Question

Lambda Computer Products competed for and won a contract to produce two prototype units of a new type of computer that is based on laser optics rather than on electronic binary bits. The company expects "learning" to occur relative to labor and also relative to the pricing of materials, equipment and supplies. The first unit produced by Lambda took 5,000 hours to produce and required $250,000 worth of material, equipment usage, and supplies. The second unit took 3,500 hours and used $200,000 worth of materials, equipment usage, and supplies. Labor is $30 per hour Use Exhibit 6 5. a. Lambda was asked to present a bid for 10 additional units as soon as the second unit was completed. Production would start immediately. What would this bid be? Cost for 10 more units S

Answer 1

Answer:

Finding the rate of learning

Learning rate in terms of production time

= (Time taken to build the second unit) / (Time taken to build the first unit prototype)

= 3500 / 5000 = 0.7 i.e 70%

=> Learning rate in terms of production time = 70%

Learning rate in terms of production cost

= (Cost to build the second unit) / (Cost to build the first unit prototype)

= 200,000 / 250,000 = 0.8 oe 80%
=>Learning rate in terms of production cost = 80%

T_N = T₁C
T_N =   Number of hours required to prepare the Nth session
T₁ =   Number of hours required to prepare the 1st session
C = learning curve coefficient.


The values for the Learning Curve Coefficient can be found on Tables available on the internet.

Or, for your easier understanding of how that Table is derived, let us proceed further to get the values here by our own using the formula for Learning Curve.
The formula for Learning Curve , C = X^{(logb/log 2)}
where,

X = Cumulative Volume of Production
b = Learning rate = 0.70
(log b/log 2) = log (Learning rate) / log 2 = Slope of learning curve

=> (Log b/ Log 2) = (Log 0.7 / Log 2) = - 0.2345
So,

value of C for the 1^st Session = C (1) = 1^{(Log0.7/Log2)}  = 1^(-0.51457)  ,
value of C for the 2^nd Session = C (2) = 2^{(Log0.7/Log2)} = 2^(-0.51457)
value of C for the 4^th Session = C (4) = 4^{(Log0.7/Log2)} = 4^(-0.51457)  ,
value of C for the 8^th Session = C (8) = 8^{(Log0.7/Log2)} = 8^(-0.51457)

		Learning Curve = 70%			Learning Curve = 80%
Computer based on Laser Optic	Unit Number	Unit Time, T =unit Number ^ LOG(0.70,2)	Time required to Build the Compters (Hrs)	Cumulative Time to Build the Compters ($)	Unit Cost, C =Session Number ^ LOG(0.70,2)	Cost of Material, Equipment, and Supplies required to Build the Compters ($)	Cumulative Cost of Material, Equipment, and Supplies required to Build the Compters ($)
1st protype	1	1.000	5000.00		1.000	250,000.00
2nd	2	0.700	3500.00		0.800	200,000.00
1st for supply	3	0.568	2840.90	2840.90	0.702	175,525.93	175,525.93
2nd for supply	4	0.490	2450.00	5290.90	0.640	160,000.00	335,525.93
3rd for supply	5	0.437	2184.23	7475.13	0.596	148,909.34	484,435.26
4th for supply	6	0.398	1988.63	9463.77	0.562	140,420.74	624,856.00
5th for supply	7	0.367	1836.98	11300.75	0.534	133,622.38	758,478.38
6th for supply	8	0.343	1715.00	13015.75	0.512	128,000.00	886,478.38
7th for supply	9	0.323	1614.14	14629.89	0.493	123,237.40	1,009,715.79
8th for supply	10	0.306	1528.96	16158.86	0.477	119,127.47	1,128,843.25
9th for supply	11	0.291	1455.79	17614.64	0.462	115,527.78	1,244,371.04
10th for supply	12	0.278	1392.04	19006.68	0.449	112,336.59	1,356,707.63

Total Material, Equiment and Supplies cost Amount to build 10 computers = $1,356,707.63

Total time to to build 10 computers = 19,006.68 hrs

Labor @ $30 per hr

Total Labor cost = 19006.68 * 30 = $570,200.4

Total cost including labor and material = $1,356,707.63 + $570,200.4 = $1,926,908.03

Thus Lamda Computers must bid (atleast) at cost of 10 more units = $1,926,908.03

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