CS 473

Homework : OoO and Cache

Solutions

OoO Execution

1. (10 points) Suppose the following code is to executed on a CDC 6600. Draw a timing chart showing the execution.

   X1 <- X2 / X3
   X7 <- X1 * X4
   X0 <- X7 * X1
   X4 <- X7 + X6
   X1 <- X0 + X4
   

Cache (30 points)

To make these problems tractable, we'll assume a byte-addressed computer with an eight bit address.

Consider the following cache access string:

0xb4, 0xb4, 0x13, 0x83, 0x84, 0xd8, 0xd9, 0xda, 0x87, 0x88, 0x6c, 0x8e, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x4c, 0x4d

For each of the following cache organizations, show the contents of the cache following each access. In each organization, the cache can hold a total of 16 bytes of data.

1. Direct-mapped, four-byte line

   Since the line size is four bytes, two bits are required for the offset into the cache line. Since there are sixteen bytes in the cache and a line is four bytes, there are four lines. Since it's direct-mapped, the number of sets is the same as the number of lines, so there are four sets, so the index field is two bits. This leaves four bits for the tag.

   Access	Fields			Tags in Blocks
   	Tag	Index	Offset	00	01	10	11
   b4	1011	01	00		1011
   b4	1011	01	00		1011
   13	0001	00	11	0001	1011
   83	1000	00	11	1000	1011
   84	1000	01	00	1000	1000
   d8	1101	10	00	1000	1000	1101
   d9	1101	10	01	1000	1000	1101
   da	1101	10	10	1000	1000	1101
   87	1000	01	11	1000	1000	1101
   88	1000	10	10	1000	1000	1000
   6c	0110	11	00	1000	1000	1000	0110
   8e	1000	11	10	1000	1000	1000	1000
   32	0011	00	10	0011	1000	1000	1000
   33	0011	00	11	0011	1000	1000	1000
   34	0011	01	00	0011	0011	1000	1000
   35	0011	01	01	0011	0011	1000	1000
   36	0011	01	10	0011	0011	1000	1000
   37	0011	01	11	0011	0011	1000	1000
   4c	0100	11	00	0011	0011	1000	0100
   4d	0100	11	01	0011	0011	1000	0100

2. Two-way set-associative, four-byte line, using LRU replacement

   There are two bits for the offset, as before. This time there are two lines in every set; since there are still four lines, we have two sets. This time we have five bits for the tag.

   Access	Fields			Tags in Blocks
   				0			1
   	Tag	Index	Offset		MRU			MRU
   b4	10110	1	00				10010	L
   b4	10110	1	00				10010	L
   13	00010	0	11	00010	L		10010	L
   83	10000	0	11	00010	R	10000	10010	L
   84	10000	1	00	00010	R	10000	10010	R	10000
   d8	11011	0	00	11011	L	10000	10010	R	10000
   d9	11011	0	01	11011	L	10000	10010	R	10000
   da	11011	0	10	11011	L	10000	10010	R	10000
   87	10000	1	11	11011	L	10000	10010	R	10000
   88	10001	0	00	11011	R	10001	10010	R	10000
   6c	01101	1	00	11011	R	10001	01101	L	10000
   8e	10001	1	10	11011	R	10001	01101	R	10001
   32	00110	0	10	00110	L	10001	01101	R	10001
   33	00110	0	11	00110	L	10001	01101	R	10001
   34	00110	1	00	00110	L	10001	00110	L	10001
   35	00110	1	01	00110	L	10001	00110	L	10001
   36	00110	1	10	00110	L	10001	00110	L	10001
   37	00110	1	11	00110	L	10001	00110	L	10001
   4c	01001	1	00	00110	L	10001	00110	R	01001
   4d	01001	1	01	00110	L	10001	00110	R	01001

3. Fully associative, one-byte line, using LRU replacement.

   Since the block is one byte, the width of the offset field is 0 (so, effectively, there is no offset field). As it's all one set, the width of the index field is also 0 (so, effectively, there is no index field). That leaves all eight bits as tag.

   In the following figure, I'm showing the blocks in LRU order, as if the data were moving in the cache. Of course, it would really remain in whatever block it got placed, and LRU counters would be maintained)

   Access	Contents
   b4	b4
   b4	b4
   13	13	b4
   83	83	13	b4
   84	84	83	13	b4
   d8	d8	84	83	13	b4
   d9	d9	d8	84	83	13	b4
   da	da	d9	d8	84	83	13	b4
   87	87	da	d9	d8	84	83	13	b4
   88	88	87	da	d9	d8	84	83	13	b4
   6c	6c	88	87	da	d9	d8	84	83	13	b4
   8e	8e	6c	88	87	da	d9	d8	84	83	13	b4
   32	32	8e	6c	88	87	da	d9	d8	84	83	13	b4
   33	33	32	8e	6c	88	87	da	d9	d8	84	83	13	b4
   34	34	33	32	8e	6c	88	87	da	d9	d8	84	83	13	b4
   35	35	34	33	32	8e	6c	88	87	da	d9	d8	84	83	13	b4
   36	36	35	34	33	32	8e	6c	88	87	da	d9	d8	84	83	13	b4
   37	37	36	35	34	33	32	8e	6c	88	87	da	d9	d8	84	83	13
   4c	4c	37	36	35	34	33	32	8e	6c	88	87	da	d9	d8	84	83
   4d	4d	4c	37	36	35	34	33	32	8e	6c	88	87	da	d9	d8	84