In: Operations Management
Kiran industries is preparing a bid to the government to produce engines for rescue boats. Direct labor costs averages $30 per hour. The company has manufactured these engines over the past 3 years on an exclusive contact and has experience the following costs:
Total Cumulative Costs |
|||
Cumulative units produced |
Materials |
Labor |
|
10 |
$60,000 |
$120,000 |
|
20 |
$120,000 |
$192,000 |
|
40 |
$240,000 |
$307,200 |
The bid request is for an addition 40 units. Kiran uses the cumulative average-time model for applying learning curve analysis. Also assume that the company has adequate capacity to produce the navy order without disrupting current operations.
Kiran’s rate for learning on the 3-year engine contract in the past, as shown above, has been (
________%
NOW assume that Kiran expects a 70% learning curve from this point forward. The number of estimated hours to complete the new order will be
___________
Please show all work !!!
Materials cost is increasing in direct proportion to cumulative units, that per unit materials cost is same for all cumulative units produced. So, its learning rate is 100 % (no effect of learning curve)
However, labor cost is not constant for cumulative units and is decreasing, indicating effect of learning curve.
Learning rate based on 10 and 20 cumulative units = (192000-120000)/120000 = 0.6
Learning rate based on 20 and 40 cumulative units = eln(2)*ln(307200/192000)/(ln(40)-ln(20)) -1 = 0.6
Average learning rate = (0.6+0.6)/2 = 0.6
Kiran’s rate for learning on the 3-year engine contract in the past, as shown above, has been 60 %
Number of hours to complete the first 10 units = 120000/30 = 4000 hours
Using Crawford's formula
X = Total Number of unit | 10 |
b = log(r) / log(2) | -0.7370 |
N1 = the first unit in the lot minus 1/2 | 0.5 |
N2 = the last unit in the lot plus 1/2 | 10.5 |
K = [X(1+b)/(N21+b - N11+b)]-1/b | 3.6028 |
K is the algebraic midpoint of the lot |
Total labor hours for the first 10 units = aXKb = a*10*3.6028-0.737 = 4000 hours
a = 4000/(10*3.6028-0.737) = 1028.69 hours
Total number of hours to complete the first 40 units = 307200/30 = 10240 hours
Total number of hours to complete the first 80 units is computed using Crawford's formula,
X = Total Number of unit | 80 |
r = learning rate = 70% = | 0.7 |
b = log(r) / log(2) | -0.5146 |
N1 = the first unit in the lot minus 1/2 | 0.5 |
N2 = the last unit in the lot plus 1/2 | 80.5 |
K = [X(1+b)/(N21+b - N11+b)]-1/b | 23.1961 |
K is the algebraic midpoint of the lot |
Total labor hours for the first 80 units = aXKb = 1028.69*80*23.1961-0.5146 = 16,322 hours
Number of estimated hours to complete the new order = 16322 - 10240 = 6,082 hours