Respuesta :
Answer:
The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.
Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.
Explanation:
Given details are below:
Length of the link = 10 meters
Bandwidth = 150 bits/sec
Size of a data packet = 100,000 bits
Size of a control packet = 200 bits
Size of the downloaded object = 100Kbits
No. of referenced objects = 10
Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.
The formula below is used to calculate the total time delay for sending and receiving packets :
d = dp (propagation delay) + dt (transmission delay)
For Parallel downloads through parallel instances of non-persistent HTTP :
Bandwidth = 150 bits/sec
No. of referenced objects = 10
For each parallel download, the bandwith = 150/10
= 15 bits/sec
10 independent connections are established, during parallel downloads, and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.
Therefore, for parallel downloads, the total time required is calculated as:
(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)
= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)
= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)
= (670 + 4Tp) + (6706 + 4Tp)
= 7377 + 8 Tp seconds
Thus, parallel instances of non-persistent HTTP makes sense in this case.
Let the speed of propogation of the medium be 300*106 m/sec.
Then, Tp = 10/(300*106)
= 0.03 micro seconds
The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.
The parallel downloads through persistent HTTP doesn't have significant gain over the non-persistent HTTP.
Given the following data:
- Packet length = 100,000 bits long.
- Transmission rate = 150 bits/sec.
- Object data = 100 Kbits.
- Control data = 200 bits.
- Distance = 10 meter.
- Number of objects = 10.
How to calculate the time needed to receive all objects.
In this scenario, we would denote the one-way propagation delay that exist between the server and the client with [tex]T_p[/tex].
For the bandwith:
[tex]Bandwidth =\frac{150}{10}[/tex]
Bandwidth = 15 bits/sec.
Since the speed of light is equal to [tex]3 \times 10^8\;m/s[/tex], then [tex]T_p[/tex] would be given by:
[tex]T_p=\frac{10}{3\times 10^8} \\\\T_p=0.03 \times 10^{-6}\;seconds[/tex]
Mathematically, the total time that is required to receive all objects through non-persistent HTTP is given by this expression:
[tex]T=(\frac{200}{150} +T_p+\frac{200}{150} +T_p+\frac{200}{150} +T_p+\frac{100000}{150} +T_p)+(\frac{200}{15} +T_p+\frac{200}{15} +T_p+\frac{200}{15} +T_p+\frac{100000}{15} +T_p)\\\\T=(\frac{200\;+\;200\;+\;200\;+100000}{150} +4T_p)+(\frac{200\;+\;200\;+\;200\;+100000}{15} +4T_p)\\\\T=(\frac{100600}{150} +4T_p)+(\frac{100600}{15} +4T_p)\\\\T=(670+4T_p)+(6706+4T_p)\\\\T=7377+8T_p[/tex]
Next, we would derive an equation for the total time needed for persistent HTTP connection:
[tex]T=[(\frac{200}{150} +T_p+\frac{200}{150} +T_p+\frac{200}{150} +T_p+\frac{100000}{150} +T_p)]+10(\frac{200+100000}{15} +2T_p)\\\\T=(670+4T_p)+10(\frac{100200}{15} +2T_p)\\\\T=(670+4T_p)+(6680 +20T_p)\\\\T=7350+24T_p[/tex]
In conclusion, we can deduce that the parallel downloads through persistent HTTP doesn't have significant gain over the non-persistent HTTP because it is less than 1 percent (1%).
Read more on persistent HTTP here: https://brainly.com/question/8156649
