1 Your company has a complex manufacturing process with three operations that ar

Question

1 Your company has a complex manufacturing process with three operations that are performed in series Because of the age 2 of the machines and nature of the process, machines frequently fall out of calibration and must be recalibrated. To maintain 3 the required production levels, two identical machines are used at each stage; thus if one fails, the other can be used while 4 the first is being repaired. 10 12 13 The reliabilites of the machines are as follows: Reliabli 0.85 0.92 0.9 Machine 15 16 17 18 19 20 21 a. analyze the system reliablity, assuming only one machine at each state(al back ups are out of operatio 23 b. how much is the reliability improved by having two machines at each stage? 24

Explanation / Answer

To be calculated:

(a) System reliability without back up machines

(b) System reliability with back up machines

Given values:

Reliability of machine A = 0.85

Reliability of machine B = 0.92

Reliability of machine C = 0.90

Solution:

(a) Scenario 1: When no back up machines are used

All the machines will be in series and the system structure will be given as;

A — B — C

Reliability of the components in series is calculated as;

System Reliability = P1 x P2 x …… x Pn

System Reliability = P(A) x P(B) x P(C)

System Reliability = 0.85 x 0.92 x 0.90

System Reliability = 0.7038

(b) Scenario 2: When back up machines are used

The two machines at each step, i.e., machine A and backup machine A, machine B and backup machine B and machine C and backup machine C will be in parallel.

Reliability of the components in parallel is calculated as;

System Reliability = 1 - [(1-P1) x (1-P2) x…..x (1-Pn)]

System Reliability of machine A = 1 - [(1-PA) x (1-PA)]

System Reliability of machine A = 1 - [(1 - 0.85) x (1 - 0.85)]

System Reliability of machine A = 0.9775

System Reliability of machine B = 1 - [(1-PB) x (1-PB)]

System Reliability of machine B = 1 - [(1 - 0.92) x (1 - 0.92)]

System Reliability of machine B = 0.9936

System Reliability of machine C = 1 - [(1-PC) x (1-PC)]

System Reliability of machine C = 1 - [(1 - 0.90) x (1 - 0.90)]

System Reliability of machine C = 0.99

Now, the three components A, B and C (with back up machines) will be in series and therefore, the overall system reliability with back up machines will be calculated as;

System reliability = P(A) x P(B) x P(C)

System reliability = 0.9775 x 0.9936 x 0.99

System reliability = 0.9615

When the back up machines are used, the system reliability is improved by;

= [(System reliability with back up - System reliability without back up) / System reliability without back up] x 100

= (0.9615 - 0.7038) / 0.7038 x 100

= 36.62%

The overall system reliability by having two machines at each stage is improved by 36.62%

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