A simple medium access control protocol would be to use a fixed assignment time

Question

A simple medium access control protocol would be to use a fixed assignment time division multiplexing (TDM) scheme. Each station is assigned one time slot per cycle for transmission. For the bus, the length of each slot is the time to transmit 100 bits plus the ene-to-end propagation delay. For the ring, assume a delay of 1 bit time per station, and assume that a round-robin assignment is used. Stations monitor all time slots for reception. Assume a propagation time of 2×10^8 m/s. For N stations, what is the throughput per station for

a. A 1-Km, 10 Mbps baseband bus

b. A 10-Mbps ring with a total length of 1Km

Explanation / Answer

Efficiency=useful time/total time

=transmission time/transmission +propagation time

For each station slot time is tx + tp(transmission time+prop. delay)

tx = 100b/10Mbps = 10s

tp= d/v= Propagation delay=1km/2*108ms = 5s

So slot time is 15s

If there are N stations then total cycle time is 15N sec

efficiency will be useful time/total time ie (transmission time/total time) = (10/15N)

throughtput is eff*bandwidth => (10/15N)*10Mbps = (2/3)Mbps.

this is for N stations by assuming N as 10

Answer to (b)

Performance of Token Rings

Parameters and Assumption:

•End-to-end propagation delay                                a

•Packet transmission time                           1

• Number of stations                                    N

• Assume that each station always has a packet waiting for transmission

• Note: The ring is used either for data transmission or for passing the token.

Define:

• T1 = Average time to transmit a frame. Per assumption, T1=1

• T2 = Average time to pass the token

Maximum Throughput = Frame Time/ Frame Time + over head =T1/T1+T2

Effect of propagation delay on throughput:

• Case 1: a < 1 (Packet longer than ring)

• T2 = time to pass token to the next station = a/N

S = 1/ (1+a/N)

• Case 2: a > 1 (Packet shorter than ring)

• Note: Sender finishes transmission after T1 = 1, but cannot release the token until the token returns.

T1 + T2 = min (1, a) + a/N

S = 1/ (a+a/N)

note- the above explanation for both a & b can help to solve the given query.

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