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 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.193.214.0/24, 10.193.215.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.193.214.0/24 … 10.193.215.0/24
p =
3.Computer networksMedium
Which protocol does the "ping" command use?
- AICMP
- BUDP
- CTCP
- DHTTP
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
- CGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
- DBoth resend the whole window from the start
5.Number systemsMedium
The 8-bit pattern 1110 0101 stores a signed integer in two's complement. What is its value N in decimal?
1110 01018 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0011 1100 XOR 1000 1101. What is the result's value R in decimal?
0011 1100XOR1000 1101
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: p = 23
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
214 = 11010110
215 = 11010111
common leading bits: 7
Prefix length
p = 16 + 7 = 23
Answer
p = /23
Summary route: 10.193.214.0/23
3.Computer networks
Answer: A · ICMP
Why this is the answer
Ping sends ICMP echo requests and measures the time until echo replies return.
Answer
A · ICMP
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 = −27
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 + 4 + 1 = −27
Answer
N = −27
6.Number systems
Answer: R = 177
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.
0011 1100
XOR 1000 1101
= 1011 0001
Value of the result
1011 0001₂ = 177
Answer
R = 177
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