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 10.26.60.0/22 is divided into equal subnets with prefix /26. How many subnets N are there?
10.26.60.0/22/26
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.225.89.0/24, 10.225.90.0/24, 10.225.91.0/24, 10.225.92.0/24, 10.225.93.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.225.89.0/24 … 10.225.93.0/24
p =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- AOffer, Discover, Acknowledge, Request
- BDiscover, Offer, Request, Acknowledge
- CDiscover, Request, Offer, Acknowledge
- DRequest, Offer, Discover, Acknowledge
4.Computer networksHard
With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?
- ANeither ever resends lost frames
- BGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
- CGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
- DBoth resend the whole window from the start
5.Number systemsMedium
Write −116 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1168 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 0010 XOR 1010 1001. What is the result's value R in decimal?
1111 0010XOR1010 1001
R =
3Solutions
1.IPv4 subnetting
Answer: N = 16
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).
Borrowed bits
26 − 22 = 4
Number of subnets
N = 24 = 16
Answer
N = 16
Each /26 subnet has 26 − 2 = 62 usable hosts.
2.IPv4 subnetting
Answer: p = 21
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
89 = 01011001
93 = 01011101
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.225.88.0/21 (it also covers 3 network(s) that are not in the list).
3.Computer networks
Answer: B · Discover, Offer, Request, Acknowledge
Why this is the answer
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer
B · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: C · 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
C · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systems
Answer: U = 140
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.
Invert and add one
116 = 0111 0100
inverted: 1000 1011
+1 → 1000 1100
Read it as unsigned
1000 1100₂ = 256 − 116 = 140
Answer
U = 140
6.Number systems
Answer: R = 91
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.
1111 0010
XOR 1010 1001
= 0101 1011
Value of the result
0101 1011₂ = 91
Answer
R = 91
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