1. Using the information below, CPU W = 4 ms (the average time to execute 1,000

Question

1. Using the information below,

CPU W = 4 ms (the average time to execute 1,000 instructions)

Drive A = 45 ms (average data access speed)

Drive B = 3 ms (average data access speed)

Drive C = 17 ms (average data access speed)

Drive D = 10 ms (average data access speed)

Calculate I/O access speed using this CPU and each of the four disk drives as they evaluate the following track requests in this order: 0, 31, 20, 15, 20, 31, 15. Then, in each case, calculate the access speeds after the track requests are reordered and rank the four disk drives before and after reordering. (See section: Role of Device Management in chapter for this and next question)

2. Using the information below,

CPU X = 0.3 ms (the average time to execute 1,000 instructions)

Drive A = 0.7 ms (average data access speed)

Drive B = 4 ms (average data access speed)

Drive C = .03 ms (average data access speed)

Calculate I/O access speed using this CPU and each of the disk drives as they evaluate the following track requests in this order: 16, 4, 9, 16, 29, 31, 5. Then, in each case, calculate the access speeds after the track requests are reordered and rank the three disk drives before and after reordering.

3. Calculate the availability of a server with the following scenarios (Show your work for all answers)

a. MTBF of 75 hours and an MTTR of 3 days (72 hours)

b. MTBF of 2,040 hours and an MTTR of 8.5 hours

c. MTBF of 50 weeks and an MTTR of 798 hours (hint: convert weeks to hours before proceeding)  

Explanation / Answer

Multiple questions : answering one.(3).

Availability = MTBF / (MTBF + MTTR) for scheduled working time.
a.
MTBF = 75 hours
MTTR = 72 hours

Availability = 75 / 147 = 0.5102 = 51.02%

b.
MTBF = 2040 hours
MTTR = 8.5 hours

Availability = 2040 / 2048.5 = 0.99585 = 99.59%

c.
MTBF = 2 weeks = 8400 hours
MTTR = 798 hours

Availability = 8400 / 2048.5 = 0.91324 = 91.32%

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