which sequence will be faster 1according to mips 2according

which sequence will be faster
1.according to mips
2.according to execution time

Solution

Compilers are being tested for 4GHz ie 4000MHz

MIPS = Millions of instructions per second

          = Number of instructions / Execution time * 10⁶

=IC / IC * CPI * clock cycle time * 10⁶

= IC * clock rate / IC * CPI * 10⁶

= clock rate / CPI * 10⁶

For first sequence,

CPI = (5*1 + 1*2 + 1*3) * 10⁶ / (5+1+1) * 10⁶  = 10 / 7

Execution time = (5+1+1) * 10⁶  * (10 / 7) * 1/(4000 * 10⁶) = 0.0025 seconds

MIPS = 4000 * 10⁶   / (10/7) * 10⁶  = 4000*7 / 10 = 2800

For second sequence,

CPI = (10*1 + 1*2 + 1*3) * 10⁶ / (10+1+1) * 10⁶  = 15 / 12

Execution time = (10+1+1) * 10⁶  * (15 / 12) * 1/(4000 * 10⁶) = 15 / 4000 = 0.00375 seconds

MIPS = 4000 * 10⁶   / (15/12) * 10⁶  = 4000*12 / 15 = 3200

So according to MIPS second sequence is faster and according to execution time first sequence is faster.