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 172.20.129.42/19. What is the third octet x of its subnet's broadcast address?
172.20.129.42/19
x =
3.Computer networksMedium
What does an MX record in DNS specify?
- AThe mail servers that accept email for a domain
- BAn alias name for another host
- CThe name servers for a zone
- DThe IPv4 address of a web server
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AIncreases the congestion window exponentially until a threshold or loss
- BHalves the congestion window every round trip until a packet loss is detected
- CIs used only by UDP
- DKeeps the window fixed at one segment
5.Number systemsMedium
Set the bits so they store −44 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −448 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 16-bit two's complement?
16 bit
N =
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 = 159
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
25 = 32
Subnet of this host
network: 172.20.128.0
broadcast: 172.20.159.255
Answer
x = 159
3.Computer networks
Answer: A · The mail servers that accept email for a domain
Why this is the answer
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer
A · The mail servers that accept email for a domain
4.Computer networks
Answer: A · Increases the congestion window exponentially until a threshold or loss
Why this is the answer
Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.
Answer
A · Increases the congestion window exponentially until a threshold or loss
5.Number systems
Answer: U = 212
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
44 = 0010 1100
inverted: 1101 0011
+1 → 1101 0100
Read it as unsigned
1101 0100₂ = 256 − 44 = 212
Answer
U = 212
6.Number systems
Answer: N = −32768
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.
Most negative value
−215 = −32768
Answer
N = −32768
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