Calculating Network Link Utilization and Efficiency

Network Link Utilization Analysis

Suppose you want to convey 100,000,000 bits via a dedicated link at a speed of 10 Megabits/sec. The distance between the transmitter and receiver is 20,000 meters, and the signal propagation speed is 200,000 km/sec. Information is sent in blocks of 100,000 bits.

a) Stop-and-Wait ARQ Utilization

Calculate the exact link utilization, given that every time the computer receives a complete block, it takes 5 milliseconds to check for errors and save the block to disk before sending the ACK response. Utilization is defined as the total time taken to transmit useful information divided by the total time for transmission, error control, and flow set for that link. The number of bits in the ACK is negligible. Assume Stop-and-Wait ARQ (the transmitter awaits a response from the receiver before sending the next block).

Parameters:

• Link Speed: 10 Mbps
• Propagation Speed (Vp): 200,000 km/sec
• Distance: 20,000 meters
• Block Size: 100,000 bits
• Processing Time (Tproc): 5 msec

Calculation:

Ttx = Block / Speed = 0.01 sec = 10 msec
Tp = Distance / Vp = 0.0001 sec = 0.1 msec
Tack = 0 msec (negligible)
Total Time = Ttx + Tproc + Tack + 2Tp = 10 + 5 + 0 + 0.2 = 15.2 msec

Utilization (U) = Ttx / Total Time = 10 / 15.2 = 0.657 or 65.7%

b) Continuous ARQ and Window Size

Assume Continuous Selective Repeat ARQ with error-free transmission. Assume both the ACK time and processing time are negligible. Calculate the window size for maximum utility and determine if continuous ARQ is superior to Stop-and-Wait ARQ in this scenario.

Calculation:

Window Size (K) = 1 + 2(Tp / Ttx) = 1 + 2(0.1 / 10) = 1.02. In practice, the window size is 1.

Conclusion: Since the window size is effectively 1, the performance is identical to Stop-and-Wait ARQ. This occurs because the propagation delay is very small compared to the transmission time; therefore, continuous transmission provides no practical improvement in link utilization.

c) Efficiency with Packet Loss

If using Continuous Selective Repeat ARQ and one plot fails in every 80 plots, what is the maximum utility?

Calculation: Efficiency = (1 - Pf) = 1 - (1/80) = 1 - 0.0125 = 0.9875 or 98.75%