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 1689 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1689 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.165.23/30. What is the fourth octet x of its network address?
10.20.165.23/30
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- A::1
- Bfe80::1
- C::
- Dff02::1
4.Computer networksHard
With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?
- AGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
- BNeither ever resends lost frames
- CBoth resend the whole window from the start
- DGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
5.Number systemsMedium
The 8-bit pattern 1111 1010 stores a signed integer in two's complement. What is its value N in decimal?
1111 10108 bit
N =
6.Number systemsHard
A system must give every one of 4880 items its own binary code of the same length. What is the smallest number of bits b needed?
4880
b = bit
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 < 1689
211 − 2 = 2046 ≥ 1689 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 20
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
22 = 4
Subnet of this host
network: ⌊23/4⌋·4 = 20
broadcast: 20 + 4 − 1 = 23
usable hosts: 21 … 22
Answer
x = 20
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: A · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
Why this is the answer
A Go-Back-N receiver discards out-of-order frames, so the sender must repeat everything after the loss. Selective Repeat buffers them and asks only for the missing one.
Answer
A · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systems
Answer: N = −6
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 + 8 + 2 = −6
Answer
N = −6
6.Number systems
Answer: b = 13 bit
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.
Powers of two
212 = 4096 < 4880
213 = 8192 ≥ 4880
Answer
b = 13
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