Consider an electric part with two components in series comp

Consider an electric part with two components in series (component A and component B). Component has a reliability of 88%, while component B has a reliability of 94%. It is known that if the part fails, it will cost the company $1000.

You are trying to reduce the expected failure cost per part, and you have three improvement options. Find the system reliability and total (improvementmplus failure) cost per part for each of the three options, and then suggest which of the three (if any) improvements you would recommend. (round all reliability calculatioms to 4 decimals)

improvement 1: add an 80% reliable backup for component A. The backup costs $50 per part.

improvement 2: add 90% reliable backup for component B. The backup costs $50 per part.

improvement 3: add a backup for component A and add a backup for component B. Backups have same reliabilites and costs as in improvements 1 and 2

Solution

Component A = 88% Reliability, Component B = 94% Realiability

We have to calculate system reliability Rs = R1 + (1 - R1) (R2) = 0.88 + (1 - 0.88)(0.94) = 0.88 + 0.1128 = 0.9928

Improvement 1 : 80% Reliable backup for Component A

(0.88)(0.80) * (0.9928) * (0.94) = 0.704 * 0.9928 * 0.94 = 0.6570

Improvement 2 : 90% Reliable backup for Component B

(0.88) * (0.9928) * (0.94)(0.90) = 0.88 * 0.9928 * 0.846 = 0.7391

Improvement 3 : Backup have same reliabilities

(0.88)(0.5) * (0.9928) * (0.94)(0.5) = 0.44 * 0.9928 * 0.47 = 0.2053

From the above calculation Improvement 2 is better option because there is lesser chances to parts fail.