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 255.255.255.252. How many usable host addresses H does it have?
255.255.255.252
H =
2.IPv4 subnettingHard
The network 192.168.21.0/24 is split with VLSM for these LANs: A: 24 hosts, B: 126 hosts, C: 3 hosts, D: 28 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: 24B: 126C: 3D: 28
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A10 ms
- B100 ms
- C1 ms
- D0.1 ms
4.Computer networksHard
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
- AResponses can arrive in any order on one stream, so slow replies do not block others
- BThe server can push updates over UDP
- CThe connection stays encrypted without TLS
- DSeveral requests reuse one TCP connection, avoiding repeated handshakes
5.Number systemsMedium
Set the bits so they store −49 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −498 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0110 XOR 0010 1001. What is the result's value R in decimal?
1001 0110XOR0010 1001
R =
3Solutions
1.IPv4 subnetting
Answer: H = 2
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
255.255.255.252 = /30
Usable hosts
22 − 2 = 2
Answer
H = 2
2.IPv4 subnetting
Answer: x = 128
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
B: 126 hosts → 27 = 128 (/25)
D: 28 hosts → 25 = 32 (/27)
A: 24 hosts → 25 = 32 (/27)
C: 3 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
D: .128 – .159
A: .160 – .191
C: .192 – .199
Answer
x = 128
LAN D: 192.168.21.128/27.
3.Computer networks
Answer: A · 10 ms
Why this is the answer
2 × 10⁶ m / 2 × 10⁸ m/s = 0.01 s = 10 ms. Propagation delay depends on distance and medium, not packet size.
Answer
A · 10 ms
4.Computer networks
Answer: D · Several requests reuse one TCP connection, avoiding repeated handshakes
Why this is the answer
Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.
Answer
D · Several requests reuse one TCP connection, avoiding repeated handshakes
5.Number systems
Answer: U = 207
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
49 = 0011 0001
inverted: 1100 1110
+1 → 1100 1111
Read it as unsigned
1100 1111₂ = 256 − 49 = 207
Answer
U = 207
6.Number systems
Answer: R = 191
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
XOR gives 1 where the two bits differ.
1001 0110
XOR 0010 1001
= 1011 1111
Value of the result
1011 1111₂ = 191
Answer
R = 191
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