Test

Computer Design

CPE 432

Homework 1

1) Chip Fabrication Cost

There are many factors involved in the price of a computer chip. New, smaller technology gives a boost in performance and a drop in required chip area. In the smaller technology, one can either keep the small area or place more hardware on the chip in order to get more functionality.

Figure 1.22 in your textbook gives the relevant chip statistics that influence the cost of several current chips. In the next few exercises, you will be exploring the effect of different possible design decisions for the IBM Power5.

a. What is the yield for the IBM Power5?

Yield = (1 + (Defect per unit area * die area) / 4.0) ^ -4.0

= (1 + (0.3 * 389/100) / 4.0) ^ -4.0

= (1 + 0.29) ^ -4.0

= (1.29) ^ -4.0

= 0.36

b. What is the yield for the Sun Niagra?

Yield = (1 + (Defect per unit area * die area) / 4.0) ^ -4.0

= (1 + (0.75 * 380/100) / 4.0) ^ -4.0

= (1 + 0.71) ^ -4.0

= (1.71) ^ -4.0

= 0.12

c. Why does the IBM Power5 have a lower defect rate than the Niagara?

Because it uses different process technology, it uses larger transistor feature size.

2) Power Consumption in Computer Systems

Power consumption in modern systems is dependent on a variety of factors, including the chip clock frequency, efficiency, the disk drive speed, disk drive utilization, and DRAM.

Figure 1.23 presents the power consumption of several computer system components. In this exercise, we will explore how the hard drive affects power consumption for the system.

a. Assuming the maximum load for each component, and a power supply efficiency of 80%, what wattage must the server’s power supply deliver to a system with an Intel Pentium 4 chip, 4 GB 240-pin Kingston DRAM, and one 7200 rpm hard drive?

Peak Power Consumption = 1* 66 + 4 * 2.3 + 1 * 7.9

= 66.0 + 9.2 + 7.9

= 83.1 W

Power Supply Wattage = 83.1 / 0.8

= 103.9 W...