Ten 9600-bps lines are to be multiplexed using TDM. a. Ignoring overhead bits in the TDM frame, what is the total capacity required for synchronous TDM? b. Assuming that we wish to limit average TDM link utilization to 0.8, and assuming that each TDM link is busy 50% of the time, what is the capacity required for statistical TDM?

a.

Given Data :

Data rate of each Line (Channel) = 9600 bps

Total lines need to be multiplexed = 10

Solution:

Total Capacity of Synchronous TDM = Data rate of each line x Total Lines

                                                            =  9600 bps x 10

                                                            = 96,000 bps

                                                            = 96 Kbps

Explanation:

In time division multiplexing, multiple channels (Lines) can transmit data over the medium on different time slots. In synchronous TDM, equal time has been allocated to each channel to transmit their data over the medium. so the total capacity required for 10 lines synchronous TDM is 96Kbps.

b.

Given Data:

Data rate for each line (Channel) = 9600 bps

Total lines need to be multiplexed = 10

Average TDM link utilization = 80% = 0.8

Each link busy for the time = 50% = 0.5

Solution:

Capacity of Statistical TDM = data rate x number of Lines x each channel duration/Average utilization of TDM

Capacity of Statistical TDM = 9600 bps x 10 x 0.5/0.8

                                               =60,000 bps

                                               = 60 kbps

Explanation:

In TDM, if we want to limit the capacity of TDM and each channel that need to be multiplexed. we can limit the utilization of the TDM by limiting its transmission time of each channel and also the capacity of TDM.

the answer would be d

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