given the program below llll tt1 0 question 2 10 points llll tt2 10 ll bbbbbb tt1 tt2 eeeelltt a

Given the program below llll tt1 0 question 2 10 points llll...

Given the program below, 𝑙𝑙𝑙𝑙 $𝑡𝑡1, 0 Question 2 (10 points). 𝑙𝑙𝑙𝑙 $𝑡𝑡2, 10 𝐿𝐿: 𝑏𝑏𝑏𝑏𝑏𝑏 $𝑡𝑡1, $𝑡𝑡2, 𝐸𝐸𝐸𝐸𝑙𝑙𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎𝑙𝑙 $𝑡𝑡1, $𝑡𝑡1, 1 jL Exit: 1). How many static instructions in the program? How many dynamic instructions in the program? [“Exit” is a label not an instruction] 2). Compare the performance of the following two processors: P1: 1GHz, with a CPI of 1.2 for the given program P2: 2GHz, with a CPI of 1 for the same program Calculate the CPU execution time of these two processors on the given program. Which processor is faster in executing the given program? How much faster?

Transcript

00:01 We need to calculate average cpi every cpi that is cycle per instruction for program a on computer.

00:12 C1 and computer c2 as well as the execution time.

00:17 We need to calculate.

00:18 So first let's calculate the average cpi.

00:21 So first for average cpi for program a or computer c1.

00:29 So the formula is and we will put the values average cpi is equals to percentage of type 1 instruction, which is 40 % multiply by cycle of type 1 2 plus percentage of type 2 instruction, which is 60 % multiply by type of type cycle of type 2 3 divided by 100.

00:51 So when we sort this out, we got 0 .8 plus 1 .8 divided by 100, which is 2 .6 divided by 100.

01:02 So here it is 0 .026.

01:06 Now, we need to calculate the average cpi for program a on computer c2.

01:15 The formula is same.

01:17 So for the first percentage type instruction 40 % multiply by cycle 2 plus percentage of type 2 instruction 60 % and cycle of type 2 is also true divided by 100.

01:30 Here we got 0 .8 plus 1 .2 divided by 100 2 .0 divided by 100.

01:40 So it will be 0 .02.

01:43 Now, let's calculate the execution time for program a or both both the computers.

01:49 So here now we need to calculate the execution time for program a on computer c1.

01:58 So execution time for c1 will be.

02:02 Dynamic instructions multiply by average cpi divided by frequency.

02:06 So dynamic instruction will be 8 multiply by 10 to the power 9 average cpi for c1 is 0 .026 divided by frequency.

02:17 So that will will be 4 .0 gets when we sort this out.

02:22 We got 208 triple 0 triple 0 divided by 4 triple 0 triple 0 triple 0 by calculating this we got 0 .052 seconds.

02:40 Now execution time for program a on compute for program b.

02:48 Basically for program b on computers c1...

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