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 6 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/246 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.216.100/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.216.100/28
x =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A10.2.2.3
- B10.1.200.5
- C10.0.2.3
- D11.1.2.3
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AThe size of each IP packet header
- BThe retransmission time-out, calculated from the measured round-trip time of each segment
- CHow much unacknowledged data the sender may send, based on the receiver's buffer
- DThe number of open ports
5.Number systemsMedium
Set the bits so they store −50 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −508 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 32-bit two's complement?
32 bit
N =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
23 = 8 ≥ 6 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 110
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: ⌊100/16⌋·16 = 96
broadcast: 96 + 16 − 1 = 111
usable hosts: 97 … 110
Answer
x = 110
3.Computer networks
Answer: B · 10.1.200.5
Why this is the answer
With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.
Answer
B · 10.1.200.5
4.Computer networks
Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer
Why this is the answer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer
C · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 206
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
50 = 0011 0010
inverted: 1100 1101
+1 → 1100 1110
Read it as unsigned
1100 1110₂ = 256 − 50 = 206
Answer
U = 206
6.Number systems
Answer: N = −2.147 × 10⁹
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
−231 = −2147483648
Answer
N = −2147483648
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