Question 2488 – KVS 30-12-2018 Part B
January 10, 2024UGC NET CS 2014 Dec-Paper-2
January 10, 2024Pipelining
| Question 14 |
We have two designs D1 and D2 for a synchronous pipeline processor. D1 has 5 pipeline stages with execution times of 3 nsec, 2 nsec, 4 nsec, 2 nsec and 3 nsec while the design D2 has 8 pipeline stages each with 2 nsec execution time How much time can be saved using design D2 over design D1 for executing 100 instructions?
| 214 nsec | |
| 202 nsec | |
| 86 nsec | |
| -200 nsec |
Question 14 Explanation:
k = total no. of stages
n = no. of instructions
Total execution time = (k+n-1) * maximum clock cycle
In case of D1:
k = 5
n = 100
Maximum clock cycle = 4 ns
Total execution time = (5+100-1) * 4 = 416
In case of D2:
k = 8
n = 100
Maximum clock cycle = 2 ns
Total execution time = (8+100-1) * 2 = 214
Starved time D2 over D1 = 416 – 214 = 202
n = no. of instructions
Total execution time = (k+n-1) * maximum clock cycle
In case of D1:
k = 5
n = 100
Maximum clock cycle = 4 ns
Total execution time = (5+100-1) * 4 = 416
In case of D2:
k = 8
n = 100
Maximum clock cycle = 2 ns
Total execution time = (8+100-1) * 2 = 214
Starved time D2 over D1 = 416 – 214 = 202
Correct Answer: B
Question 14 Explanation:
k = total no. of stages
n = no. of instructions
Total execution time = (k+n-1) * maximum clock cycle
In case of D1:
k = 5
n = 100
Maximum clock cycle = 4 ns
Total execution time = (5+100-1) * 4 = 416
In case of D2:
k = 8
n = 100
Maximum clock cycle = 2 ns
Total execution time = (8+100-1) * 2 = 214
Starved time D2 over D1 = 416 – 214 = 202
n = no. of instructions
Total execution time = (k+n-1) * maximum clock cycle
In case of D1:
k = 5
n = 100
Maximum clock cycle = 4 ns
Total execution time = (5+100-1) * 4 = 416
In case of D2:
k = 8
n = 100
Maximum clock cycle = 2 ns
Total execution time = (8+100-1) * 2 = 214
Starved time D2 over D1 = 416 – 214 = 202