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 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/248 subnets
H =
2.IPv4 subnettingHard
A host has the address 172.20.208.14/23. What is the third octet x of its network address?
172.20.208.14/23
x =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A2
- B0
- C1
- D4
4.Computer networksHard
Why does the side that closes a TCP connection first stay in the TIME_WAIT state for twice the maximum segment lifetime?
- ASo delayed segments die out and the final ACK can be resent if it was lost
- BTo wait for the DNS cache to expire
- CTo let the routers along the path update their routing tables for the next connection
- DTo keep the encryption keys for later reuse
5.Number systemsMedium
Set the bits so they store −28 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −288 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 0010 OR 1010 0011. What is the result's value R in decimal?
1010 0010OR1010 0011
R =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
23 = 8 ≥ 8 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 208
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: 172.20.208.0
broadcast: 172.20.209.255
Answer
x = 208
3.Computer networks
Answer: B · 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
B · 0
4.Computer networks
Answer: A · So delayed segments die out and the final ACK can be resent if it was lost
Why this is the answer
TIME_WAIT ensures a lost final ACK can be repeated and old duplicate segments expire before the same address and port pair is reused.
Answer
A · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: U = 228
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
28 = 0001 1100
inverted: 1110 0011
+1 → 1110 0100
Read it as unsigned
1110 0100₂ = 256 − 28 = 228
Answer
U = 228
6.Number systems
Answer: R = 163
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.
1010 0010
OR 1010 0011
= 1010 0011
Value of the result
1010 0011₂ = 163
Answer
R = 163
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