Computer Architecture (2 Problems)

[JackFrank]

need help answering these 2 questions

Q1.
A hypothetical disk drive has the specifications shown below. Find the average access time for a block (sector) of data.

I want the answer in milliseconds, and you need to reduce your answer to a simple final value.

• 32 sectors per track
• 3600 RPM speed
• 7 millisecond average seek time
________________________
Q2.
Transaction processing is chiefly concerned with I/O rate: number of disk accesses per
second, as opposed to data rate, and the I/O speed versus CPU speed in terms of Transactions Per Second (TPS). Assume the following information:

• CPU has the performance of 800 MIPS.
• Each transaction requires 2 disk reads plus 2 disk writes.
• The database software executes 40000 instructions to process a transaction.
• The operating system uses 15000 instructions for each disk read or write access.
• The I/O bus and memory are capable of sustaining 1000 MB/sec of data transfer without interfering with the CPU.
• Each read/write access transfers 100 bytes.

a. What is the peak performance of the CPU in terms of TPS?
b. What is the peak performance of the I/O bus system in terms of TPS?

Thank you in advance.

 

[MATLABdude]

Okay, so what are your thoughts on these questions?

EDIT: Welcome to Physics forums, by the way. However, the way these forums works is that you need to show some work before we're able to help you out (i.e. we don't give out answers / do your homework for you)

 

[Mene]

Q1. To find the average access time for a block of data, we can use the formula:
Average Access Time = Average Seek Time + Average Rotational Latency + Average Transfer Time

Average Seek Time = 7 milliseconds
Average Rotational Latency = (1/2) * (1 / (RPM / 60)) = (1/2) * (1 / (3600 / 60)) = 0.0083 milliseconds
Average Transfer Time = (1 / (RPM / 60)) = (1 / (3600 / 60)) = 0.0083 milliseconds

Therefore, the average access time = 7 + 0.0083 + 0.0083 = 7.0166 milliseconds

Q2.
a. To find the peak performance of the CPU in terms of TPS, we can use the formula:
TPS = (CPU Speed / (Number of instructions per transaction * Number of transactions)) * 1000

Number of instructions per transaction = 40000 + 2 * 15000 = 70000
Number of transactions = 2 (since each transaction requires 2 disk reads and 2 disk writes)

Therefore, TPS = (800 * 10^6 / (70000 * 2)) * 1000 = 57142.857 TPS

b. To find the peak performance of the I/O bus system in terms of TPS, we can use the formula:
TPS = (Data transfer rate / (Number of bytes per transaction * Number of transactions)) * 1000

Number of bytes per transaction = 2 * 100 = 200 (since each transaction transfers 100 bytes for each read/write access)
Number of transactions = 2 (since each transaction requires 2 disk reads and 2 disk writes)

Therefore, TPS = (1000 * 10^6 / (200 * 2)) * 1000 = 2500000 TPS