Which plant had the highest labor productivity? Japan in 2013 Japan in 2014 USA
ID: 416710 • Letter: W
Question
Which plant had the highest labor productivity?
Japan in 2013
Japan in 2014
USA in 2013
USA in 2014
Which plant had the highest total productivity (include all inputs)?
Japan in 2013
Japan in 2014
USA in 2013
USA in 2014
Which of the following is closest to the value for percent change in labor productivity for Japan from 2013 to 2014?
3%
-2%
5%
1%
Which plant experienced the best change in total productivity from 2013 to 2014?
Japan
USA
A manufacturer has designed a product. This product has four components (Components A, B, C, and D). The reliability probabilities associated with each product are:
A: 0.9
B: 0.75
C: 0.8
D: 0.6
a) What is the probability that this product will be reliable?
b) The manufacturer can afford to modify the product to incorporate redundancy in one of the components. Which of the components should be selected and why?
c) If component C were to receive a redundant backup with a reliability probability of 0.6, what would be the reliability of the product? What would be the failure probability?
d) If each component were to receive a redundant backup with a reliability probability of 0.5, what would be the reliability of the product? What would be the failure probability?
Explanation / Answer
Answer (a) The probability that this product will be reliable = p(A)*p(B)*p(C)*p(D)
= (0.90) *(0.75) *(0.80) *(0.60) = 0.32
Answer (b) Component D should be selected as it has the lowest reliability probability value and most likely to be unsuccessful.
Answer (c) =>Finding the new reliability of component C:
For Component C to be reliable then the probability = 1 – (probability of component C getting unsuccessful or fail) = 1 – (probability of component C and failed redundant backup)
= 1 – [(1- 0.8) *(1- 0.6)] = 1 – [(0.2) *(0.4)] = 1 – 0.08 = 0.92
Hence, probability for the reliable product = (0.9) *(0.75) *(0.92) *(0.6) = 0.37
Therefore, the failure probability = 1 – 0.37 = 0.63
Answer (d) => Finding the new reliabilities for each component:
For component A => 1- [(1-0.90) * 0.5] = 1 – (0.1) *(0.5) = 0.95
For component B => 1- [(1-0.75) * 0.5] = 1 – (0.25) *(0.5) = 0.875
For component C => 1- [(1-0.8) * 0.5] = 1 – (0.2) *(0.5) = 0.9
For component D => 1- [(1- 0.6) * 0.5] = 1 – (0.4) *(0.5) = 0.8
Hence, the reliability of product = (0.95) * (0.875) * (0.9) * (0.8) = 0.599
Therefore, the failure probability = 1 – 0.599 = 0.401
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For the questions mentioned on labour productivity, Input values are not available in the question.
However, Labor productivity should be evaluated using output and input ( = Output / Input)
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