CS501 ASSIGNMENT NO. 5, DISCUSSION AND SOLUTION( DUE DATE: 3RD JULY, 2013)

[RaJa AhMeD(MSCS)]

Plz Discuss this assignment

	Assignment No. 05 Semester: Spring 2013 CS501: Advanced Computer Architecture	Total Marks: 20 Due Date:  03-07-13
Instructions Please read the following instructions carefully before assignment submission. It should be clear that your assignment will not get any credit if: The assignment is submitted after the due date. The submitted assignment does not open or file is corrupt. The assignment is found to be copied from the internet. The assignment is found to be copied from other student. The assignment submitted is not according to required file format (.doc). Objective The objective of this assignment is: To assess your overall understanding of Computer Architecture and Organization To assess your overall understanding of Computer processing To assess your overall understanding of DMA, Polling and interrupts Note: The assignment should be in .doc format. Assignment .05 covers lecture 25-31. You can also consult reference books for help. Students are advised to submit their assignment as early as possible in order to avoid any sort of inconvenience like Load shedding etc.

 

 

Question No 1:                                                                                                     (10 marks) Suppose we have a benchmark that executes in 100 seconds of elapsed time, where 90 seconds is CPU time and the rest is I/O (Input/output) time. If CPU time improves by 50% per year for the next five years but I/O time doesn’t improve, how much faster will our program run at the end of five years? You are required to calculate the CPU improved performance and improved elapsed time.     After n years CPU/ time I/O time Elapsed time % I/O time 0 (Current Year)         1         2         3         4         5         NOTE: Theoretical answer will not be considered ______________________________________________________________________ Question No 2:                                                                                                      (4 marks) Consider a 20 MIPS (Microprocessor without Interlocked Pipeline Stages) processor with several input devices attached to it, each running at 1000 characters per second. Assume that it takes 17 instructions to handle an interrupt. If the hardware interrupt response takes 1msec, what is the maximum number of devices that can be handled simultaneously? NOTE: Theoretical answer will not be considered ______________________________________________________________________ Question No 3:                                                                                                        (6 Marks) Scenarios discussion: If we want the lowest latency for an I/O operation to a single I/O device; while in terms of lowest impact on processor utilization from a single I/O device then what will be the orders/arrangements of Interrupt driven, DMA(Direct Memory Access) and polling in both scenarios? Explain reasons. NOTE: Give answer within 3-5 lines. Otherwise answer will not be considered. ______________________________________________________________________ GOOD LUCK J

 

[RaJa AhMeD(MSCS)]

Thanx Sis BFSS. is main to same Q ha assignment ka.

Kindly share any link related to Q # 3??

[B FOR FAZAL]

welcome hasnat ahmed brother

 i am searching  q 2 , 3 what about question 2

[RaJa AhMeD(MSCS)]

BFSS sis i thnk book wali example he iss ka answer ha. sir nay just confuse krnay k liay MIPS ka poora name likha ha??

[Muhammad zubair]

yes u are right.... aisa hi he...

[Muhammad Arshad Mahmood]

elapsed time = cpu time + i/o time . as the time changed the performance of cpu increased by the designers but they did not take any action for the performance improvement of i/o devices. so all the time i/o time will remain same but the cpu performance increased. if current year it was 180 sec then next year it reduces 40 % or 40 % performance increased. 

[Muhammad Arshad Mahmood]

This is may be or may be not help full.

This information unit discusses how to measure CPU utilization by workload. Our approach is to convert application time for each service class into overall percent of CPU used for that workload. The total of all these numbers is the base for the APPL% CP value in the report.

The concept of capture ratio (CR) must be understood in order to do this. Simply put, the CPU time reported for the sum of all your workloads never adds up to the total CPU time used by the system. The total CPU time reported for all your workloads will typically account for 85-90% of the total CPU time used. This is not an error. It is the best level of accuracy that the reporting tools can achieve, in a consistent, repeatable manner.

The uncaptured time is sometimes, misleadingly, called system overhead. This is incorrect, because most of it is genuine user work — it is either a political or philosophical view whether activities like paging are seen as genuine work or just overhead.

Thus the capture ratio is:

      Captured CPU Time CR =  -----------------        Total CPU Time

The question now is: how do you account for the uncaptured, though real, CPU time? The answer is to distribute the uncaptured time among your workloads, so that the total CPU time is accounted for. Several approaches how to do this are discussed.

How to compute CPU time per workload type

Following are the basic steps for computing CPU time by workload. A graphical view of CPU consumption over time (for example, by hour or by day) can help you to understand your workload flow, as shown in Figure 7 as an example of a Spreadsheet Reporter macro.

1. Calculate total CPU time in seconds:

   Total CPU time = MVS Busy % * Interval * #CPs

   If your system is running in an PR/SM environment, the calculation has to be performed using LPAR BUSY and the number of logical processors.

2. Get the captured time for all address spaces.

   The Workload Activity report shows the percentage of application time. It is the sum of task control block and preemptible-class SRB (CPU) time, non-preemptible service request block (SRB) time, region control task (RCT) time, I/O interrupt (IIT) time, and hiperspace service (HST) time. Subtract zAAP and zIIP time, because it is contained in CPU time.

              (CPU + SRB + RCT + IIT + HST - AAP - IIP) APPL% CP = ----------------------------------------- * 100                            Interval

   Captured time = (APPL% CP / 100) * Interval

   This results in the uncaptured CPU time:

   Uncaptured CPU time = Total CPU time - Captured time

3. Calculate the capture ratio:

         Captured CPU Time CR =  -----------------        Total CPU Time

4. Get application (captured) percentage for one workload you are interested in:

   This APPL% CP value represents the percentage based on one processor. You get the percentage based on the total capacity by:

                      APPL% CP Total Captured % = --------                      #CPs

5. Distribute uncaptured time based on the capture ratio:

             Total Captured % Total % = ----------------                 CR

[Sana Rajpoot]

Q#1

[Sana Rajpoot]

Q2 is on page 312 lecture#30 example#4...!

[RaJa AhMeD(MSCS)]

Ok. Then Sana Sis wot abt Q # 3??? Sub nay Q 1,2 to solve kr lia ha, Q 3 ka problem ha???

[Muhammad zubair]

Instructor's Reply:

Dear Student, The 3rd question of 5th assignment is itself a question not a part of 1st and 2nd questions.  Regards

[2p4dl922mawrk]

Salam

Sana ap and galt aa raha hai 1.5 kahan se aya aur kese aya aur ap muje ye bato k book main 180 se to 129 per kese gay 1.5 wahan pe use karne to  ni ata ap ka ans galt haa raha hai ap ka ans is trah

[zeemal]

zabardast!!Sana