Central Pollution Control Board (CPCB), New Delhi, has ordered 20 sophisticated equipments for gathering and analyzing the emission data at 12 critical locations in Delhi. The supplier of the equipment took 360 hours to build the equipment. If the direct and indirect costs average around ₹2,000 per

Question

Central Pollution Control Board (CPCB), New Delhi, has ordered 20 sophisticated equipments for gathering and analyzing the emission data at 12 critical locations in Delhi. The supplier of the equipment took 360 hours to build the equipment. If the direct and indirect costs average around ₹2,000 per hour and about 75% learning rate is assumed, then estimate (a) the time needed to complete unit 3 and 12 and (b) the total labor cost of unit 12.

Accepted Answer

(a) Equation (15.9) is used to find the learning curve slope “m” for 75% learning as:

[latex]m=\frac{\log (0.75)}{\log 2}=-0.415[/latex]

[latex]m=\frac{\log ext { (learning rate) }}{\log 2}[/latex] (15.9)

Equation (15.8) is used to estimate the time for a specific unit.

[latex]\begin{aligned}T_{3} &=60(3)^{-0.415}=228.19 ext { hours } \T_{12} &=360(12)^{-0.415}=128.36 ext { hours }\end{aligned}[/latex]

[latex]T_{n}=T_{1}(n)^{m}[/latex] (15.8)

(b) The total cost estimate for all 12 units can be obtained by using Equation (15.12). But for this, individual time estimates for all 12 units are made first and then multiplied by ₹ 2,000.

[latex]T_{1}=360(1)^{-0.415}=360 ext { hours } ; \quad T_{2}=360(2)^{-0.415}=270 ext { hours }[/latex]

[latex]T_{3}=360(3)^{-0.415}=228.19 ext { hours } \quad T_{4}=360(4)^{-0.415}=202.51 ext { hours }[/latex]

[latex]T_{5}=360(5)^{-0.415}=184.59 ext { hours } ; \quad T_{6}=360(6)^{-0.415}=171.15 ext { hours }[/latex]

[latex]T_{7}=360(7)^{-0.415}=160.54 ext { hours } ; \quad T_{8}=360(8)^{-0.415}=151.88 ext { hours }[/latex]

[latex]T_{9}=360(9)^{-0.415}=144.64 ext { hours } ; \quad T_{10}=360(10)^{-0.415}=138.45 ext { hours }[/latex]

[latex]T_{11}=360(11)^{-0.415}=133.08 ext { hours } ; \quad T_{12}=360(12)^{-0.415}=128.36 ext { hours }[/latex]

[latex]C_{T}=C\left(T_{1}+T_{2}+\ldots+T_{n} ight)[/latex] (15.12)

Using Equation (15.12):

[latex]\begin{array}{l}C_{T}=C\left(T_{1}+T_{2}+\ldots+T_{n} ight) \C_{T}=₹ 2,000\left(T_{1}+T_{2}+T_{3}+T_{4}+T_{5}+T_{6}+T_{7}+T_{8}+T_{9}+T_{10}+T_{11}+T_{12} ight) \C_{T}=₹ 2,000 imes 2,273.39=₹ 45,46,780\end{array}[/latex]

[latex]C_{T}=C\left(T_{1}+T_{2}+\ldots+T_{n} ight)[/latex] (15.12)

Question 15.6: Central Pollution Control Board (CPCB), New Delhi, has order...

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