Chapter 11, System Management, Ida A. Flynn and Ann McIver McHoes, Understanding Operating Systems, Second Edition, PWS Publishing Company (1993)
Lesson A: Evaluating an Operating System, The Operating System's Four Components, Measuring System Performance, Measurement Tools, Feedback Loops, Monitoring
1, 2, 3, 10, 11, 12, 13
Problem 1. Describe how you would use a negative feedback loop to avoid eating too much food during each course of a very large meal.
See the Feedback Loops tutorial on the course web site.
Eat food until you are comfortably satisfied. Wait a while. Repeat.
When you are stuffed, stop eating.
The input to the feedback loop is your estimate of how satisfied or stuffed you are.
The control signal is your decision to reduce the rate at which you are eating.
Problem 2. Describe how a bank manager might use a positive feedback loop to direct waiting customers to five loan officers, being careful to
Include a description of how you would monitor the system and measure its success.
Solution:
Assume you are not permitted to hire more loan officers. You minimize the waiting time for customers if you maximize the speed of loan processing. You maximize the speed of loan processing by paying the loan processing employees commission on loans they process. You motivate quality work by the loan officers by paying additional bonus when a loan is completely repaid.
Presumably, the larger the loan, the more certain you need to be that the customer can repay the loan. The system is a success if the amount of money per unit time processed exceeds a predetermined threshold.
Problem 3. Using the information given in Table 11.1, calculate I/O access speed using CPU 1 and CPU 3 and each of the four disk drives as they evaluate the following track requests: 0, 31, 20, 15, 20, 31, 15. In each case, determine whether reordering the requests would be advantageous.
Note: If you reorder the track requests, you can get: 0, 15, 15, 20, 20, 31, 31. This cuts the number of tracks the head moves from 79 tracks in 6 moves to 31 tracks in 3 moves.
| No Reorder | CPU #1 30 ms / Reorder |
CPU #2 1.2 ms / Reorder |
CPU #3 0.2 ms / Reorder |
||
| Disk Drive | Average Access Time (ms) | Total Average Time (ms) 6*Col 2 |
Average Access Time (ms) 30 + 3*Col 2 |
Average Access Time (ms) 1.2 + 3*Col 2 |
Average Access Time (ms) 0.2 + 3*Col 2 |
| A | 35 | 210 | 135 | 106.2 | 105.2 |
| B | 10 | 60 | 60 | 31.2 | 30.2 |
| C | 5 | 30 | 45 | 16.2 | 15.2 |
| D | 50 | 300 | 180 | 151.2 | 150.2 |
Conclusions: For slow disk drives, A and D, reordering was appropriate for all the CPUs. For the faster disk drives, B and C, reordering was beneficial only for the faster CPUs, numbers 2 and 3.
Problem 10. Calculate the availability of a magnetic tape cartridge with a MTBF of 80 hours and a MTTR of 3 days (72 hours).
Availability = MTBF / (MTBF + MTTR) = 80 / (80 + 72) = 80 / 152 = 0.526 = 52.6%
Problem 11. Calculate the reliability of a hard disk drive with a MTBF of 1050 hours during the last 40 hours of this month. Use Nickel's formula, shown on page 250, where e = 2.71828.
The formula called Nickel's formula is known in statistics as the survival function. It has been applied to reliability much earlier than 1975.
When examining the last 40 hours, either the hard disk drive failed during that time or it did not fail during that time. If it did not fail during the past 40 hours, its reliability during that period is 1. If it did fail during that time, then the reliability is less than 1.
Assume the problem is asking about the next 40 hours, with a MTBF of 1050 hours. Then reliability is computed according to Nickel's formula as: R(t) = exp( - T / MTBF ) = exp( - 40 / 1050 ) = exp ( - 0.0380952381 ) = 0.9626212583 = 96.3%
Problem 12. Calcualte the availability of a hard disk drive with an MTBF of 1050 and an MTTR of 8 hours.
Availability = MTBF / (MTBF + MTTR) = 1050 / (1050 + 8) = 1050 / 1058 = 0.9924385633 = 99.2%
Problem 13. Assuming you had sufficient funds to upgrade only one component for a system with which you are familiar, explain which component you would choose to upgrade to improve overall system performance and why.
Answers will vary.
My bottleneck is responsiveness on Internet. This is primarily due to using a "free" ISP service. I would prefer to get a dedicated ISDN line with a no-frills ISP.
I would upgrade to Windows 2000. While UNIX is great, some software I must use requires Windows NT. Hopefully, Windows 2000 will satisfy the requirement.
What would you do?