TTTK 2833 - Assignment 2 (Answer)

Answer

1. (a) 1 / 500 mhz = 1sec / 500 x 10⁶ cycles = 2 x 10^-9 sec per cycle = 2ns per cycle

(b) 1 / 10.5 ns = 1cycle / 10.5 x 10^-9 sec = 0.9524 x 10⁸ cycles per sec = 95.24 x 10⁶= 95.24 MHz

2. (a) 221 = 2 ¹ x 2²⁰ = 2048 KWords

(b) 512 M = 2 ⁵⁶ , Main memory address is 56 bits.

3. (a) 64K = 2¹⁶, 16 bits in a main memory address.

(b) A memory block is equal to 8 words, and 216 ÷ 23 = 216-3 = 213. Then is 213 = 8K blocks in

main memory.

(c)

Tag	Line	Word
11	2	3

(d)

	Tag	Line	Word
i. 73B2	01110011101	10	010	line 2 word 2
ii. DF45	11011111010	00	101	line 0 word 5
iii. 8C7C	10001100011	11	100	line 3 word 4
iv. 07E9	00000111111	01	001	line 1 word 1
v. 318B	00110001100	01	011	line 1 word 3

4. The Main memory is 128 × 4K words = 2⁷x 2¹² = 2¹⁹ words. Means main memory address is 19 bits long. A cache line is the size of a memory block, then a cache line is 128 = 27 words and the word field is 7 bits. If there are 64 cache lines is divided into 4 line sets, then is 64 ÷ 4 = 16 sets. The set field is 4 bits. This leaves 8 bits for the tag field.

Tag	Line	Word
8	4	7

5. (a) The word = 4 bits. Each cache line = 16 bytes: a line = 4 words; a word = 4 bytes.

The cache is 4-way set associative, 4 lines to aset. Then a set is 4 lines × 16 bytes per line = 64 bytes, and 16KB ÷ 64 bytes = 2¹⁴ ÷ 2⁶ = 2⁸ sets. The set field, therefore is 8 bits. The main

memory address is 32 bits ,then the tag field is 32 - (8 + 4) = 20 bits.

Tag	Line	Word
20	8	4

6 (b)

3	A	2	C	6	F	0	D
0011	1010	0010	1100	0110	1111	0000	1101

Word: 1101b = 13d = Dh. If tmachine is byte-addressable, means byte 1 from word 3.

Set: 11110000 = 240d = F0h

Tag: 00111010001011000110b = 238278d = 3A2C6h