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 192.168.239.0/24 is divided into /28 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 11?
192.168.239.0/24/28subnet 11 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.70.66/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.70.66/28
x =
3.Computer networksMedium
Why is UDP often preferred for live video calls?
- AUDP is required for all video
- BUDP encrypts data automatically
- CUDP guarantees every packet arrives, so the picture never freezes
- DLow delay matters more than retransmitting every lost packet
4.Computer networksHard
A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?
- A50 kB
- B4 kB
- C500 kB
- D2.5 MB
5.Number systemsMedium
Set the bits so they store −103 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1038 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1110 1101 << 3. What is the result's value R in decimal?
1110 1101<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: x = 175
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
24 = 16
Subnet number 11
network: (11 − 1)·16 = 160
broadcast: 160 + 16 − 1 = 175
Answer
x = 175
2.IPv4 subnetting
Answer: x = 78
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
24 = 16
Subnet of this host
network: ⌊66/16⌋·16 = 64
broadcast: 64 + 16 − 1 = 79
usable hosts: 65 … 78
Answer
x = 78
3.Computer networks
Answer: D · Low delay matters more than retransmitting every lost packet
Why this is the answer
Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.
Answer
D · Low delay matters more than retransmitting every lost packet
4.Computer networks
Answer: C · 500 kB
Why this is the answer
The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.
Answer
C · 500 kB
5.Number systems
Answer: U = 153
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
103 = 0110 0111
inverted: 1001 1000
+1 → 1001 1001
Read it as unsigned
1001 1001₂ = 256 − 103 = 153
Answer
U = 153
6.Number systems
Answer: R = 104
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
Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1110 1101 << 3
= 0110 1000
Value of the result
0110 1000₂ = 104
Answer
R = 104
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