49. A channel has a bit rate of 20 kbps. The stop and wait protocol with frame size 4500 bits is used. The delay for error detection and sending ACK by the receiver is 0.25 seconds because of a fault. Find the maximum efficiency of the channel if the destination is 30000km away and the speed of the propagation of the signal is 2.8X108 m/s. Find the decrease in efficiency due to the fault
Question
- A channel has a bit rate of 20 kbps. The stop and wait protocol with frame size 4500 bits is used. The delay for error detection and sending ACK by the receiver is 0.25 seconds because of a fault. Find the maximum efficiency of the channel if the destination is 30000km away and the speed of the propagation of the signal is 2.8X108 m/s. Find the decrease in efficiency due to the fault
Solution
To solve this problem, we first need to calculate the transmission time (Tt), propagation time (Tp), and the total time (Ttotal).
- Calculate the Transmission Time (Tt): The transmission time is the time required to push all of the packet's bits into the link. This can be calculated using the formula: Tt = Frame size / Bit rate
Given: Frame size = 4500 bits Bit rate = 20 kbps = 20,000 bits per second
So, Tt = 4500 bits / 20,000 bits per second = 0.225 seconds
- Calculate the Propagation Time (Tp): The propagation time is the time required for a bit to travel from the sender to the receiver. This can be calculated using the formula: Tp = Distance / Speed of light
Given: Distance = 30000 km = 30,000,000 meters Speed of light = 2.8 x 10^8 m/s
So, Tp = 30,000,000 meters / 2.8 x 10^8 m/s = 0.107 seconds
- Calculate the Total Time (Ttotal): The total time is the sum of the transmission time, propagation time, and the time for error detection and sending ACK by the receiver.
So, Ttotal = Tt + Tp + Time for error detection and sending ACK
Given: Time for error detection and sending ACK = 0.25 seconds
So, Ttotal = 0.225 seconds + 0.107 seconds + 0.25 seconds = 0.582 seconds
- Calculate the Efficiency: The efficiency of the Stop and Wait protocol can be calculated using the formula: Efficiency = Tt / Ttotal
So, Efficiency = 0.225 seconds / 0.582 seconds = 0.386 or 38.6%
- Calculate the Decrease in Efficiency due to the Fault: The decrease in efficiency due to the fault is the difference between the efficiency without the fault and the efficiency with the fault.
Without the fault, the total time would be Tt + Tp, so the efficiency would be Tt / (Tt + Tp).
So, Efficiency without fault = 0.225 seconds / (0.225 seconds + 0.107 seconds) = 0.677 or 67.7%
So, the decrease in efficiency due to the fault = Efficiency without fault - Efficiency with fault = 67.7% - 38.6% = 29.1%
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