# Andrew S. Tanenbaum Edition 5th Exercise 6 Question 39 (Page No. 609)

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To get around the problem of sequence numbers wrapping around while old packets
still exist, one could use 64-bit sequence numbers. However, theoretically, an optical
fiber can run at 75 Tbps. What maximum packet lifetime is required to make sure that
future 75-Tbps networks do not have wraparound problems even with 64-bit sequence
numbers? Assume that each byte has its own sequence number, as TCP does.

To avoid getting the same sequence numbered packet at the destination, the Wrap Around Time should be at least the lifetime of the packet, i.e WAT >= LT ,so the max LT can be WAT

Now bandwidth given= 75 Tbps ( I am assuming b means bits)

1s ------------  you can number (75* 2^40)/8 bytes

so time required to assign sequence number for 2^64 bytes is (8* 2^64)/ (75 * 2^40) sec which is 1789569.7 sec

So the maximum life time of the packet can be ~ 498 hours long !!!!!!!!

## Related questions

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A CPU executes instructions at the rate of 1000 MIPS. Data can be copied 64 bits at a time, with each word copied costing 10 instructions. If an coming packet has to be copied four times, can this system handle a 1-Gbps line? For simplicity, assume that all instructions, even those instructions that read or write memory, run at the full 1000-MIPS rate.