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 192.168.172.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
192.168.172.0/24/27subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.81.150/20. What is the third octet x of its network address?
172.20.81.150/20
x =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A1
- B4
- C2
- D0
4.Computer networksHard
TCP congestion control's "slow start" phase:
- AKeeps the window fixed at one segment
- BIncreases the congestion window exponentially until a threshold or loss
- CIs used only by UDP
- DHalves the congestion window every round trip until a packet loss is detected
5.Number systemsMedium
Set the bits so they store −15 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −158 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1101 0011 XOR 0001 1110. What is the result's value R in decimal?
1101 0011XOR0001 1110
R =
3Solutions
1.IPv4 subnetting
Answer: x = 32
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
25 = 32
Subnet number 2
network: (2 − 1)·32 = 32
broadcast: 32 + 32 − 1 = 63
Answer
x = 32
2.IPv4 subnetting
Answer: x = 80
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
24 = 16
Subnet of this host
network: 172.20.80.0
broadcast: 172.20.95.255
Answer
x = 80
3.Computer networks
Answer: D · 0
Why this is the answer
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer
D · 0
4.Computer networks
Answer: B · Increases the congestion window exponentially until a threshold or loss
Why this is the answer
Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.
Answer
B · Increases the congestion window exponentially until a threshold or loss
5.Number systems
Answer: U = 241
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
15 = 0000 1111
inverted: 1111 0000
+1 → 1111 0001
Read it as unsigned
1111 0001₂ = 256 − 15 = 241
Answer
U = 241
6.Number systems
Answer: R = 205
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.
1101 0011
XOR 0001 1110
= 1100 1101
Value of the result
1100 1101₂ = 205
Answer
R = 205
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