

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The solutions to the practice set questions from chapter 7 of the cs433g computer system organization course for the fall 2005 semester. The practice set involves calculating the maximum transactions per second rate and system cost per transaction for two different disk organization configurations, as well as determining the number of disks required for a faster cpu.
Typology: Assignments
1 / 2
This page cannot be seen from the preview
Don't miss anything!


CS433g: Computer System Organization – Fall 2005 Practice Set (Chapter 7) – Solution Due Date: None
The I/O bus and a memory system of a computer are capable of sustaining 1000MB/sec without interfering with the performance of an 800-MIPS CPU (costing $50,000). Here are the assumptions about the database software that must run on this computer:
You have a choice of two types of disk organizations:
Either disk system can support on average 30 disk reads or writes per second. You may approximate 1 MB as 1,000,000 bytes and 1 GB as 1000 MB.
The system must support 10 gigabytes of data in the system. Assume that disk requests can be spread evenly to all the disks, in a round robin fashion (i.e., I/O request 1 goes to disk 1, request 2 goes to disk 2, etc.).
Part A What is the maximum transactions per second (TPS) rate possible with each disk organization? For this question, you should compute the peak TPS that would be allowed by each system component (CPU, bus, disk) and then determine the TPS from the limiting component.
Solution : We need 10GB of disk space so this requires either 20 small disks or 8 large disks. Each transaction requires (2 reads + 2 writes) * 15,000 + 40,000 = 100,000 instructions per transaction.
Unit Performance # of Units Demand/Transaction TPS Limit CPU 800 MIPS 1 1 MIPS 8, Bus 1000 MB/s 1 1000 Bytes 250, Small Disk
30 I/Os per second
20 4 I/Os 150
Large Disk
30 I/Os per second
8 4 I/Os 60
Part B What is the system cost per transaction for each configuration, assuming the peak transaction rate possible?
Solution : Since we know the TPS limit from the first part, we just divide by the cost of each system (which is $52,000 for both systems). Therefore it costs $346 per transaction for the small disk configuration and $866 per transaction for the large disk configuration.
Part C Given a CPU that is twice as fast as the original CPU but costs the same amount, what will be the cost of a system with enough small disks so that they do not limit the number of transactions the system can handle?
Solution : Since the new CPU is twice as fast, it can perform 16,000 MIPS. This means that there must be 16,000 * 4 / 30 = 2,134 small disks to have enough bandwidth to keep the processor busy. Therefore, the cost of the small disk system would be $50,000 + (2,134 * $100) = $263,400.
Part D Now assume you have the same number of small disks as in the original design. How fast must a new disk be (I/Os per second) to achieve the same TPS rate with the new CPU as in part d?
Solution : In this question we still use 20 disks, but now we want to increase the number of I/Os per second each disk can perform in order to keep up with the CPU in part d. Since the CPU can do 16,000 TPS and each transaction requires 2 reads and 2 writes, the disk system must be able to perform 64,000 operations per second. This means each disk must perform 64,000/20 = 3,200 I/Os per second.
666,667s + 17,970s = 684,637s = 7.92 days