uestion 1.d. What is the latest finish time for task A5? Ch allenge 8. Your comp
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uestion 1.d. What is the latest finish time for task A5? Ch allenge 8. Your company erects windmills for major power companies. During the last 3 the company has erected 512 windmills. The first windmill costed $50,000 and the last one (20% cost reduction). The city of Orlando issued a request for proposals (RFP) to erect years $40,000 512 new windmills at the lowest cost. You are teaming with the chief engineer and the marketing manager to define the cost per windmill to use in the proposal. The chief engineer argues the cost should be $40,000 to be consistent with the latest experience. The marketing manager believes the company can reduce current costs by another 20% based on their previous experience, ie. $32,000. The proposer with the lowest unitary cost per windmill will be awarded with a contract but if it can potentially lose money if the actual cost exceeds the proposed one. Question 8. Based on the concept of learning curve, what unitary cost per windmill would you propose? Challenge 8. Two companies are bidding for a 1,000 square meter warehouse construction job. The contract will be awarded to the company submitting the lowest price bid. It is well known erew of 8 people will require thirty 10-hour working days to complete the job. Labor costs are $10/hour, material costs are $50/square meter, finishing costs are $40,000 and overhead costs a Question 8.a. Your investigative team found your competitor wants a 15% profit for the proje uestion 8.c. Your research and development team has created new technologies that allow y ns, would you maintain the % profit used in Question 8.b? If not, what % profit woul What price per square meter will your rival submit? Ouestion 8.1. What % profit would you select to receive the award? you complete the work with a 4 people crew instead of the typical 8. Under theseExplanation / Answer
Cost of first unit=$50,000
Cost for the nth unit=C*(L^n)
C=unit cost of the first unit
L=Learning curve rate
n=number of time quantity of production is doubled
If learning curve is 80%( or L=0.8), The second unit will cost 0.8 times the cost of first unit and 4th unit will cost (0.8^2)=0.64 times the second unit
In this case 512=2^9
n=number of times the manufactured quantity doubled=9
Cost of first unit=C=$50,000
Cost of 512th unit=$40,000
$40,000=$50,000*(L^9)
L^9=40000/50000=0.8
L=(0.8^(1/9))= 0.975511
We can use the average cost equation:
Learning Curve Equation:
Y=a*(X^b)
Where,
Y=Cumulative average cost per unit
a=Cost of initial unit
X=Cumulative units of production
b=The learning index=log(Learning curve percentage)/log2
In this case learning curve percentage =0.9755
512=2^9
Average cost of 512 units of new order=40000*(512^b)
b=log0.9755/l0g2= -0.03577
Average cost of 512 units of new order=40000*(512^(-0.03577))=40,000*0.8=$32,000
Based on Learning Curve:
$32,000 per unit cost is achievable
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