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 subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?
255.255.255.248
H =
2.IPv4 subnettingHard
A host has the address 10.20.168.243/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.168.243/29
x =
3.Computer networksMedium
What is the main difference between the HTTP methods GET and POST?
- AGET retrieves a resource, POST submits data in the request body
- BGET is encrypted, POST is not
- CGET uses UDP, POST uses TCP
- DGET deletes a resource, POST reads it from the server's database
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo encrypt the payload of each segment
- BTo compress the data more efficiently
- CTo avoid confusion with old segments and make spoofed segments harder to inject
- DTo select the best route to the destination for the segments of that connection
5.Number systemsMedium
The 8-bit pattern 1100 0000 stores a signed integer in two's complement. What is its value N in decimal?
1100 00008 bit
N =
6.Number systemsHard
A system must give every one of 1960 items its own binary code of the same length. What is the smallest number of bits b needed?
1960
b = bit
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).
Prefix length
255.255.255.248 = /29
Usable hosts
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 246
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
23 = 8
Subnet of this host
network: ⌊243/8⌋·8 = 240
broadcast: 240 + 8 − 1 = 247
usable hosts: 241 … 246
Answer
x = 246
3.Computer networks
Answer: A · GET retrieves a resource, POST submits data in the request body
Why this is the answer
GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.
Answer
A · GET retrieves a resource, POST submits data in the request body
4.Computer networks
Answer: C · 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
C · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: N = −64
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 = −64
Answer
N = −64
6.Number systems
Answer: b = 11 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
210 = 1024 < 1960
211 = 2048 ≥ 1960
Answer
b = 11
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