hw1 009389081
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Problems: P5, P6, P9, P18, P24, P28, P30.
P5:
(a) Given,The caravan has 10 cars
The propagation speed (R) is 100 km/hr.
Number of tolls are 3. A tollbooth services a car at a rate of one car every 12 seconds.
The caravan travels 150 km (L).
The propagation delay is L/R = 150/100 = 1.5 hours = 90 minutes.
The transmission delay is 12 * 10 = 120 seconds for every tollbooth
So, for 3 tollbooths it is 120 seconds * 3 = 6 minutesThe transaction delay is 6 minutes.End-to-End delay is Propagation delay + Transaction delay = 90 + 6 = 96 minutes
789 Given,
The caravan has 8 cars
The propagation delay is the same, which is 90 minutes.
The transmission delay is 12 * 8 = 96 seconds for every tollbooth.So, for 3 tollbooths it is 96 seconds * 3 = 288 seconds = 4 minutes 48 seconds.
End-to-End delay is Propagation delay + Transaction delay = 90 + 4 min 48 sec = 94 minutes and 48
sec
100 km 100 km
Ten-carcaravan
Tollbooth
Tollbooth
100 km 100 km
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P6:
(a)
The Propagation delay dprop=!
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seconds
(b)
The Transmission delay dtrans=!
! seconds
(c)
End to End delay is!
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! Seconds
(d) At t=0, A begins to transmit the packet. At t= dtrans, last bit of the packet is just leaving the host
A.
(e) If dprop > dtrans, at t= dtrans, the first bit will be in the link and has not reached the host B.
(f) If dprop < dtrans, at t= dtrans, the first bit has reached host B.
(g) Given that dprop = dtrans, so,!
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P9:
The example is provided with a 1 Mbps link, the users are generating data at a rate of 100
kbps when busy, but are busy generating data only with probability p = 0.1. Now 1Mbps is replaced
by 1Gbps link.
(a)
The maximum number of users that can be supported simultaneously under circuit switching
are N which is
For 1Mbps=10 Users are supported, for 1Gbps(1000Mb approx.)=10^4 are supportedwhich is equal to 10,000.
(b)
Probability that N users are sending data
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Probability that more than N users are sending data.
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So the probability that more than N users sending data is
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!!!!! !!!!!
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