ECE 6913 Midterm Exam Practice Test Questions And Correct Answers
(Verified Answers) 2026 Q&A - NYU Tandon School of Engineering
1. The results of the SPEC CPU2006 bzip2 benchmark running on an AMD Barcelona has an
instruction count of 2.389E12, an execution time of 750 s, and a reference time of 9650 s
a. Find the CPI if the clock cycle time is 0.333 ns.
IC = 2.389T, ET=750s, RefET = 9650s
CPI = (ET * clock rate)/IC = 750s x 3.00GHz / 2.389T = 0.94
b. Find the SPECratio
SPECratio = T_ref/T = 9650s /750s = 12.86
c. Find the increase in CPU time if the number of instructions of the benchmark is increased by
10% without affecting the CPI
ET = CPI x IC x Tcycle = 0.94 x 2.389T x 1.1 x 0.333ns = 822s or an increase in ET of 9.67%
d. Find the increase in CPU time if the number of instructions of the benchmark is increased by
10% and the CPI is increased by 5%.
ET = CPI x IC x Tcycle = 0.94 x 1.05 x 2.389T x 1.1 x 0.333ns = 863s or an increase in ET of 15%
e. Find the change in the SPECratio for this change
SPECratio = T_ref/T = 9650s /863s = 11.18
f. Suppose that we are developing a new version of the AMD Barcelona processor with a 4 GHz
clock rate. We have added some additional instructions to the instruction set in such a way that
the number of instructions has been reduced by 15%. The execution time is reduced to 700 s and
the new SPECratio is 13.7. Find the new CPI.
ETnew = 700s = CPIX x 2.389T x 0.85 x 0.250ns
So, CPIX = 700/507.66 = 1.38
g. This CPI value is larger than obtained in part a. as the clock rate was increased from 3 GHz to 4
GHz. Determine whether the increase in the CPI is similar to that of the clock rate. If they are
dissimilar, why?
CPI increased by 1.38/0.94 = 1.46
Clock rate increased by 4/3 = 1.33
ECE 6913 Midterm Exam
, Both – Instruction Count and the Clock Cycle time were lowered as a result of adding instructions
to the instruction set.
Since each instruction, on average, is accomplishing more work, the reduced cycle time is likely
to require the average instruction to complete its task using more cycles and hence an increase
in the average CPI of the Benchmark
h. By how much has the CPU time been reduced?
700s / 750s = 0.933 or by 6.67%
2. Assume a program requires the execution of 50 ×106 FP instructions, 110 ×106 INT
instructions, 80 ×106 L/S instructions, and 16 ×106 branch instructions. The CPI for each type of
instruction is 1, 1, 4, and 2, respectively. Assume that the processor has a 2 GHz clock rate.
a. By how much must we improve the CPI of FP instructions if we want the program to run two
times faster?
The total number of clock cycles for the program is given by:
Total Clock Cycles = (CPIfp × #FP instructions) + (CPIint × #INT instructions) + (CPIl/s × #L/S
instructions) + (CPIbranch × #branch instructions)
Substituting the values given:
Total Clock Cycles = (1 × 50 × 10^6) + (1 × 110 × 10^6) + (4 × 80 × 10^6) + (2 × 16 × 10^6)
= 50 × 10^6 + 110 × 10^6 + 320 × 10^6 + 32 × 10^6 = 512 × 10^6 cycles
To run the program twice as fast, we need to reduce the total clock cycles to:
New Total Clock Cycles = 512 × 10^ = 256 × 10^6 cycles
If we improve only the FP instructions, we would still need to account for the other instructions
(INT, L/S, and Branch), which use a total of:
Non-FP Clock Cycles = (1 × 110 × 10^6) + (4 × 80 × 10^6) + (2 × 16 × 10^6) = 110 × 10^6 + 320 ×
10^6 + 32 × 10^6 = 462 × 10^6 cycles
This exceeds the target of 256 million cycles, even without considering FP instructions.
Therefore, improving the CPI of FP instructions alone cannot reduce the total clock cycles by
half. Hence, the program cannot run two times faster by improving the CPI of FP instructions
alone.
ECE 6913 Midterm Exam
(Verified Answers) 2026 Q&A - NYU Tandon School of Engineering
1. The results of the SPEC CPU2006 bzip2 benchmark running on an AMD Barcelona has an
instruction count of 2.389E12, an execution time of 750 s, and a reference time of 9650 s
a. Find the CPI if the clock cycle time is 0.333 ns.
IC = 2.389T, ET=750s, RefET = 9650s
CPI = (ET * clock rate)/IC = 750s x 3.00GHz / 2.389T = 0.94
b. Find the SPECratio
SPECratio = T_ref/T = 9650s /750s = 12.86
c. Find the increase in CPU time if the number of instructions of the benchmark is increased by
10% without affecting the CPI
ET = CPI x IC x Tcycle = 0.94 x 2.389T x 1.1 x 0.333ns = 822s or an increase in ET of 9.67%
d. Find the increase in CPU time if the number of instructions of the benchmark is increased by
10% and the CPI is increased by 5%.
ET = CPI x IC x Tcycle = 0.94 x 1.05 x 2.389T x 1.1 x 0.333ns = 863s or an increase in ET of 15%
e. Find the change in the SPECratio for this change
SPECratio = T_ref/T = 9650s /863s = 11.18
f. Suppose that we are developing a new version of the AMD Barcelona processor with a 4 GHz
clock rate. We have added some additional instructions to the instruction set in such a way that
the number of instructions has been reduced by 15%. The execution time is reduced to 700 s and
the new SPECratio is 13.7. Find the new CPI.
ETnew = 700s = CPIX x 2.389T x 0.85 x 0.250ns
So, CPIX = 700/507.66 = 1.38
g. This CPI value is larger than obtained in part a. as the clock rate was increased from 3 GHz to 4
GHz. Determine whether the increase in the CPI is similar to that of the clock rate. If they are
dissimilar, why?
CPI increased by 1.38/0.94 = 1.46
Clock rate increased by 4/3 = 1.33
ECE 6913 Midterm Exam
, Both – Instruction Count and the Clock Cycle time were lowered as a result of adding instructions
to the instruction set.
Since each instruction, on average, is accomplishing more work, the reduced cycle time is likely
to require the average instruction to complete its task using more cycles and hence an increase
in the average CPI of the Benchmark
h. By how much has the CPU time been reduced?
700s / 750s = 0.933 or by 6.67%
2. Assume a program requires the execution of 50 ×106 FP instructions, 110 ×106 INT
instructions, 80 ×106 L/S instructions, and 16 ×106 branch instructions. The CPI for each type of
instruction is 1, 1, 4, and 2, respectively. Assume that the processor has a 2 GHz clock rate.
a. By how much must we improve the CPI of FP instructions if we want the program to run two
times faster?
The total number of clock cycles for the program is given by:
Total Clock Cycles = (CPIfp × #FP instructions) + (CPIint × #INT instructions) + (CPIl/s × #L/S
instructions) + (CPIbranch × #branch instructions)
Substituting the values given:
Total Clock Cycles = (1 × 50 × 10^6) + (1 × 110 × 10^6) + (4 × 80 × 10^6) + (2 × 16 × 10^6)
= 50 × 10^6 + 110 × 10^6 + 320 × 10^6 + 32 × 10^6 = 512 × 10^6 cycles
To run the program twice as fast, we need to reduce the total clock cycles to:
New Total Clock Cycles = 512 × 10^ = 256 × 10^6 cycles
If we improve only the FP instructions, we would still need to account for the other instructions
(INT, L/S, and Branch), which use a total of:
Non-FP Clock Cycles = (1 × 110 × 10^6) + (4 × 80 × 10^6) + (2 × 16 × 10^6) = 110 × 10^6 + 320 ×
10^6 + 32 × 10^6 = 462 × 10^6 cycles
This exceeds the target of 256 million cycles, even without considering FP instructions.
Therefore, improving the CPI of FP instructions alone cannot reduce the total clock cycles by
half. Hence, the program cannot run two times faster by improving the CPI of FP instructions
alone.
ECE 6913 Midterm Exam
