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Murljashka [212]
4 years ago
9

Consider a DHT with a mesh overlay topology (that is, every peer tracks all peers in the system). What are the advantages and di

sadvantages of such a design? What are the advantages and disadvantages of a circular DHT (with no shortcuts)?
Computers and Technology
1 answer:
Vladimir79 [104]4 years ago
6 0

Answer:

Advantages of DHT with mesh overlay topology

  • A single hop is used to route a message to the peer, the nearest key is used route message between points
  • There are  bi-directional links between each pair of peers other than the broadcasting peer therefore creating a multiple delivery paths from source to other peers

Disadvantages of DHT with mesh overlay topology

  • complexity of the design of a Mesh overlay Topology
  • Consuming process ( tracking of all peers by each peer )

Advantages of circular DHT ( with no shortcuts )

  • less consuming process ( each peer tracks only two peers )

Disadvantages of Circular DHT ( with no short cuts )

  • The number of messages sent per query is minimized
  • 0(N) hopes are required to route message to a peer responsible for the key  

Explanation:

Advantages of DHT with mesh overlay topology

  • A single hop is used to route a message to the peer, the nearest key is used route message between points
  • There are  bi-directional links between each pair of peers other than the broadcasting peer therefore creating a multiple delivery paths from source to other peers

Disadvantages of DHT with mesh overlay topology

  • complexity of the design of a Mesh overlay Topology
  • Consuming process ( tracking of all peers by each peer )

Advantages of circular DHT ( with no shortcuts )

  • less consuming process ( each peer tracks only two peers )

Disadvantages of Circular DHT ( with no short cuts )

  • The number of messages sent per query is minimized
  • 0(N) hopes are required to route message to a peer responsible for the key  

Distributed hash table (DHT) is a distributed system that provides a lookup service similar to a hash table. in the DHT key-value are stored in it

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Explanation:

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Arun is 5 years older than Anu. five years ago, the ratio of their ages was 3:2 . find their present age​
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Answer: Arun is 15 years old and Anu is 10 year old

Explanation:

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#include <iostream> using namespace std; int isPrimeNumber(int); int main() { bool isPrime; for(int n = 2; n < 100; n++) { // isPrime will be true for prime numbers isPrime = isPrimeNumber(n); if(isPrime == true) cout<<n<<" "; } return 0; } // Function that checks whether n is prime or not int isPrimeNumber(int n) { bool isPrime = true; for(int i = 2; i <= n/2; i++) { if (n%i == 0) { isPrime = false; break; } } return isPrime; }

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3 years ago
It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

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Answer:

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Explanation:

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