CS 422 - HOMEWORK 3 SOLUTION			Page 1
			----------------------------

Version 2, Nov. 18, 1998

Problem 1
---------
Let	C = cache hit ratio (probability of a cache hit) = 0.9
	M = memory hit ratio = 0.6
	Teff = average time to load a word (effective read time)
	Tcache = time to load a word from cache to CPU = 20 ns
	Tmemory = time to load a word from memory to cache = 60 ns
	Tdisk = time to load a word from disk to memory = 12 ms
	TmemoryEff = effective memory load time
Then,

	Teff	=  C Tcache + (1-C) TmemoryEff
		=  C Tcache + (1-C) [ M Tmemory + (1-M) Tdisk ]
		=  0.9 (20 ns) + 0.1 [ 0.6 (60+20) ns
					+ 0.4 (12 ms + 60 ns + 20 ns) ]
		=  18 ns + 0.1 [ 48 ns + 4.8 ms ]  =  18 ns + 4.8 ns + 480 us
		=  480 us

Problem 2
---------
								    10
Number of entries in a 1-page page table  =  4 KB / 4  =  1,024  = 2

Number of bytes accessible from a 1-page page table = 1 K (4 KB) = 4 MB

							      20
Number of entries which can be stored in 4 MB = 4 MB / 4 B = 2

Each additional level in the page table increases the number of entries
		10                                    64
by a factor of 2  .  A 64-bit address space contains 2   bytes.  So,

we need to find the minimum integer L such that:

	 10L       64           52
	2     >=  2  / 4 K  =  2

So, L = 6.

Problem 3
---------
a)  Define	R0
		R1

		R0 <-- 0
		R1 <-- value of n
		R2 <-- address of A
		R3 <-- address of B
		R4 <-- address of C
	L2:	if (R0 >= R1) goto L1
		R5 <-- *R3		// load indirect B[i]
		R5 <-- R5 + *R4		// add indirect C[i]
		R5 --> *R2		// store indirect A[i]
		R0 <-- R0 + 1
		R2 <-- R2 + 4		// update pointers
		R3 <-- R3 + 4
		R4 <-- R4 + 4
		goto L2
	L1:	...

b)  Suppose the code is in page 0 and the arrays are in consecutive pages
    in the order A, B, and then C.  Since each array is 1 page long, A, B,
    and C are in pages 1, 2, and 3 respectively.

    1)	Each instruction fetch references a page
    2)	Each indirect memory access referenes a page

    So, the approximate sequence of page references once we enter the loop
    is:  0, 2, 0, 3, 0, 1, 0, 0, 0, 0, 0, ... repeats ...

Problem 4
---------
a)
    1)	Optimal:  Unknown because we don't know the future page references
    2)	FIFO:  Frame 3 because it has been in memory the longest (loaded first)
    3)	LRU:  Frame 1 because it is the least recently referenced (oldest
		reference time)
    4)	Clock:  It's not clear how we got to this state since all frames have
		been referenced in the interval 160-163.  The use bits must
		have been reset at time 160.  Either frame 0 or 1 since the
		use bit is 0.  But is probably frame 0 since it has been
		loaded the longest.
    5)	Enhanced Second Chance:  Frame 1 since its use and modify bits are both
		0.

b)  The reference string is:  1, 2, 0, 3, 4, 0, 0, 0, 0, 2, 4, 2, 1, 0, 3, 2
    where the first 4 are from the table.  Below, we show the window as it
    shifts to the right.

    1, 2, 0, 3, 4, 0, 0, 0, 0, 2, 4, 2, 1, 0, 3, 2
    ----------
       ---------F
          ----------
             ----------
                ----------
                   ----------
                      ---------F
                         ---------F
                            ----------
                               ---------F
                                  ---------F
                                     ---------F
                                        ----------
   So, 6 faults.