ComSci 222 Computer Architecture
Homework #2 - 01 February 1999
Instructor: Michael O'Donnell Ryerson 257A e-mail: odonnell@cs
Student: Nick Russo e-mail: n-russo@uchicago.edu

1.1

(a)

$Speedup(x) = \frac{1}{(1-x) + x/20}$

(d)

Suppose that a representative program takes 100 time units to run without vector mode. If computations which comprise $70\%$ of that time can be made to use vector computation (which is 20 times faster), then the total time used will be reduced to $(100-70) + \frac{70}{20} = 33.5$ time units.

The hardware design group can double the speed of vector mode. This would reduce the total time to $(100-70 + \frac{70}{40} = 31.75$ time units for the same job.

To achieve this same improvement, suppose we can make $X\%$ of the computation use vector mode. Then $(100-X) + \frac{X}{20}$ must be equal to 31.75; X must be 71.84. This is only a $3\%$ increase in the amount of computation which uses vector mode over our original enhancement.

I would recommend increasing the use of vector mode. Because vector mode runs 20 times faster, non-vector mode computations become the dominant factor in time consumption. Further improvements to the speed of the vector mode are hardly noticeable compared to the conversion of more computation to vector computation.

1.6

My scheme to reduce the number of loads and stores is $6\%$ faster. Suppose there are 100 instructions to be performed. On the original computer, this would take 100 fast clock cycles. With my enhancement, only 70 regular instructions and 20 (of the original 30) load-store instructions are necessary. This would take 90 slow clock cycles. The fast clock cycles are $5\%$ faster, which means they take the same time as 95.24 cycles on the slow machine. My version will be done first.

1.8

(a)

Microprocessor	die area	pins	wafer cost	die yield	good chips/wafer
MIPS 4600	77	208	$3200	0.479	171.07
PowerPC 603	85	240	$3400	0.449	144.47
HP 71x0	196	504	$2800	0.210	27.02
Digital 21064A	166	431	$4000	0.253	39.23
SuperSPARC/60	256	293	$4000	0.149	14.17

(b)

Microprocessor	wafer cost	good chips/wafer	cost/good die
MIPS 4600	$3200	171.07	$18.71
PowerPC 603	$3400	144.47	$23.54
HP 71x0	$2800	27.02	$103.62
Digital 21064A	$4000	39.23	$101.96
SuperSPARC/60	$4000	14.17	$282.29

(c)

Microprocessor	cost/good die	final cost
MIPS 4600	$18.71	$31.54
PowerPC 603	$23.54	$44.42
HP 71x0	$103.62	$175.29
Digital 21064A	$101.96	$153.26
SuperSPARC/60	$282.29	$313.18

(d)

Microprocessor	defect density	final cost
SuperSPARC/60	0.006	$184.17
SuperSPARC/60	0.012	$398.57

(e)

Microprocessor	$\alpha$	final cost
Digital 21064A	3.00	$132.99
Digital 21064A	4.50	$138.40

1.12

(a)

Suppose some program takes 100 time units. If enhancements 1 and 2 are both used for $30\%$ of the original program, then the time in those modes totals 30/30 + 30/20 = 2.5 time units. If we are to achieve an overall speedup of 10, then we must take 10 time units for the whole program, which means we have 7.5 time units to perform the remaining computations. Using enhancement 3 for $X\%$ of the time to achieve a local speedup of 40/7.5 means that X can be found by the following equation.

$\frac{40}{7.5} = \frac{1}{(1-X)+X/10}$ X = .90


Thus enhancement 3 must be used for $90\%$ of the remaining computations, or $36\%$ of the original program time.

(b)

Enhancements 1, 2, and 3 account for $80\%$ of the original time, and only take $4.5\%$ as long as the whole program originally took. The remaining $20\%$ is not enhanced, so it still takes $20\%$ of the original time. 20/24.5, or $82\%$ of the reduced execution time finds no enhancement in use.

(c)

If only one enhancement can be implemented, it should be enhancement 3. It takes $37\%$ as long as the unenhanced version.

If two can be used, they should be 1 and 3. This system would only take $22.5\%$ of the original time.

The following represents the calculations I used to do problem 1.8

#include <iostream.h>
#include <math.h>
#include <string>

enum package { PQFP, PGA };

class microprocessor {
public:
  microprocessor(){};
  microprocessor(char* n,
		 int d,
		 int p,
		 int w,
		 package pk) {
    Name = n;
    Die_Area = d;
    Pins = p;
    Wafer_Cost = w;
    Package = pk;
  };
  char* Name;
  int Die_Area;
  int Pins;
  int Wafer_Cost;
  package Package;
};

double wafer_size = 200; // 20cm == 200mm
double defect_density = .01; // 1/cm^2 == 1/100mm^2
double wafer_yield = .95;
double alpha = 3;

double die_yield (microprocessor m){
  double temp =(1+ (defect_density * m.Die_Area / alpha));
  return wafer_yield * 1/(exp(alpha*log(temp)));
};

double good_chips_per_wafer (microprocessor m){
  double a = (M_PI * (wafer_size*wafer_size/4) / m.Die_Area );
  double b = (M_PI * wafer_size) /sqrt(m.Die_Area*2);
  return (a-b)*die_yield(m);
};

double cost_per_good_die (microprocessor m){
  return m.Wafer_Cost / good_chips_per_wafer(m);
};

double final_cost (microprocessor m){
  if (m.Package==PQFP) {
    if (m.Pins<220) { return 12+(10*300.0/3600.0)+cost_per_good_die(m); }
    else { return 20+(10*320.0/3600.0)+cost_per_good_die(m); }
  };
  if (m.Package==PGA) {
    if (m.Pins<300) { return 30+(10*320.0/3600.0)+cost_per_good_die(m); }
    else if (m.Pins<400) { return 40+(12*340.0/3600.0)+cost_per_good_die(m); }
    else if (m.Pins<450) { return 50+(13*360.0/3600.0)+cost_per_good_die(m); }
    else if (m.Pins<500) { return 60+(14*380.0/3600.0)+cost_per_good_die(m); }
    else if (m.Pins>500) { return 70+(15*400.0/3600.0)+cost_per_good_die(m); }
  };
};

int main() {
  
  microprocessor list[5];
  list[0] = microprocessor("MIPS 4600",77,208,3200,PQFP);
  list[1] = microprocessor("PowerPC 603",85,240,3400,PQFP);
  list[2] = microprocessor("HP 71x0",196,504,2800,PGA);
  list[3] = microprocessor("Digital 21064A",166,431,4000,PGA);
  list[4] = microprocessor("SuperSPARC/60",256,293,4000,PGA);
  
  printf("%-14s\t%4s\t%4s\t%6s\t%6s\t%16s\n","Microprocessor",
	 "Area","Pins","W Cost","dYield","Good chips/wafer");
  for (int i=0; i<5; i++){
    microprocessor* a = & list[i];
    printf("%-14s\t%4d\t%4d\t$%-6d\t%6.3f\t%16.2f\n",a->Name,
	   a->Die_Area,
	   a->Pins,
	   a->Wafer_Cost,
	   die_yield(*a),
	   good_chips_per_wafer(*a)
	   );
  }
  cout << endl << endl;
  
  printf("%-14s\t%6s\t%16s\t%13s\n","Microprocessor",
	 "W Cost","Good chips/wafer","cost/good die");
  for (int i=0; i<5; i++){
    microprocessor* a = & list[i];
    printf("%-14s\t$%-6d\t%16.2f\t$%-13.2f\n",a->Name,
	   a->Wafer_Cost,
	   good_chips_per_wafer(*a),
	   cost_per_good_die(*a)
	   );
  }
  cout << endl << endl;
  
  printf("%-14s\t%13s\t%10s\n","Microprocessor",
	 "cost/good die","final cost");
  for (int i=0; i<5; i++){
    microprocessor* a = & list[i];
    printf("%-14s\t$%-13.2f\t$%-10.2f\n",a->Name,
	   cost_per_good_die(*a),
	   final_cost(*a)
	   );
  }
  cout << endl << endl;

  microprocessor* a = & list[4];
  printf("%-14s\t%14s\t%10s\n","Microprocessor","defect density","total cost");
  defect_density = .006;
  printf("%-14s\t%14.3f\t$%-10.2f\n",a->Name,defect_density,final_cost(*a));
  defect_density = .012;
  printf("%-14s\t%14.3f\t$%-10.2f\n",a->Name,defect_density,final_cost(*a));
  cout << endl << endl;

  defect_density = .008;
  microprocessor D = microprocessor("Digital 21064A",166,431,4000,PGA);
  printf("%-14s\t%5s\t%10s\n","Microprocessor","alpha","total cost");
  printf("%-14s\t%5.2f\t$%-10.2f\n",D.Name,alpha,final_cost(D));
  alpha = 4.5;
  printf("%-14s\t%5.2f\t$%-10.2f\n",D.Name,alpha,final_cost(D));
  cout << endl << endl;
  
  return 0;
}