Question

In: Accounting

Suppose that you are the estimator who is assigning costs to a major project to be...

Suppose that you are the estimator who is assigning costs to a major project to be undertaken this year by your firm, Acme Software Developers. One particular software development process involves many labor-hours, but the work is highly redundant. You anticipate a total of 100,000 labor-hours to complete the first iteration of the software development process and a learning curve rate of 80%. Assume you are going to use the cumulative average time in your calculations to determine the time it takes for each iteration.

You are attempting to estimate the cost of the tenth iteration of this repetitive process. Based on this information and a $60 per hour labor rate, what would you expect to budget as

A. The cost of the tenth iteration?

B. The cost of the twentieth iteration?

Solutions

Expert Solution

Cummulative average time = Y = aXb

Where Y =cumulative average time per unit or batch

a = time taken to produce initial quantity

X = the cumulative units of production or, if in batches, the cumulative number of batches.

b = the learning index or coefficient, which is calculated as: log learning curve percentage ÷ log 2. So b for an 80 per cent curve would be log 0.8 ÷ log 2 = – 0.322

Cumulative average time for 10 iterations

Y = 100000*10-0.322

= 47643.10

TOTAL TIME TO PRODUCE 10 ITERATIONS = 47643.10*10 = 476431 hours.

Cumulative average time for 9 iterations

Y = 100000*9-0.322

= 49287.17

TOTAL TIME TO PRODUCE 9 ITERATIONS = 49287.17*9 = 443584.53 hours.

time taken for 10th iteration = 476431 hours - 443584.53 hours. = 32846.47

Cost of 10th iteration = 32846.47*60 = $1970788.2

Cumulative average time for 20 iterations

Y = 100000*20-0.322

= 38112.58

TOTAL TIME TO PRODUCE 20 ITERATIONS = 38112.58*20 = 762251.59 hours.

Cumulative average time for 19 iterations

Y = 100000*19-0.322

= 38747.29

TOTAL TIME TO PRODUCE 19 ITERATIONS = 38747.29*19 = 736198.52 hours.

time taken for 20th iteration = 762251.59 hours - 736198.52 hours = 26053.07

Cost of 10th iteration = 26053.07*60 = $1563184.06


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