How to calculate end-to-end delay in this scenario

Question

Here is my question: We want to send a message of 150,000 bytes in size over the network. There are four jumps, each 20 km long, and operates at a speed of 100 Mbps. However, before sending, we divided the message into 1500 byte packets. What is the end-to-end message delay? Use the speed of light in copper = c = 2 * 10 ^ 8 m / s and round your answer to the nearest whole millisecond.

What I still have is 48 ms, which I calculated as follows: 1500 bytes = 12,000 bits 12,000 bits / (100x10 ^ 6) [100 Mbps] = 0.00012 with 20 km / (2 * 10 ^ 8) [speed of light in this equation] = 1e-7 Add them and you get 0.0001201s per hop x 4 hps = 0.0004804s → 48ms rounded to the nearest integer.

This seems to be wrong, and I cannot understand my mistake. My online course has no help, so I turn to SO. If someone can point me in the right direction, I would appreciate help.

Answer 1

Edit:

I think I finally understood. The network topology looks like this:

source - link#1 - router#1 - link#2 - router#2 - link#3 - router#3 - link#4 - dest

Let us first consider this from the point of view of the source. We send message package by package. As soon as we put the first packet in the wire, we send the second, then the third, etc. How long does it take to put all 100 packets in a wire?

100 * 1500B * 8(b/B) / 100 Mb/s = 12 ms

- , dest. , , # 100 №1, .

20km/(2 * 10^8 m/s) = 0.1 ms - to get to the router#1
1500B * 8(b/B) / 100 Mb/s = 0.12 ms - to put it on the link#2
20km/(2 * 10^8 m/s) = 0.1 ms - to get to the router#2
1500B * 8(b/B) / 100 Mb/s = 0.12 ms - to put it on the link#3
20km/(2 * 10^8 m/s) = 0.1 ms - to get to the router#3
1500B * 8(b/B) / 100 Mb/s = 0.12 ms - to put it on the link#4
20km/(2 * 10^8 m/s) = 0.1 ms - to get to the dest

0,76 , , # 1 . :

12 ms + 0.76 ms = 12.76 ms ~= 13 ms

:

, -, , , 4, , 4 . - , .

13 . :

4 * (1500B * 8(b/B) / 100 Mb/s + 20km/2 * 10^8 m/s)
+ (150,000/1500 - 1) * (1500 * 8 (b/B) / 100Mb/s) = 12.76 ms

Answer 2

4

S->r1->r2->r3->D

No. of packets is =15*10^4 / 1500 = 100 packets
T.D=> 15*10^2*8 / 100* 10^6 (this is for 1 packet) = 0.12ms
P.D=> 20*10^3 / 2* 10^8 = 0.1ms
Total is =>0.1+0.12 = 0.22ms
4 hops takes => 0.22*4=0.88ms
0.88ms is for 1st packet only

0.88 D. D, . , D 1 0,12 R2.

, 99 0,12 * 99 .// (N-1)

= > 0,88 + 0,12 * 99 = 12,76

→

= . (1- ) + TransmissionN - ( + + )

( = 0)

= . (1- ) + * (N-1)

PPT

Answer 3

"store-and-forward", .. , , .

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Answer 4

(p/r + l/c):

: 1500 byte * 8 = 12000 bit
: 100 Megabit * 1 000 000 = 100 000 000 bit/sec

12 000 bit/100 000 000 bit/sec = 0.00012 sec (0.12 ms)

: 20 000/(2 * 10 ^ 8) = 0.0001 sec (0.1 ms)

: 0.12 ms + 0.1 ms = 0.22 ms

4 , total time in ms = 4 * 0.22 ms = 0.88 ms

, , (, , )

= 150 000 byte - 1 500 byte = 148 500byte
(148500 * 8 =) 1 188 000 bit

, 1 188 000 bit/100 000 000 bit/sec = 0.01188 sec (**11.88ms**)

11.88 ms + 0.88 ms = 12.76 (13 - )

Answer 5

. - " ".

Answer 6

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Answer 7

, , . , 150 000 100 1500 , (150 000 * 8)/(100 * 10 ^ 6) , . . , , (1500 * 8)/(100 * 10 ^ 6) , , 1/100 . , , . , , . , (150 000 * 8)/(100 * 10 ^ 6) . (1500 * 8)/(100 * 10 ^ 6) , , , , , , . 99 , ((150 000 - 1500) * 8)/(100 * 10 ^ 6) . , 4 * (150 000 * 8)/(100 * 10 ^ 6) + 4 * (20 * 10 ^ 3)/(2 * 10 ^ 8) = 48 , 3 * (1500 * 8)/(100 * 10 ^ 6) + (150000 * 8)/(100 * 10 ^ 6) + 4 * (20 * 10 ^ 3)/(2 * 10 ^ 8) = 12,76 , , .