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SHOW ALL WORK FOR THE QUESTIONS ASKED! WHEN IT SAYS CREATE A CHART, DO CREATE A

ID: 357767 • Letter: S

Question

SHOW ALL WORK FOR THE QUESTIONS ASKED! WHEN IT SAYS CREATE A CHART, DO CREATE A CHART! AND WHEN IT SAYS TO EXPLAIN AN ANSWER, EXPLAIN THAT ANSWER! ANSWER ALL THE QUESTIONS ASKED IN GREAT DETAIL!

Your company has a complex manufacturing process with three operations that are performed in series. Because of the age of the machines and nature of the process, machines frequently fall out of calibration and must be recalibrated. To maintain the required production levels, two identical machines are used at each stage; thus, if one fails, the other can be used while the first is being repaired.

The reliabilities of the machines are as follows:

Machine

Reliability

A

0.85

B

0.92

C

0.9

1. Analyze the system reliability, assuming only one machine at each state (all backups are out of operation)

2. How much is the reliability improved by having two machines at each stage?

Machine

Reliability

A

0.85

B

0.92

C

0.9

Explanation / Answer

Answer to question 1 :

Assuming one machine at each stage , we consider machines A, B and C re connected in series .

The system reliability under such scenario

= Reliability of A x Reliability of B x Reliability of C

= 0.85 x 0.92 x 0.9

= 0.7038

SYSTEM RELIABILITY= 0.7038

Answer to question 2 :

When two identical machines are connected in parallel with reliability of each = r, reliability of the subsystem = 1 – ( 1 – r)^2

Accordingly , reliability of the subsystem with 2 machine As connected in parallel= 1 – ( 1 – 0.85)^2 = 1-0.0225 = 0.9775

Reliability of subsystem with 2 machine Bs connected in parallel = 1 – ( 1- 0.92)^2 = 1- 0.0064 =0.9936

Reliability of subsystem with 2 machine Cs connected in parallel = 1 – ( 1- 0.9)^2 = 1 – 0.01 = 0.99

Reliability of the system when 3 subsystems are connected in parallel

= Reliability of subsystem A x reliability of subsystem B x Reliability of subsystem C

= 0.9775 x 0.9936 x 0.99

= 0.9615

Therefore , improvement in reliability = 0.9615 – 0.7038 =0.2577

RELIABILITY IMRPOVED BY = 0.2577

SYSTEM RELIABILITY= 0.7038