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 1850 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1850 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.129.218/22. What is the third octet x of its network address?
172.20.129.218/22
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CD
- BCSMA/CA
- CToken passing
- DTDMA
4.Computer networksHard
What does TCP fast retransmit do?
- AResends a segment after three duplicate ACKs, without waiting for the timeout
- BDoubles the window after every loss
- CSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
- DResends every segment twice to be safe
5.Number systemsMedium
Set the bits so they store −78 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −788 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 0000 << 1. What is the result's value R in decimal?
0100 0000<< 1
R =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1850
211 − 2 = 2046 ≥ 1850 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 128
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
22 = 4
Subnet of this host
network: 172.20.128.0
broadcast: 172.20.131.255
Answer
x = 128
3.Computer networks
Answer: A · 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
A · CSMA/CD
4.Computer networks
Answer: A · Resends a segment after three duplicate ACKs, without waiting for the timeout
Why this is the answer
Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.
Answer
A · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: U = 178
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
78 = 0100 1110
inverted: 1011 0001
+1 → 1011 0010
Read it as unsigned
1011 0010₂ = 256 − 78 = 178
Answer
U = 178
6.Number systems
Answer: R = 128
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 left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0100 0000 << 1
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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