Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Im not sure how to use those operating voltage and ratio of capacity load. pleas

ID: 3802982 • Letter: I

Question

Im not sure how to use those operating voltage and ratio of capacity load.

please let me know how voltage and capacity load can influence the CPU time

thank you

Hw.1.pdf Adobe Acrobat Reader DC SET Hw.1.pdf x Microelectronic Cir. PDF/A. 3. You are the chief architect of Galaxy S9, which is your company's next smartphone. The CPU of Galaxy S9 adopt a new mode, called dynamic voltage/frequency scaling, where CPU can select one of the two operating modes with different voltage and frequency setting by switching two different modules where capacity load is different. The table below summarizes the performance analysis of two operation modes on the processor. Clock Operating Ratio of Instruction Clock Cycles Rate(GHz) Voltage(V) Capacity load Instruction Count per Type (millions) Instruction(CPD High Low High Low High Low ALU 15 3 Load&Store; 2 1 1.2 0.8 Branch 10 (a) What are the CPU times of this application on Both high and Low voltage mode? (Please include time units) A 4:48 2017-03-23 Search Windows

Explanation / Answer

Here ,

     The Clock rate and the Operating Voltage are directly proportional and

           Operating Voltage and capacity load are directly proportional.

CPU Time of the application for high voltage :-

           Here we have to select clock rate is = 2 GHZ

                                    hence time period = 1 / 2 * 109 = 0.5 nsecs

          Total Number of cycles is = ALU Cycles + L&S Cycles + Branch Cycles

                                               = 15*106 * 1 + 3*106 * 5 + 2*106 * 10

                                               = 15*106 + 15*106 + 20*106

                                                               = 50*106

            Hence total number of cycles is = 50*106.

CPU Time is = Total number of cycles * cycle time

                     = 50*106 * 0.5 * 10-9

                            = 25 * 10-3

                            = 25 msecs

Hence Total CPU Time is = 25 msecs.

CPU Time of the application for low voltage :-

           Here we have to select clock rate is = 1 GHZ

                                    hence time period = 1 / 1 * 109 = 1 nsecs

          Total Number of cycles is = ALU Cycles + L&S Cycles + Branch Cycles

                                               = 15*106 * 1 + 3*106 * 5 + 2*106 * 10

                                               = 15*106 + 15*106 + 20*106

                                                               = 50*106

            Hence total number of cycles is = 50*106.

CPU Time is = Total number of cycles * cycle time

                     = 50*106 * 1 * 10-9

                            = 50 * 10-3

                            = 50 msecs

Hence Total CPU Time is = 50 msecs.