A B C .97 .91 .98 D=.98 E=.98 You\'ve been hired as an operations manager for a

Question

A B C

.97 .91 .98

D=.98

E=.98

You've been hired as an operations manager for a local hospital and your director has let you know your first task is to improve the reliability

D=.98

E=.98

You've been hired as an operations manager for a local hospital and your director has let you know your first task is to improve the reliability

of the helium distribution system (Liquid Helium is essential for the MRI magnet as well as other applications). That the main problem is with the cooling tank and plumbing to a manifold (Process B). As a result, you've ordered back up equipment (Processes D and E) to increase the reliability of the delivery and cooling. a. What is the reliability of the original system (Processes A, B and C, each must work for the system to work)? Based on a 24 hour day, how many hours per day is helium available? (round to 2 decimal places) b. With D and E as backups, only one of those systems needs to work to provide helium to the hospital. What is the reliability of the the system after adding processes D and E? (round to 4 decimal places) Based on a 24 hour day, how many hours per day is helium available?

Explanation / Answer

a) With only A, B & C available, these are in series.

Ra = 0.97, Rb = 0.91, Rc = 0.98

Reliability of overall system R = Ra * Rb * Rc

So, R = 0.97 * 0.91 * 0.98 = 0.865046

In a 24-hour day, the Availability of Helium = 24 * R = 24*0.865046 = 20.76

So, Helium will be available for 20.76 hours

b) Now, subsystems D & E added as backup to B

B, D, E are in parallel.

The reliability of Sub-system will be

Rbde = 1- (1-Rb) * (1-Rd) * (1-Re) = 1- (1-0.91) * (1-0.98) * (1-0.98)

Rbde = 1 - 0.09 * 0.02 * 0.02 = 1- 0.000036 = 0.999964

Now, Overall system Reliability is calculated as :

R = Ra * Rbde * Rc = 0.97 * 0.999964 * 0.98 = 0.950565778

In a 24-hour day, the Availability of Helium = 24 * R = 24 * 0.950565778 = 22.8136

So, Helium will be available for 22.8136 hours

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