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 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2412 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.58.118.0/24, 10.58.119.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.58.118.0/24 … 10.58.119.0/24
p =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- AToken passing
- BCSMA/CD
- CTDMA
- DCSMA/CA
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AHow much unacknowledged data the sender may send, based on the receiver's buffer
- BThe retransmission time-out, calculated from the measured round-trip time of each segment
- CThe size of each IP packet header
- DThe number of open ports
5.Number systemsMedium
Set the bits so they store −108 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1088 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 1001 XOR 1101 1101. What is the result's value R in decimal?
1010 1001XOR1101 1101
R =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 12 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: p = 23
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
118 = 01110110
119 = 01110111
common leading bits: 7
Prefix length
p = 16 + 7 = 23
Answer
p = /23
Summary route: 10.58.118.0/23
3.Computer networks
Answer: B · 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
B · CSMA/CD
4.Computer networks
Answer: A · 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
A · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 148
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
108 = 0110 1100
inverted: 1001 0011
+1 → 1001 0100
Read it as unsigned
1001 0100₂ = 256 − 108 = 148
Answer
U = 148
6.Number systems
Answer: R = 116
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
XOR gives 1 where the two bits differ.
1010 1001
XOR 1101 1101
= 0111 0100
Value of the result
0111 0100₂ = 116
Answer
R = 116
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