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 
D_{i+1} = f * D_{i} 
where D_{i} is the size of fragment i
D_{1} + D_{2} + ... + D_{K} = D 
Suppose:
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:
is solved.
you will see that there can be a time gap between successive fragment transmission of fragments that belong to the same video object !!!
Example:
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 preload fragment 3 to harddisk and consume it later.
Receiver start consuming the fragment at this time and will also start looking for fragment D_{21}
This fragment D_{23} 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 D_{23} is downloaded before the receiver finishes displaying fragment D_{22}.
A receiver is able to display the entire video object
without gaps if:
(That is because the receiver starts looking for fragment D_{i+1} when it has started consuming fragment D_{i}) 
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).
The video object has length D (D is a proper parameter of the system)
We know that the whole video object is devided into K objects 1, 2, ..., K where their lengths are related as:
D_{i} = f * D_{i1} ......... (1)
So:
The total length of these K frgaments must be equal to the length of the original video object, so:
D_{1} + D_{2} + D_{3} + ... + D_{K} = D
D_{1} + f * D_{1} + f^{2} * D_{1} + ... + f^{K1} * D_{1} = D
D_{1} * (1 + f + f^{2} + f^{K1}) = D ........................................... (2)
From Calculus 101, you may still remember that
f^{K}  1 1 + f + f^{2} + f^{K1} =  ...... (3) f  1If not, you can derive it by multiplying both sides of the equation above by f and add them up.
Substitute equation (3) in equation (2), we get that:
f^{K}  1 D_{1}  = D f  1Or:
f  1 D_{1} = D  ......... (4) f^{K}  1
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 f^{K}  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 K1 and K
So the maximum amount of data that need to be stored is when the receiver is downloading fragments K1 and K simultaneously.
Since fragment K1 is downloaded first, we have this situation:
f  1 D_{K1} = f^{K2} D  f^{K}  1and
f  1 D_{K} = f^{K1} D  f^{K}  1
Therefore: