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 6 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/246 subnets
H =
2.IPv4 subnettingHard
The network 192.168.89.0/24 is split with VLSM for these LANs: A: 14 hosts, B: 53 hosts, C: 109 hosts, D: 6 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 14B: 53C: 109D: 6
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 uses UDP, POST uses TCP
- CGET deletes a resource, POST reads it from the server's database
- DGET is encrypted, POST is not
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- APath MTU discovery fails because "fragmentation needed" messages are blocked
- BTLS certificates are sent over ICMP
- CDNS needs ICMP to resolve names
- DTCP cannot open connections without ICMP
5.Number systemsMedium
Set the bits so they store −6 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −68 bit
U =
6.Number systemsHard
A system must give every one of 4187 items its own binary code of the same length. What is the smallest number of bits b needed?
4187
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
23 = 8 ≥ 6 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 208
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
C: 109 hosts → 27 = 128 (/25)
B: 53 hosts → 26 = 64 (/26)
A: 14 hosts → 24 = 16 (/28)
D: 6 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .127
B: .128 – .191
A: .192 – .207
D: .208 – .215
Answer
x = 208
LAN D: 192.168.89.208/29.
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: A · 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
A · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: U = 250
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
6 = 0000 0110
inverted: 1111 1001
+1 → 1111 1010
Read it as unsigned
1111 1010₂ = 256 − 6 = 250
Answer
U = 250
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 < 4187
213 = 8192 ≥ 4187
Answer
b = 13
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