Question

D Question 5 1 pts A tailor must make 7 of the same size suits for a client. The first suit took 13.5 hours to make and the last suit took 6.9 hours to make. What was the tailor's learning curve % of this run of suits assuming a steady state of production is never reached? Express your answer in 96 to the nearest 0.1%

Answer 1

Before answering this question well let us discuss in brief the overall concept of the learning curve:-

A learning curve is a representation in graph form of the state of learning something over time or repeated experiences. On the Learning curve, the x-axis depicts the time or experience and the y-axis depicts the percentage of learning. The percentage of learning shows how certain things can be mostly learned quickly while some different aspects may remain. Alternatively, the learning curve effect is the effect of increased efficiency with production volume.

There are 3 assumptions of the learning curve effect which are as follows:-

(i) The time required to complete a given task is performed.

(ii) The decrease will decrease at a decreasing rate.

(iii) The decrease will follow a predictable pattern.

However, the most effective way of a learning curve percentage calculation is using an exponential decay function. Thus the

formula of calculating the learning curve effect is as follows:-

T_n = T₁nb

Where n = number of units

T₁= Amount of time to produce the first unit

T_n = Amount of time to produce unit n

b=Learning curve factor which can be calculated as follows:-

ln (p)/ln(2), where ln/(x) is the natural logarithm of x and p is the learning percentage.

It is given in the question that:-

T₁ = 13.5 Hours (i.e the amount of time required to produce the first suit)

T_n = 6.9 Hours (i.e the amount of time required to produce the last suit)

n= Number of suits i.e. 7 in this case

Now, putting these values in the formula of learning curve effect we get:-

T_n =T₁ nb

6.9=13.5*7*b

solving this we get b =0.0728

Now, we know that b =ln(p)/ln(2)

Putting the value of b in the above equation we get,

0.0728 = ln(p)/0.3010 { since the value of log 2 is 0.3010}

so, ln(p) = 0.0728/0.3010 = 0.02191

So, log₁₀ (x) =0.02191

Applying log rule we get:-

a = log_b (b^a)

0.02191 = log₁₀ (10^0.02191)

log ₁₀ (x) = log₁₀ (10^0.02191)

x= 10^0.02191 =1.05%

Thus , the learning curve % comes out to be 1.05%.