Suppose we have a simple circuitswitched network consisting

Suppose we have a simple circuit-switched network consisting of four switches and four links. All links in the network have a capacity of 6 Mbps. Each user requires a fixed rate R=150 Kbps when transmitting. However, a given user is active only 10% of the time. How many users can be supported with this network? Assume FDM is used to support the maximum number of users. If the time to set up a circuit is 50msec, how long would it take a user to transmit a file of size 640KB? If there are a total of 100 users, and if packet switching is used, what is the probability that 20 users are transmitting simultaneously?

Solution

Cicuit switching uses dedicated links between communication end points . eg . Telephones .

As a cicuit swiched network uses dedicated channels for transmission therefore the network would remain idle

for 90% of the time for all users .

a. Assuming uniform distribution of users over a period of time : ( Approximating : 1 Mbps = 1000 Kbps )

Over a link of 6 Mbps using FDM of 150 kbps for each user : 6000 / 150 = 40 users can be supported simultaneously

For each node two links are available so available bandwidth = 6 * 2 = 12 Mbps , So 40 * 2 = 80 simultaneous users

( If Statistical TDM is used then 80 * 10 = 800 users can be supported as users are active only 10% of time )

80 users   ( with FDM )

b. Total time taken = Time taken to setup connection + Time taken to transmit data .

Time taken to setup connection = 50 msec .

Time taken to transmit data = Total data / Transfer speed = 640 / 150 = 4.267sec

Total time taken = 4.267 + 0.05 sec = 4.272 sec

c . Using Binomial Distribution : ¹⁰⁰C₂₀ * (0.1)²⁰ * (1 0.1)^100-20  . Calculating would give the probability value .

where 0.1 is the given probability of each user being active .