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 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
A host has the address 172.20.69.39/23. What is the third octet x of its subnet's broadcast address?
172.20.69.39/23
x =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A0.125 ms
- B8 ms
- C1 ms
- D12 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
Set the bits so they store −74 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −748 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0100 << 2. What is the result's value R in decimal?
1100 0100<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 69
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
21 = 2
Subnet of this host
network: 172.20.68.0
broadcast: 172.20.69.255
Answer
x = 69
3.Computer networks
Answer: C · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
C · 1 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: U = 182
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
74 = 0100 1010
inverted: 1011 0101
+1 → 1011 0110
Read it as unsigned
1011 0110₂ = 256 − 74 = 182
Answer
U = 182
6.Number systems
Answer: R = 16
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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 0100 << 2
= 0001 0000
Value of the result
0001 0000₂ = 16
Answer
R = 16
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