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 1088 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1088 hosts
p =
2.IPv4 subnettingHard
The network 192.168.63.0/24 is split with VLSM for these LANs: A: 11 hosts, B: 28 hosts, C: 26 hosts, D: 3 hosts, E: 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 A?
A: 11B: 28C: 26D: 3E: 6
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A10 ms
- B1 ms
- C100 ms
- D0.1 ms
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AThe size of each IP packet header
- BThe number of open ports
- CThe retransmission time-out, calculated from the measured round-trip time of each segment
- DHow much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systemsMedium
The 8-bit pattern 0010 1001 stores a signed integer in two's complement. What is its value N in decimal?
0010 10018 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 1010 XOR 0010 0000. What is the result's value R in decimal?
0110 1010XOR0010 0000
R =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1088
211 − 2 = 2046 ≥ 1088 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 64
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: 28 hosts → 25 = 32 (/27)
C: 26 hosts → 25 = 32 (/27)
A: 11 hosts → 24 = 16 (/28)
E: 6 hosts → 23 = 8 (/29)
D: 3 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .31
C: .32 – .63
A: .64 – .79
E: .80 – .87
D: .88 – .95
Answer
x = 64
LAN A: 192.168.63.64/28.
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 · How much unacknowledged data the sender may send, based on the receiver's buffer
Why this is the answer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer
D · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: N = 41
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.
32 + 8 + 1 = 41
Answer
N = 41
6.Number systems
Answer: R = 74
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.
0110 1010
XOR 0010 0000
= 0100 1010
Value of the result
0100 1010₂ = 74
Answer
R = 74
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