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 192.168.118.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?
192.168.118.0/24/26subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.109.64/18. What is the third octet x of its network address?
172.20.109.64/18
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
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ADNS needs ICMP to resolve names
- BTCP cannot open connections without ICMP
- CTLS certificates are sent over ICMP
- DPath MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systemsMedium
Set the bits so they store −4 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −48 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 0100 AND 0011 1100. What is the result's value R in decimal?
1010 0100AND0011 1100
R =
3Solutions
1.IPv4 subnetting
Answer: x = 127
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
26 = 64
Subnet number 2
network: (2 − 1)·64 = 64
broadcast: 64 + 64 − 1 = 127
Answer
x = 127
2.IPv4 subnetting
Answer: x = 64
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
26 = 64
Subnet of this host
network: 172.20.64.0
broadcast: 172.20.127.255
Answer
x = 64
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: D · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
D · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 252
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
4 = 0000 0100
inverted: 1111 1011
+1 → 1111 1100
Read it as unsigned
1111 1100₂ = 256 − 4 = 252
Answer
U = 252
6.Number systems
Answer: R = 36
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
AND gives 1 only where both bits are 1.
1010 0100
AND 0011 1100
= 0010 0100
Value of the result
0010 0100₂ = 36
Answer
R = 36
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