1 HW Solutions

1 EGR 321 --- Computer Organization Homework 1 Solutions California Baptist University Please typeset your answers to the following homework problems. All problems are from the ARM Edition of the text. 1. Exercise 1.1 in text (page 54). Laptop computers, tablet computers, supercomputers, embedded microcontrollers 2. Exercise 1.3 in text (page 55). Step 1. The compiler translates a program written in a high-level programming language into instructions in assembly language. Step 2. The assembler then translates the symbolic version of the instructions in assembly language into the binary machine language program. 3. Exercise 1.5 in text (page 55). (a) P2 has the highest performance. The following is the reason. Assuming that we use instructions per second as an indication of performance, we have performance P1 = 3 × 10 9 /1.5 = 2 × 10 9 (instructions/sec) performance P2 = 2.5 × 10 9 /1.0 = 2.5 × 10 9 (instructions/sec) performance P3 = 4 × 10 9 /2.2 = 1.818 × 10 9 (instructions/sec) Note: We may also first calculate the CPU time and then note performance P1 / performance P2 = CPU time P2 / CPU time P1 By using the formula CPU Time = No. of instr. x CPI x Clock cycle time = No. of instr. x CPI / Clock rate We may calculate the CPU time for each processor (assuming No. of instr. as a given constant I). (b) Note that CPU time = No. of cycles x Clock cycle time = No. of cycles / Clock rate and hence No. of cycles = CPU time x Clock rate No. of cycles P1 = 10 × 3 × 10 9 = 30 × 10 9 No. of cycles P2 = 10 × 2.5 × 10 9 = 25 × 10 9 No. of cycles P3 = 10 × 4 × 10 9 = 40 × 10 9 Next note that CPU Time = No. of Instr. x CPI x Clock cycle time = No. of instr. x CPI / Clock rate and hence No. of instr. = No. of cycles/CPI No. of instr. P1 = 30 × 10 9 /1.5 = 20 × 10 9 No. of instr. P2 = 25 × 10 9 /1.0 = 25 × 10 9 No. of instr. P3 = 40 × 10 9 /2.2 = 18.182 × 10 9 (c) CPU time new = CPU time old × 0.7 = 7 s CPI new = CPI old × 1.2, then the new CPIs are CPI P1 = 1.5 x 1.2 = 1.8, CPI P2 = 1.0 x 1.2 = 1.2, CPI P3 = 2.2 x 1.2 = 2.64

2 Now using the formula Clock rate = No. of instr. x CPI / CPU time we have Clock rate P1 = 20 × 10 9 × 1.8/7 = 5.143 GHz Clock rate P2 = 25 × 10 9 × 1.2/7 = 4.286 GHz Clock rate P3 = 18.182 × 10 9 × 2.64/7 = 6.857 GHz 4. Exercise 1.6 in text (page 55). P2 is faster. The following is the reason. Class A: 10 5 instr. Class B: 2 × 10 5 instr. Class C: 5 × 10 5 instr. Class D: 2 × 10 5 instr. CPU time = No. of instr. × CPI / Clock rate For P1: CPU time for class A = 10 5 x 1 / 2.5 x 10 9 = 0.4 × 10 −4 s CPU time for class B = 2 x 10 5 x 2 / 2.5 x 10 9 = 1.6 × 10 −4 s CPU time for class C = 5 x 10 5 x 3 / 2.5 x 10 9 = 6 × 10 −4 s CPU time for class D = 2 x 10 5 x 3 / 2.5 x 10 9 = 2.4 × 10 −4 s Total CPU time for P1 = 10.4 × 10 −4 s For P2: CPU time for class A = 10 5 x 2 / 3 x 10 9 = 0.667 × 10 −4 s CPU time for class B = 2 x 10 5 x 2 / 3 x 10 9 = 1.333 × 10 −4 s CPU time for class C = 5 x 10 5 x 2 / 3 x 10 9 = 3.333 × 10 −4 s CPU time for class D = 2 x 10 5 x 2 / 3 x 10 9 = 1.333 × 10 −4 s Total CPU time for P2 = 6.667 × 10 − 4 s (a) CPI = CPU time × Clock rate/No. of instr. CPI P1 = 10.4 × 10 −4 × 2.5 × 10 9 /10 6 = 2.6 CPI P2 = 6.667 × 10 −4 × 3 × 10 9 /10 6 = 2.0 (b) No. of clock cycles P1 = 10 5 × 1 + 2 × 10 5 × 2 + 5 × 10 5 × 3 + 2 × 10 5 × 3 = 26 × 10 5 No. of clock cycles P2 = 10 5 × 2 + 2 × 10 5 × 2 + 5 × 10 5 × 2 + 2 × 10 5 × 2 = 20 × 10 5 5. Exercise 1.7 in text (page 56) (for part (b), you may assume the average CPIs found in part (a)). CPU time = No. of instr. x CPI x Clock cycle time = No. of instr. x CPI / Clock rate (a) Compiler A CPI Compiler B CPI With compiler A CPI = CPU time / (No. of instr. x Clock cycle time) = 1.1 / (1 x 10 9 x 10 -9 ) = 1.1 With compiler B CPI = CPU time / (No. of instr. x Clock cycle time) = 1.5 / (1.20 x 10 9 x 10 -9 ) = 1.25 (b) Assuming that the execution time (i.e., CPU time) are both equal to T seconds. Clock rate A = No. of instr. A x CPI A / CPU time = 1 x 10 9 x 1.1 / T = 1.1 x 10 9 / T (cycles/s) Clock rate B = No. of instr. B x CPI B / CPU time = 1.20 x 10 9 x 1.25 / T = 1.5 x 10 9 / T (cycles/s) Clock rate A / Clock rate B = (1.1 x 10 9 / T) / (1.5 x 10 9 / T) = 0.73

4 1.11.8 The CPU time has been reduced by (750 – 700) / 750 x 100% = 6.67%. 7. Exercise 1.12 in text (page 58). 1.12.1 Consider the CPU time as follows: For P1: CPU time = No. of instr. x CPI / Clock rate = 5.0 x 10 9 x 0.9 / (4 x 10 9 ) =1.125 s For P2: CPU time = No. of instr. x CPI / Clock rate = 1.0 x 10 9 x 0.75 / (3 x 10 9 ) = 0.25 s Though P1 has a larger clock rate, the CPU time for P2 is smaller and hence P2 has higher performance. 1.12.2 For P1, the CPU time for executing 1.0 x 10 9 instructions is CPU time = No. of instr. x CPI / Clock rate = 1.0 x 10 9 x 0.9 / (4 x 10 9 ) =0.225 s The number of instructions P2 can execute in 0.225 s is No. of instr. = CPU time x Clock rate / CPI = 0.225 x 3 x 10 9 / 0.75 = 9 x 10 8 1.12.3 MIPS = Clock rate / CPI For P1: MIPS = 4.0 x 10 9 / 0.9 = 4.44 x 10 9 = 4.44 x 10 3 millions of instructions per second For P2: MIPS = 3.0 x 10 9 / 0.75 = 4 x 10 9 = 4 x 10 3 millions of instructions per second Though P1 has a higher MIPS, the CPU time for P2 is smaller and hence P2 has higher performance (see 1.12.1). 1.12.4 MFLOPS = No. FP operations / (execution time x 10 6 ) For P1: MFLOPS = 0.4 x 5.0 x 10 9 / (1.125 x 10 6 )= 1.78 x 10 3 millions of floating-point operations per second For P2: MFLOPS = 0.4 x 1.0 x 10 9 / (0.25 x 10 6 )= 1.6 x 10 3 millions of floating-point operations per second Though P1 has a higher MFLOPS, the CPU time for P2 is smaller and hence P2 has higher performance (see 1.12.1). 8. Exercise 1.13 in text (pages 58-59). ( Correction: It should be: Consider a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executing INT instructions, 55 s executing L/S instructions, and 40 s executing Branch instructions.) 1.13.1 T fp = 70 × 0.8 = 56 s, total time T = 56 + 85 + 55 + 40 = 236 s. Reduction: 5.6% 1.13.2

5 Total time T = 250 × 0.8 = 200 s, T fp + T l/s + T branch = 165 s, T int = 200-165 = 35 s. Reduction time INT: (85-35)/85*100% = 58.82% 1.13.3 Total time T = 250 × 0.8 = 200 s, T fp + T int + T l/s = 210 s > 200 s. No.