffic to other networks
- CIt is the DNS server
- DIt is the switch port the host is connected to
4.Computer networksHard
A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?
- ANagle's algorithm
- BSlow start
- CThe three-way handshake
- DFast retransmit
5.Number systemsMedium
Set the bits so they store −119 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1198 bit
U =
6.Number systemsHard
A system must give every one of 3875 items its own binary code of the same length. What is the smallest number of bits b needed?
3875
b = bit
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 = 207
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
24 = 16
Subnet of this host
network: ⌊201/16⌋·16 = 192
broadcast: 192 + 16 − 1 = 207
usable hosts: 193 … 206
Answer
x = 207
3.Computer networks
Answer: B · It is the router used for traffic to other networks
Why this is the answer
The host sends packets for non-local destinations to the default gateway, usually its local router.
Answer
B · It is the router used for traffic to other networks
4.Computer networks
Answer: A · Nagle's algorithm
Why this is the answer
Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.
Answer
A · Nagle's algorithm
5.Number systems
Answer: U = 137
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
119 = 0111 0111
inverted: 1000 1000
+1 → 1000 1001
Read it as unsigned
1000 1001₂ = 256 − 119 = 137
Answer
U = 137
6.Number systems
Answer: b = 12 bit
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Powers of two
211 = 2048 < 3875
212 = 4096 ≥ 3875
Answer
b = 12
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