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Andrew S. Tanenbaum Edition 5th Exercise 6 Question 37 (Page No. 609)

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Suppose that you are measuring the time to receive a segment. When an interrupt
occurs, you read out the system clock in milliseconds. When the segment is fully processed,
you read out the clock again. You measure 0 msec 270,000 times and 1 msec
730,000 times. How long does it take to receive a segment?
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Total Time= 270000 + 730000 = 1000000

T1 = 270000

T2= 730000

Average = T1+T2/Total Time

= (270,000*0 + 730,000*1 msec)/1,000,000

= 730 micro Seconds
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Andrew S. Tanenbaum Edition 5th Exercise 6 Question 36 (Page No. 609)
In a network whose max segment is 128 bytes, max segment lifetime is 30 sec, and has 8-bit sequence numbers, what is the maximum data rate per connection?
In a network whose max segment is 128 bytes, max segment lifetime is 30 sec, andhas 8-bit sequence numbers, what is the maximum data rate per connection?
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