Question

itby important Q. Disc uss the features of RISC and CISC Architecture. dware implementations Mt &t M2 of the same instruction set. There are three classes F, I & N of instructions in the instruction The average CPl for the three instructi Class set. Miclock rate is 600 MHz, M2's clock cycle is 2ns. on classes on M1 & M2 are as follows: Comments Floating Point Integer Arithmetic Non-arithmetic CPI for M 5.0 2.0 2.4 CPI for M2 4.0 3.8 2.0 (a) What are the peak performances of Mi & M2 in MIPS? (b) If 50% of all instructions executed in a certain program are form class N, and the rest are divided equally among F and I, which machine is faster and by what factor? Designers of Mi plan to redesign the machine for better performance. With the assumptions of part (b), which of the following redesign options has the greatest performance impact and why? Using a faster floating point unit with double the speed (Class F CPI 2.5) Adding a second integer ALU to reduce the integer CPI to 1.20 Using faster logic that allows a clock rate of 750MHz with the same CPls. (c) 1. 2. 3.

Please Solve 1(c).

Answer 1

1 (C) . before starting the answer of this let us find the average CPI before new implementation

Type	CPI	PART	product
F	5	0.25	1.25
I	2	0.25	0.5
N	2.4	.50	1.2

Total Cpi = 2.95 with 600 MHz clock rate

now we will use the speed up factor to compare which new strategy will be good

$Speed up S =\frac{old execution time}{new execution time}$

old execution time = 2.95* 1/ (600 *10^-6)

1.) now just modify CPI of F to 2.5 so

Type	CPI	PART	product
F	2.5	0.25	0.625
I	2	0.25	0.5
N	2.4	.50	1.2

New Cpi = 2.325 new clock rate =600

new execution time = 2.325 * 1 / ( 600 * 10^-6)

Speed up S = 2.95 * 600 * 10^-6 / 2.325* 600 * 10^-6 = 1.27

2 .) now just modify CPI of I to 1.2 so

Type	CPI	PART	product
F	5.0	0.25	1.25
I	1.2	0.25	0.3
N	2.4	.50	1.2

New Cpi = 2.75 new clock rate =600

new execution time = 2.75 * 1 / ( 600 * 10^-6)

Speed up S = 2.95 * 600 * 10^-6 / 2.75* 600 * 10^-6 = 1.07

3.) now just update the clock speed so

Type	CPI	PART	product
F	5	0.25	1.25
I	2	0.25	0.5
N	2.4	.50	1.2

New Cpi = 2.95 with 750 MHz clock rate

new execution time = 2.95 * 1 / ( 750 * 10^-6)

Speed up S = 2.95 * 750 * 10^-6 / 2.95* 600 * 10^-6 = 1.25

so considering all the speedups we get largest speed up i.e. S =1.27 for the first approach why?

because in original the F class has largest CPI also the contribution in the CPI by F is 1.25 which is largest so if we anyhow modify this parameter then we may get the best possible result and which is done in the first approach that is why impact of first approach is maximum which leads us to a largest speedup

Also if we consider that in the 3rd approach we got 1.25 speedup which is near to the largest because the speed of all the instructions are increased .

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