09/30/2006

A student wrote, asking about problem 2(b) on the last problem set.
BUT, I was heading out the door to catch a flight, so I asked Veynu
to respond.  As we always do, if the response carries information
that could be useful to others, we prefer to share it with you all.

Yale Patt

        > From narasima@ece.utexas.edu Thu Sep 28 03:05:36 2006
        > Subject: Re: Problem Set
        > Date: Thu, 28 Sep 2006 03:05:37 -0500
        >
        > Hello ********, I am responding on behalf of Dr. Patt 
        > since he is out of town for a few days.
        >
        > > Prof. Patt,
        > >
        > > On question #2 letter 'b', should it read from 12 to 60 
        > > instead of 4 to 60?
        >
        > No, it should stay as it is, 4 to 60 is correct.
        >
        > > 4 is the number of address lines, but there are 
        > > 12 locations since it has an addressability of 3 bits.
        >
        > No, 4 is NOT the number of address lines.  The number of 
        > address lines refers to how many bits wide the address is.  
        > In the example memory from figure 3.21 in the book, the number 
        > of address lines is 2 which is represented by A[1:0].  So, 
        > with 2 bits of address we can uniquely identify 2^2 = 4 memory 
        > locations.  Another way of saying this is that the "address
        > space" of the memory is 4 locations.  So in this example memory, 
        > there are 4 memory locations, not 12 as you said above.  Within 
        > each of the 4 memory locations, 3 bits are stored, so we would 
        > say the memory has an "addressability" of 3 bits.
        >
        > I hope this clears things up, if it is still unclear please 
        > see a TA in office hours.
        >
        > > Thanks for yout time,
        > > <<name withheld ...>>
        >
        > You are welcome.
        >
        > --Veynu