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 980 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
980 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.215.133/20. What is the third octet x of its network address?
172.20.215.133/20
x =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ADiscover, Offer, Request, Acknowledge
- BDiscover, Request, Offer, Acknowledge
- CRequest, Offer, Discover, Acknowledge
- DOffer, Discover, Acknowledge, Request
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo avoid confusion with old segments and make spoofed segments harder to inject
- BTo select the best route to the destination for the segments of that connection
- CTo encrypt the payload of each segment
- DTo compress the data more efficiently
5.Number systemsMedium
Set the bits so they store −84 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −848 bit
U =
6.Number systemsHard
A system must give every one of 4099 items its own binary code of the same length. What is the smallest number of bits b needed?
4099
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 22
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
29 − 2 = 510 < 980
210 − 2 = 1022 ≥ 980 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 208
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
24 = 16
Subnet of this host
network: 172.20.208.0
broadcast: 172.20.223.255
Answer
x = 208
3.Computer networks
Answer: A · Discover, Offer, Request, Acknowledge
Why this is the answer
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer
A · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: A · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
A · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 172
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
84 = 0101 0100
inverted: 1010 1011
+1 → 1010 1100
Read it as unsigned
1010 1100₂ = 256 − 84 = 172
Answer
U = 172
6.Number systems
Answer: b = 13 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
212 = 4096 < 4099
213 = 8192 ≥ 4099
Answer
b = 13
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