made easy test series

Question

Consider Amit lives in Delhi, connected to the internet via a 100 Mbps connection retrieve a 250 KB webpage from server in Bangalore, where page contain 3  images of 500 KB each. Assume one way propagation delay is 75 ms and Amit’s access link is the bandwidth bottleneck for this connection. If T₁ is the time taken for the page with images to appear on Amit’s screen using nonpersistent HTTP and T₂ is time using persistent connection, then T = T₁ – T₂ will be ________  msec.

Comments

Refer this link to know about HTTP Non-Persistent & Persistent Connection:
https://www.geeksforgeeks.org/http-non-persistent-persistent-connection/

Answer 1

For non persistent we have to make the connection request after every object while in Persistent we can view all object in one go.
Here the total transmission time is nothing but the transmission time of page with all three images. 

Comments

awesome explaination .

Thanks.

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In non  persistent connection we have to close the connection also after receiving the object . Why we are not considering here that time . Also for connection establishment 3 way handshake is done which requires time of 1.5 RTT .

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I also think that this solution is not correct , for connection establishment it should take 1.5 RTT.

Also in diagram for non persistent connection , for better clarity all four connection establishment should be drawn beside one other and not after one other , if you are drawing after one other where does connection closure time goes?

so it should be like

connection 1 establishment

webpage transfer

connection 1 closure                    connection 2 establishment

                                                     image 1 transfer

                                                     connection 2 closure                     connection 2 establishment

                                                                                                          image 2 transfer

                                                                                                           connection 2 closure     

similarly for image 3

Answer 2

Given data rate $= 100\space Mbps$

One way propagation delay  $= 75\space ms$
Round Trip Time $RTT = 2$ x One way propagation delay  $= 2$ x $75 ms\space = 150\space ms$

 

In $non-persistent\space HTTP$, we need one TCP connection each for $each\space object$ sent. Here, we have to send $4$ objects $\rightarrow\space$ $1\space$ base file (web page) and $3$ images. 

Client Action	Server Action	Time requred
Request Connection	Accept Connection	1 RTT
Request Webpage	Accept Request Send Web page	1 RTT
Receive Webpage (Finds 3 objects)	Close connection	Transmission Time
Request Connection	Accept Connection	1 RTT
Request Image 1	Accept Request Send Image1	1 RTT
Receive Image 1	Close connection	Transmission Time
Request Connection	Accept Connection	1 RTT
Request Image 2	Accept Request Send Image 2	1 RTT
Receive Image 2	Close connection	Transmission Time
Request Connection	Accept Connection	1 RTT
Request Image 3	Accept Request Send Image 3	1 RTT
Receive Image 3	Close Connection	Transmission Time

Thus, the total time is $T1 = 4* 2 RTT +$ Total Data Transmission time

Transmission time $= \frac{Amount\space of\space data}{Data\space Rate} = \frac{250 KB + 3 * 500 KB}{100\space Mbps }= \frac{(250*8 + 3*500*8) Kb}{100\space Mbps}$
    $=  \frac{(2000 + 12000) . 10^3 }{ 100 * 10^6} \space s = 140\space ms$

Now $T_1 = 4*2*150\space  ms+ 140\space ms = 1340\space ms$

 

$Persistent\space HTTP$ is the idea of using a single TCP connection to send and receive multiple HTTP requests/responses. All objects will be sent with a single request here. 

Client Action	Server Action	Time Required
Request Connection	Accept Connection	1 RTT
Request Web page	Accept Request Send Web page	1 RTT
Receive Web page (Finds 3 objects)	Close connection	Transmission Time
Request all objects	Accept Request Send all objects	1 RTT
Receive all objects	Close connection	Transmission Time

So, total time here is $T_2 = 3* RTT +$ Total Data Transmission Time
        $= 3* 150\space ms + 140\space ms = 590\space ms$

 

Therefore $T_1 - T_2 = 1340\space ms - 590\space ms = 750\space ms$.

Comments

Connection termination takes only 1 RTT and the message is piggybacked with the last fragment(file in this case). If your doubt is not cleared, tell me specifically when and for which communication (persistent or non-persistent) so that i can reply clearly.

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Anyways valid point,  i should have included that in the table. Assumed that you know msg piggybacking.

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Yes,  but it wont make any significant difference if the data is not too large. For data involving less than 2 MB and times greater than or equal to milliseconds,  the difference is not obvious.

For example., in the given question data is 250 KB

250 x 10$^{3}$ x 8 = 2,000,000 bits

Where the actual calculation

250 x 2$^{10}$ x 8 = 2,048,000 bits

Yes, we will get error but it is not signifcant.

Options in those questions won't be that close in case of networking problems. (Involving transmission, propagation etc.). The convention is to take 10$^{3}$ for Kb, 10$^{6}$ for Mb etc in networking problems.

So it is your choice.