Lazer Technologies Inc. (LTI) has produced a total of 20 high-power laser system

Question

Lazer Technologies Inc. (LTI) has produced a total of 20 high-power laser systems that could be used to destroy any approaching enemy missiles or aircraft. The 20 units have been produced, funded in part as private research within the research and development arm of LTI but the bulk of the funding came from a contract with the U.S. Department of Defense (DoD) Testing of the laser units has shown that they are effective defense weapons, and through redesign to add portability and easier field maintenance, the units could be truck-mounted DoD has asked LTI to submit a bid for 100 units The 20 units that LTI has built so far cost the following amounts and are listed in the order in which they were produced: Use Exhibit 64 and Exhibit 65. NUMBER (S MILLIONS) NUMBER (S MILLI 52 2 513.0 7.5 12 13 5.8 4.5 4.0 3.5 3.1 1.8 15 18 18 2.4 a. Based on past experience, what is the learning rate? (Do not round intermediate calculations. Round your answer to the nearest whole percent.) 61% eaming rate b. What bld should LTI submit for the total order of 100 units, assuming that learning continues? cost for 100 more units S[ c. What is the cost expected to be for the last unit under the learning rate you estimated? Expected cost for last unit L

Explanation / Answer

b.

Using Crawford's incremental cost model ,

we have ,

Y = aKb

where: Y = the incremental unit time (or cost) of the lot midpoint unit.
           K = the algebraic midpoint of a specific production batch or lot.

a = time (or cost) required to produce the first unit.

b = slope of the function when plotted on log-log paper.
= log of the learning rate/log of 2.

and ,

K = [L(1+b)/(N21+b - N11+b)]-1/b
where:  L = the number of units in the lot.
             b = log of learning rate / log of 2
            N1 = the first unit in the lot minus 1/2.
            N2 = the last unit in the lot plus 1/2.

K =214.622

The cost of the midpoint unit is

Y=aK^b

=13*214.622^-0.7132

=0.282

The total cost for the lot of 100 units =100*Y =100*0.282=$28.2 millions

c.Cost expected for the last unit is

Y120 =aK^b

K can be replaced by the cummulative number of units produced .

So,for 120th unit ,the cummulative number of units produced =7260 (taking cummulative value of 120th unit from 1 to 120 )

Y120 =13*(7260)^-0.7132

=$0.023 millions

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