PDiligently show all work including graphs, charts, tables, and formulas. Write
ID: 354254 • Letter: P
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PDiligently show all work including graphs, charts, tables, and formulas. Write legibly and state decisions/recommendations in complete sentences. > Use the space provided below each problem/question wisely. No additional sheet may be attached. A computer has three main modules that have individual reliabilities of .80, .90, and.90. Because of recent failures, management is now considering adding redundancy. Determine the reliability of the system with backups under these conditions: 1. a) Each module has a backup with reliability equal to its own and a backup switch with a reliability of 1.00 The backup consists of an identical computer that operates as a whole rather than backing up individual sections. The single switch for the backup computer has a reliability of.96. b)Explanation / Answer
Answer to question 1.a :
Reliability of the back up with reliability of back up switch being 1 and back up with reliability R = 1 x R = R
Reliability of this subsystem is now connected in parallel with main unit with reliability R .
Reliability of the whole thing i.e. main unit with back up subsystem in parallel
= 1 – ( 1 – R ) ^ 2
Reliability of the 1st module with reliability of 0.80 of the main module with back up system as described = 1 – ( 1- 0.80) ^2 = 1 – 0.2x0.2 = 1 – 0.04 = 0.96
Reliability of the 2nd module with reliability of 0.90 of the main module with back up system as described= 1 – ( 1 – 0.90)^2 = 1 – 0.1 x 0.1 = 1 – 0.01 = 0.99
Reliability of the 3rd module with reliability of 0.90 of the main module with back up system as described= 1 – ( 1 – 0.90)^2 = 1 – 0.1 x 0.1 = 1 – 0.01 = 0.99
Therefore , reliability of the system with module 1 , module 2 and module 3 in series = 0.96 x 0.99 x 0.99 = 0.9408
Answer to question 1.b
Reliability of the main computer system = 0.80 x 0.90 x 0.90 = 0.648
Reliability of the back up subsysytem with same main computer system configuration with reliability of 0.648 and switch with reliability 0.96 will be
= 0.648 x 0.96 = 0.622
Therefore, reliability of the whole system
= 1 – ( 1 – reliability of main computer system ) x ( 1 – reliability of backup up subsystem )
= 1 – ( 1 – 0.648) x ( 1 – 0.622)
= 1 – 0.352 x 0.378
= 1 – 0.133
= 0.867
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