asked 89.7k views
3 votes
A time standard was set as 0.20 hour per unit based on the 100th unit produced. Assume the task has a 75 percent learning curve. Refer to Exhibit 6.4. What would be the expected time of the 200th, 400th, and 800th units? (Do not round intermediate calculations. Round your answers to 2 decimal places.)

asked
User Adiana
by
8.5k points

1 Answer

1 vote

To calculate the expected time for the 200th, 400th, and 800th units using the learning curve, you can use the following formula:

T_n = T_1 \times (n)^{\log(0.75) / \log(2)}T

n

=T

1

×(n)

log(0.75)/log(2)

Where:

T_nT

n

is the expected time for the nth unit.

T_1T

1

is the initial time (0.20 hours per unit).

nn is the unit number.

Let's calculate:

For the 200th unit:

T_{200} = 0.20 \times (200)^{\log(0.75) / \log(2)}T

200

=0.20×(200)

log(0.75)/log(2)

T_{200} ≈ 0.20 \times (200)^{-0.415} ≈ 0.20 \times 0.653 ≈ 0.1306 \text{ hours}T

200

≈0.20×(200)

−0.415

≈0.20×0.653≈0.1306 hours

For the 400th unit:

T

400

=0.20×(400)

log(0.75)/log(2)

T_{400} ≈ 0.20 \times (400)^{-0.415} ≈ 0.20 \times 0.413 ≈ 0.0827 \text{ hours}T

400

≈0.20×(400)

−0.415

≈0.20×0.413≈0.0827 hours

For the 800th unit:

T_{800} = 0.20 \times (800)^{\log(0.75) / \log(2)}T

800

=0.20×(800)

log(0.75)/log(2)

T_{800} ≈ 0.20 \times (800)^{-0.415} ≈ 0.20 \times 0.206 ≈ 0.0413 \text{ hours}T

800

≈0.20×(800)

−0.415

≈0.20×0.206≈0.0413 hours

So, the expected times for the 200th, 400th, and 800th units are approximately:

0.13 hours for the 200th unit.

0.08 hours for the 400th unit.

0.04 hours for the 800th unit.

answered
User Nemin
by
7.1k points
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