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 department needs 1309 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1309 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.57.6/27. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.57.6/27
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- A::1
- Bff02::1
- Cfe80::1
- D::
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?
- ATo let the routers along the path update their routing tables for the next connection
- BTo wait for the DNS cache to expire
- CTo keep the encryption keys for later reuse
- DSo delayed segments die out and the final ACK can be resent if it was lost
5.Number systemsMedium
The 8-bit pattern 1111 0110 stores a signed integer in two's complement. What is its value N in decimal?
1111 01108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 0110 OR 0001 1010. What is the result's value R in decimal?
0110 0110OR0001 1010
R =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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).
Host bits needed
210 − 2 = 1022 < 1309
211 − 2 = 2046 ≥ 1309 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 31
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 of this host
network: ⌊6/32⌋·32 = 0
broadcast: 0 + 32 − 1 = 31
usable hosts: 1 … 30
Answer
x = 31
3.Computer networks
Answer: A · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
A · ::1
4.Computer networks
Answer: D · 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
D · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: N = −10
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 32 + 16 + 4 + 2 = −10
Answer
N = −10
6.Number systems
Answer: R = 126
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.
0110 0110
OR 0001 1010
= 0111 1110
Value of the result
0111 1110₂ = 126
Answer
R = 126
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