Question

Ima Neworker requires 30 minutes to produce her first unit of output. If her learning curve rte is 65%, how many units will be produced before the output rate exceeds 12 units per hour?

Answer 1

To find when Ima Neworker's rate will exceed 12 units per hour, given a learning curve rate of 65%, we analyze the improvement in production rate from the initial 2 units per hour up to the target, using the learning curve concept.

Step-by-step explanation:

The question relates to the concept of a learning curve, which represents how new workers or processes improve in efficiency as experience is gained. Ima Neworker can produce her first unit in 30 minutes (which is half an hour), so when she starts, her production rate is 2 units per hour. The question asks how many units will be produced before her production rate exceeds 12 units per hour, given a learning curve rate of 65%. This means that each time the cumulative production doubles, the time taken to produce each unit falls to 65% of the previous time.

Since the initial production rate is 2 units per hour, we want to know how many units she has to produce before her production rate exceeds 12 units per hour. 12 units per hour is 6 times faster than her initial rate, and we can reference a learning curve table or use the formula to calculate the necessary doubling periods required to achieve this.

Answer 2

Approximately 18 units will be produced before the output rate exceeds 12 units per hour.

Step-by-step explanation:

The learning curve formula is given by:

`Y = aX^(b)`

In which:

Y is the average time per unit.

X is the cumulative number of units produced.

a is the time required to produe the first unit

b = log of learning rate/log 2

In our problem, we have:

Y = 12 units per hour. We are working in minutes, what is the average time per unit?

60 minutes - 12 units

Y minutes - 1 unit

`12Y = 60`

`Y = 5`

So Y = 5.

X is the value we want to find

a = 30

b =
`(log 0.65)/(log 2)=-0.6215`

So

`Y = aX^(b)`

`5 = 30X^(-0.6215)`

`(1)/(6) = (1)/(X^(0.6215))`

`X^(0.6215) = 6`

`\sqrt[0.6215]{X^(0.6215)} = \sqrt[0.6215]{6}`

`X = 17.86`

Approximately 18 units will be produced before the output rate exceeds 12 units per hour.