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Alekssandra [29.7K]
3 years ago
11

Consider a DASH system for which there are N versions of a video file (in an increasing order of quality) and N versions of a co

rresponding audio file (in an increasing order of quality). Suppose we want to allow the player to choose at any time any chunk of the N video versions and any of the N audio versions. a. If we create files so that the audio is mixed in with the video in a the same file, so server sends only one media stream at given time, how many files will the server need to store? b. If the server instead sends the audio and video streams separately and has the client synchronize the streams, how many files will the server need to store?
Computers and Technology
1 answer:
Vilka [71]3 years ago
8 0

Answer: a) N² files. b) 2N files.

Explanation:

if an user can choose freeely both the video and the audio quality, if they are stored separately, but the user can download any mix of them as a single file, the server must store one audio version for each video version, so it will need to store NxN = N² files.

If, instead, the server can send the audio and video streams separately, so the user can choose one of the N video available, and one of the N audio versions, the server will need to store only N video files + N audio audio files, i.e. , 2 N files.

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// C code

// This code will compute the values of the sales ticket sales for concerts

#include <stdio.h>

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#define MAXC 3 //max categories

char group [MAXG][MAXN];

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void switchRows (int m, int n) {

char tc;

int ti;

float v;

// printf ("Switching %d with %d\n", m, n);

for (int i = 0; i < MAXN; i++) {

tc = group [m][i];

group [m][i] = group [n][i];

group [n][i] = tc;

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for (int i = 0; i < MAXC; i++) {

ti = fans [m][i];

fans [m][i] = fans [n][i];

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int target;

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int minFans (int b) {

float min = fans [b][0];

int target = b;

for (int i = b; i < count; i++)

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printArray ();

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