Device FIFO queue problem

A device boots with an empty FIFO queue. In the first 400 ns period after startup, and in each subsequent 400 ns period, a maximum of 80 words will be written to the queue. Each write takes 4 ns. A worker thread requires 3 ns to read a word, and 2 ns to process it before reading the next word. What is the shortest depth of the FIFO such that no data is lost?

While a perfectly optimal solution is complex, an interviewer is most interested in how you approach the problem.
First, note that writes do not have to be evenly distributed within a period. Thus a likely worst case is 80 words are written at the end of the first period, followed by 80 more at the start of the next.
Note that the maximum write rate for a full period is exactly matched by a full period of processing (400 ns / ((3 ns + 2 ns)/process) = 80 processed words/period).
As the 2nd period in our example is fully saturated, adding writes from a 3rd period would not add additional stress, and this example is a true worst case for the conditions.
For an estimate of maximum queue size, notice that these 160 writes take 640 ns (160 writes * 4 ns / write = 640 ns), during which time only 128 words have been read (640 ns / ((3 ns + 2 ns) / word) = 128 words). However, the first read cannot start until the first write has finished, which fills an extra slot in the queue.
Also, depending on the interactions between read and write timing, a second additional slot may be necessary to ensure a write does not trash the contents of a concurrently occurring read. Thus, a safe estimate is that the queue must be at least 34 words deep (160 – 128 + 1 + 1 = 34) to accommodate the unread words.


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