This document discusses two C code implementations of nested loops to calculate c(J) = c(J) + a(I, J)*b(J), compares their performance, and draws a conclusion. Code A has the outer loop iterating over I and the inner over J, while Code B is the opposite. The author implemented both codes, used a memory profiler, and found Code A was almost twice as fast as Code B. They conclude Code A is the faster of the two implementations.

1. CSE 661 QUESTION 8
For the given two pseudo codes, I have written C codes which are as follows:
LOOP A:
For I = 1, 1000
For J = 1, 1000
c(J) = c(J) + a(I, J)*b(J)
CODEA:
LOOP B:
For J = 1, 1000
For I = 1, 1000
c(J) = c(J) + a(I, J)*b(J)
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2. CSE 661 QUESTION 8
CODE B:
Also, I have used a memory profiler for the memory analysis of these codes:
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3. CSE 661 QUESTION 8
And the corresponding outputs are:
Conclusion:
We can conclude that Code A is faster than Code B. The execution time of Code A is almost half of
what Code B takes to execute.
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