D Access Time(Round Robin) = M --- B
|The parameter K is the number of fragments that a video object will be divided into.|
This parameter (f) is derived from the parameter K (number of fragments)
The parameter f will be explained later, for now, I will just simply tell you that:
B f = ---- MK
Di+1 = f * Di
where Di is the size of fragment i
D1 + D2 + ... + DK = D
Suppose we select K = 3
Then, according to the formula above, we must divide each video object into this many segments:
B 12 f = ---- = ----- = 2 MK 3*2
The following figure shows the division of one video object into K = 3 where each subsequent segment if f = 2 times as large as the previous one:
We have that:
(And will therefore transmit the video object in K channels, since one video object is partitioned into K fragments)
(Recall that K is the number of fragments.
Pyramid Broadcasting will send one fragment in each channel)
So the problem:
you will see that there can be a time gap between successive fragment transmission of fragments that belong to the same video object !!!
|Is the receiver always able to obtain the next stream before the current video fragment finishes playing ???|
|Number of (sub)channels is (also) equal to 3 (1 fragment from each video is sent on one channel.)|
B 12 f = ---- = ----- = 2 MK 3*2
|Because the video is transmitted at 4 times the consumption rate, every minute of downloaded video will take 4 minutes to view !|
Note that it takes 4 times as long to consume the video object than transmitting it !
Therefore, we must pre-load fragment 3 to harddisk and consume it later.
Receiver start consuming the fragment at this time and will also start looking for fragment D21
This fragment D23 is NOT downloaded (skipped) because the next fragment is only downloaded after consumption of the previous segment has begun.
You see that the start of D23 is downloaded before the receiver finishes displaying fragment D22.
A receiver is able to display the entire video object
without gaps if:
(That is because the receiver starts looking for fragment Di+1 when it has started consuming fragment Di)
The figure depicts:
We saw that the "naive" way of broadcasting M video objects - each of size D - on a channel of bandwidth B resulted in an access time of:   MD/B (see: click here)
What is the access time of the pyramid broadcast method ?
(Hint: it better be smaller than MD/B or we have gained nothing :-))
B f = --- M*Kthen the receiver will be able to obtain the whole video object continuously (without interruption).
We know that the whole video object is devided into K objects 1, 2, ..., K where their lengths are related as:
Di   =   f * Di-1 ......... (1)
The total length of these K frgaments must be equal to the length of the original video object, so:
D1 + D2 + D3 + ... + DK   =   D
D1 + f * D1 + f2 * D1 + ... + fK-1 * D1   =   D
D1 * (1 + f + f2 + fK-1)   =   D ........................................... (2)
From Calculus 101, you may still remember that
fK - 1
1 + f + f2 + fK-1 = ------- ...... (3)
f - 1
If not, you can derive it by multiplying both sides of
the equation above by f and add them up.
Substitute equation (4) in the expression for access time above ( click here ), and we get:
It looks like that the access time is dependent on f and K.
However, since K = B/(M * f) (B (bandwidth) and M (# of video objects) are fixed system parameters), the access time is really only dependent on f
Decreases linearly when bandwidth B increases
(According to the formula above for the access time, the access time decreases exponentially when f increases. f increases linearly when B increases, therefore, access time decreases exponentially when B increases)
D f - 1 Access Time = --- ------- ......... (4) f fK - 1
The total size of the video object is D * b bits.
If you have space to store the whole video object, you certainly have enough space :-)
The two largest fragments are fragments K-1 and K
So the maximum amount of data that need to be stored is when the receiver is downloading fragments K-1 and K simultaneously.
Since fragment K-1 is downloaded first, we have this situation:
f - 1 DK-1 = fK-2 D ------- fK - 1and
f - 1 DK = fK-1 D ------- fK - 1