Solution to Homework #3: SMIL based multimedia presentation, VoD
Part 2:
Assume 3 Mbps instead of 1.5Mbps
is used to provide the transport of a 100 minute NTSC movie stored in the copier
memory. Here we assume a fixed rate encoding. We plan to use a Ultra Wide SCSI-2
disk with 4 heads (each head with
40 Mbps data transfer rate) and 40
MBps SCSI channel transfer rate to deliver the movie. Note that the disk can produce a maximum of 40Mbps*4/8=20MBps. With two disks connected throught the SCSI channel we will be still ok.
How many bytes of disk space
are needed to store the movie?
Ans: diskSpace=3000000*100*60/8=2.25GBytes.
Assume a fixed rate encoding,
i.e., each frame is encoded into the same amount of bits, how many bits
are used to encode a frame?
Ans: This is a NTSC movie with 30 frames per second.
Assume 3Mbps.
Therefore each frame is encoded in 3000000/30=100000 bits.
How many phases can each head
produce?
Ans: Each phase requires 3Mbps data transfer speed.
The disk head has 40Mbps.
Floor(40M/3M)=13 phases.
With 4 heads per movie, how
long is a phase?
Ans: Total number of phases per movie=4*13=52 phases.
Each phase is 100*60/52=115.38 seconds.
How many frames will be read
by each head?
Ans: 100min movie has 100*60*30=180000 frames. To
make the frame number dividable by 4 and 13, add 24 frame to make it 180024. Each
head reads 180024/4=45006 frames.
Let the answer in 3) be n.
The frames read by each head will be grouped into blocks and each block
contains n frames. How many blocks will be read by a head?
Ans: Each head reads 180024/4=45006 frames. Each block
consists of 13 frames.
Therefore each head reads 45006/13=3462 blocks.
Starting with frame 0, show
all the frame ID numbers in the first block. Similar to that in Figure 4
of Sincoskie’s paper.
Ans:
0
3462
6924
10386
13848
17310
20772
24234
27696
31158
34620
38082
41544
Without additional buffer,
what is the longest time that a customer has to wait before seeing the movie?
Ans: 115.38 seconds. Or more precisely, if we consider
the padded frames, then 45006 frames*100kbits/frame / 40 Mbps = 112.515
seconds.
