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