A a factor in the delay of a store and forward packet...

a) A factor in the delay of a store-and-forward packet-switching system is how long it takes to store and forward a packet through a switch. If switching time is (10* MAXIMUM number out of the 5 numbers of your student exam ID which is 45JSU) μ sec, is this likely to be a major factor in the response of a client-server system where the client is in New York and the server is in California (the distance is around 3928 km)? Assume the propagation speed in copper and fiber to be 2/3 the speed of light in vacuum and all communication links are 80 MBps (bytes/sec). How you could define that such a switching time could be a major factor.

Transcript

00:01 Hello students, so here are the answers of the question.

00:03 Find the propagation delay if the data travels at the speed of the light 2 .3 into 10 raised to power 8 meter per second.

00:09 So the propagation delay is the time it takes for a signal to travel from one point to another.

00:16 In this case, the signal needs to travel 6000 km to the satellite, then another 10 ,000 km to the other server.

00:25 For the total distance of 16000, the speed of the light is 2 .3 into 10 raised to power 8 meter per second.

00:32 So we can calculate the propagation delay as follows.

00:35 Propagation delay is equal to distance divided by time.

00:39 So here distance is 16000 because of 10 ,000 and 6000.

00:47 So when we combine these two, when we add these two, we get 16000.

00:52 So this is the total distance and here speed is also given that is 2 .3 into 10 raised to power 8 meter per second.

01:05 So the propagation delay will be when we calculate 16000 divided by 2 .3 into 10 raised to power 8.

01:14 So this will give this value 0 .07 second.

01:17 Therefore the propagation delay is 70 millisecond.

01:21 Now the second question is find the number of the bits in transmit during the propagation delay.

01:28 The answer here is the data is entering the network at 100 megabits per second, which means 100 million bits are being transmitted per second.

01:38 During the propagation delay of 70 milliseconds, the number of the bits, number of bits in transient is number of the bits in transient.

01:47 We can use this formula bit rate into the propagation delay.

01:51 So here the bit rate is 100 megabits per second.

01:56 So 100 into 10 raised to power 6 bits per second and time here is 0 .07 because 70 millisecond.

02:04 So by putting these value, we get 7 into 10 raised to power 6 bits.

02:07 So number of the bits in transient is 7 into 10 raised to power 7.

02:11 So therefore there are 7 million bits in the transient during the propagation delay.

02:18 Now the question is stored on a flash memory devices 200 megabyte message to be transmitted by an email from one server to another passing three nodes of a connection network.

02:29 This network forces packages to be the size of 10 kb excluding a packet header of 10 bytes.

02:37 Nodes are 400 miles apart and the servers are 50 miles apart from their corresponding node.

02:44 All transmission links of the type 100 megabits per second.

02:47 The processing time of each node is 0 .2 second.

02:52 Now find the propagation delay per packet between the server and the node between the nodes.

02:57 So the propagation delay is the time it takes for the signal to travel from one point to another...

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