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 255.255.240.0. How many usable host addresses H does it have?
255.255.240.0
H =
2.IPv4 subnettingHard
A host has the address 172.20.110.128/18. What is the third octet x of its subnet's broadcast address?
172.20.110.128/18
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A1 ms
- B100 ms
- C0.1 ms
- D10 ms
4.Computer networksHard
After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?
- ATo synchronise all stations to one clock
- BTo lower the chance of colliding again when many stations compete
- CTo give the station with the highest MAC address priority on the shared cable
- DTo make the cable carry signals further
5.Number systemsMedium
Set the bits so they store −80 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −808 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 0110 OR 0001 0111. What is the result's value R in decimal?
1111 0110OR0001 0111
R =
3Solutions
1.IPv4 subnetting
Answer: H = 4094
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
255.255.240.0 = /20
Usable hosts
212 − 2 = 4094
Answer
H = 4094
2.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 in the third octet
26 = 64
Subnet of this host
network: 172.20.64.0
broadcast: 172.20.127.255
Answer
x = 127
3.Computer networks
Answer: D · 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
D · 10 ms
4.Computer networks
Answer: B · To lower the chance of colliding again when many stations compete
Why this is the answer
Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.
Answer
B · To lower the chance of colliding again when many stations compete
5.Number systems
Answer: U = 176
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
80 = 0101 0000
inverted: 1010 1111
+1 → 1011 0000
Read it as unsigned
1011 0000₂ = 256 − 80 = 176
Answer
U = 176
6.Number systems
Answer: R = 247
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
OR gives 1 where at least one bit is 1.
1111 0110
OR 0001 0111
= 1111 0111
Value of the result
1111 0111₂ = 247
Answer
R = 247
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