E0262 -Multimedia Information Systems MULTIMEDIA DATA

E0262 -Multimedia Information Systems Problem 1 If 74 min. of audio file can be recorded on a standard 120 mm diameter CD,(i) calculate the maximum length of audio file that can be recorded if the disc space is 325 MB.(ii) Can this file be stored on a single standard 80 mm diameter CD? Solution:(i) For a standard 120 mm diameter CD, 650 MB of disc space corresponds to 74 min. audio file. So 325 MB disc space will corresponds to(74 X 325) / 650 = 34 min. length of audio file (ii) No, this audio file can not be stored on 80 mm diameter CD. This is because the maximum length of audio file that can be recorded is 21 min.

E0262 -Multimedia Information Systems Calculate the number of standard 120mm diameter CD's required to store equivalent data of a single DVD of capacity 4.7 GB. Solution:120 mm diameter CD has 650 MB storage capacity and a single DVD capacity is 4.7 GB.Therefore, the number of CD's required to store equivalent data of a single DVD= (4.7 GB) / (650 MB) = (4.7 X 1024) / (650)= 7.404~ 8 CD's

E0262 -Multimedia Information Systems Problem 2 Calculate the total number of frames required to construct a video if it contains 8 sequences, each sequence contains 5 scenes, each scene contains 7 shots and each shot contains 2 frames. Solution:Total number of frames required to construct a video= (number of sequences) X (number of scenes) X (number of shots) X (number of frames in each shot)= 8 X 5 X 7 X 2= 980 frames

E0262 -Multimedia Information Systems If an image contains 3000 horizontal pixels and 4000 vertical pixels(i) calculate the resolution of the image.(ii) Calculate the memory required to store the image file if each pixel is represented by 8 bits.(iii) Calculate the compression ratio if the image in part (ii) is compressed so that the memory required to store is reduced to 2 MB. Solution: (i) The resolution of the image= ( number of horizontal pixels) X (number of vertical pixels)= 3000 X 4000= 12 MP

E0262 -Multimedia Information Systems (ii) The memory required to store the image is= (12 X 10^6) X 8= (96 X 10^6) / (1024 X 1024 X 8) MB= MB (iii) Compression ratio= (original file size) / (file size after compression)= ( MB) / (2 MB)= : 1(note: compression ratio has to be greater than 1)

E0262 -Multimedia Information Systems Problem 3 Suppose that a packet consists of P bytes of data and 18 bytes of header. Consider sending a digitally encoded voice stream encoded at 128 kb/s. Assume that a packet payload is completely filled (at a rate of 128 kb/s) before the packet is sent. The time to fill the packet payload is called the packetization delay. a) What is the packetization delay for packets with P = 46 bytes and P = 1500 bytes?

E0262 -Multimedia Information Systems Packetization delay for P = 46 bytes = (46 * 8) / 128,000 = milliseconds Packetization delay for P = 1500 bytes = (1500 * 8) / 128,000 = milliseconds b) What is the store-and-forward delay (transmission delay) for packets with P = 56 bytes and P = 1500 bytes if the link rate is 10 Mb/s.

E0262 -Multimedia Information Systems Store-and-forward delay is the same as transmission delay. So, Store-and-forward delay for P = 46 bytes = (( ) * 8) / 10,000,000 = milliseconds Store-and-forward delay for P = 1500 bytes = (( ) * 8) / 10,000,000 = milliseconds