1. Limitations of the use of learning curves include:
◼ Any change in the process, product, or personnel, disrupts
the curve
2. Some applications of the learning curve are: Internal man-
power forecasting, scheduling, establishing costs and budgets;
external purchasing and subcontracting of items; and strategic
3. Three approaches to solving learning curve problems are by:
doubling, learning curve tables, and a formula approach.
4. We cannot determine the implications for Great Lakes Ser-
7. The doubling effect in learning curves means that each time
output is doubled (say from 4 to 8 units), the time required to
produce the doubled unit is x% of the time to produce the base
8. Pursue a steeper learning curve by:
◼ Following an aggressive pricing policy
◼ Focusing on continuing cost reduction and productivity
1. If the learning is not as good as expected and rises to 90%,
how much will the 4th boat cost?
$4,050,000
of the first 4 boats below $16,000,000?
0.81
3. How many boats need to be produced before the cost of the
next boat is at or below $4,000,000?
of each boat is below $4,000,000?
6
END–OF-MODULE PROBLEMS
E.1 1st return takes 45 minutes
(a) 2nd return takes (45 min)(0.85) = 38.25 min
(b) 4th takes (38.25)(0.85) = 32.5 min
15 0.355 = 5.325 minutes
(b) For the first 10 units, at an 80% learning factor, the
learning-curve coefficient (from Table E.3) is 6.315
(a) First 3 = (2.746)(563) = 1,546
(b) First 6 = (5.101)(563) = 2,872
(c) T18 = (30)(0.394) = 11.82 hr
1
i
=
E.6 $5,000 = cost/hr
(b) Cost10 = (14.31 hr)($5,000/hr) = $71,550
(c)
10
1
(30)(6.315) 189.45 hours total
i
Ti
=
==
Cost for all 10 = (189.45)($5,000/hr) = $947,250
E.7 T1 = 112; Rate = 90%
The cumulative factor for the 8th unit at 90%, Table E.3 is 6.574.
So the first 8 units will take 6.574(112) = 736.29. The last 7 units
will take 736.29 – 112 = 624.29 hours.
E.8 T1 = 5 hr T2 = 4 hr
(a) Learning rate = 4/5 = 80%
(b) T3 = T1 C = (5)(0.702) = 3.51
(c) T4 = 5(0.640) = 3.2
T6 = 5(0.562) = 2.81
(d)
6
1
5(4.299) 21.5
i
Ti
=
==
E.9 For the 15th unit at a 90% learning factor, the learning-
curve coefficient (from Table E.3) is 0.663.
6 0.663 = 3.978 hours
E.10 T1 = 80 hours
T4 = ?
Use the doubling effect twice to reach the 4th unit
(a) 95% L.C.
(0.95)(0.95)(80 hours) = 72.2 hours
T8 = T4(0.92) = 35.8 hours
(b) T4 = T1(0.84)(0.84) = 46 0.84 0.84 = 32.4 hours
T8 = T4(0.84) = 27.3 hours
(c) T4 = T1(0.77)(0.77) = 46 0.77 0.77 = 27.3 hours
T8 = T4(0.77) = 21.0 hours
E.12 (a) For Susan: 4th unit required 5 hours; 8th unit required
4 hours
When volume doubled from 4 to 8, learning
improved 1 hour or 20% (1 hr/5 hr = 0.2) or a learning
3 hours
When volume doubled from 3 to 6, learning im-
proved 1 hour or 25% (1 hr/4 hr = 0.25) or a learning
75
3
75
6
4
, 0.634 6.31 for the first unit
0.634
3
, 0.475 6.31 for the first unit
0.475
10th unit will require 0.385 6.31 2.43 hours
NC
NC
= 
= 
 =
Neither is expected to reach 1 hour by the 10th unit.
E.13 Using Table E.3, 6th unit at 80% = 0.562
80
620 minN=
Thus,
20
1st unit 35.58 min
0.562
==
24.98 min, or about 25 minutes.
E.14 T3 = 20,000, Cost = $40/hour, 85% learning curve,
Resolve for plane #1 N = 3, C = 0.773 for 85%
20,000 hr 25,873 hrs for plane #1
0.773 =
41
5
6
Now, 25,873 0.723:
so cost 18,706 hours $40 $748,240
25,873 0.686:
so cost 17,749 hours $40 $709,960
25,873 0.657:
so cost 16,999 hours $40 $67
T T C
T
T
=  =
=  =
=
=  =
=
= = 9,960
12 15
through Cost = 56,274 $40
$2,250,960
TT=
=
E.16 (a) Expected price at a volume of 1.4 1012 bits:
◼ From Table E.3:
Learning curve coefficient for 0.70 learning curve
factor and doubling:
C = 0.7000
Cost:
= 100 0.7000 = 70 milli-cents/bit
(b) Expected price at a volume of 89.6 1012 bits
◼ Using the general learning curve formula: Cost of
the nth unit = TN Ln where: T is the present unit
cost, L is the learning curve factor, n is the number
of times the volume will double over the present
328 BUSINESS ANALYTICS MODULE E LEARNI NG CURVES
◼ For the present problem:
Cost = 100 0.77 = 100 0.0823543 = 8.235 milli-cents/bit
3rd unit, Table E.3 = 0.702
Estimated time for turbine number 1 = 460/.702 = 655.27
Using factors from Table E.3
T4 = 655.27(0.64) = 419.37 hours = $25,162.
T5 = 655.27(0.596) = 390.54 hours = $23,432.
T6 = 655.27(0.562) = 368.26 hours = $22,096.
E.18 Time to produce eighth plane = 28,718 hours
Learning factor = 0.80
◼ Time to produce first plane:
require 6,000 4.339 = 26,034 hours, while four boats
would require 6,000 3.556 = 21,336 hours. Thus four
E.23 (a) In order to estimate the learning curve rate, we take the
ratios of the units that have doubled:
Report 2 56 0.848
Report 1 66
Report 4 48 0.857
56
Report 2
Report 6 45 0.849
53
Report 3
Report 8 41 0.854
48
Report 4
==
==
==
==
135 hours.
E.26 T1 = $448 million
T11 = $270 million
N11
270 .60
TT
where C10th unit is obtained from Table E.3:
◼ From Table E.3: Learning curve coefficient for 0.80
learning curve factor and 10th unit:
C = 0.477
330 BUSINESS ANALYTICS MODULE E LEARNI NG CURVES
0.877, or 88%.
CASE STUDY
1(b). What are the advantages and disadvantages to SMT from
this approach?
The advantage to SMT is that SMT has the business and knows
2. How does the SMT proposed learning rate compare with that
of other industries?
Refer to Table E.1. SMT is to produce a new machine to be used
in the manufacture of logic chips. This machine is a highly ad-
vanced design, and contains a network of piping of different mate-
fication, process, or personnel. Changes in any of these are likely
to adversely affect the curve and have negative consequences on
time and cost. Additionally, in this case both IBM and SMT are
1
i
=
E.6 $5,000 = cost/hr
(b) Cost10 = (14.31 hr)($5,000/hr) = $71,550
(c)
10
1
(30)(6.315) 189.45 hours total
i
Ti
=
==
Cost for all 10 = (189.45)($5,000/hr) = $947,250
E.7 T1 = 112; Rate = 90%
The cumulative factor for the 8th unit at 90%, Table E.3 is 6.574.
So the first 8 units will take 6.574(112) = 736.29. The last 7 units
will take 736.29 – 112 = 624.29 hours.
E.8 T1 = 5 hr T2 = 4 hr
(a) Learning rate = 4/5 = 80%
(b) T3 = T1 C = (5)(0.702) = 3.51
(c) T4 = 5(0.640) = 3.2
T6 = 5(0.562) = 2.81
(d)
6
1
5(4.299) 21.5
i
Ti
=
==
E.9 For the 15th unit at a 90% learning factor, the learning-
curve coefficient (from Table E.3) is 0.663.
6 0.663 = 3.978 hours
E.10 T1 = 80 hours
T4 = ?
Use the doubling effect twice to reach the 4th unit
(a) 95% L.C.
(0.95)(0.95)(80 hours) = 72.2 hours
T8 = T4(0.92) = 35.8 hours
(b) T4 = T1(0.84)(0.84) = 46 0.84 0.84 = 32.4 hours
T8 = T4(0.84) = 27.3 hours
(c) T4 = T1(0.77)(0.77) = 46 0.77 0.77 = 27.3 hours
T8 = T4(0.77) = 21.0 hours
E.12 (a) For Susan: 4th unit required 5 hours; 8th unit required
4 hours
When volume doubled from 4 to 8, learning
improved 1 hour or 20% (1 hr/5 hr = 0.2) or a learning
3 hours
When volume doubled from 3 to 6, learning im-
proved 1 hour or 25% (1 hr/4 hr = 0.25) or a learning
75
3
75
6
4
, 0.634 6.31 for the first unit
0.634
3
, 0.475 6.31 for the first unit
0.475
10th unit will require 0.385 6.31 2.43 hours
NC
NC
= 
= 
 =
Neither is expected to reach 1 hour by the 10th unit.
E.13 Using Table E.3, 6th unit at 80% = 0.562
80
620 minN=
Thus,
20
1st unit 35.58 min
0.562
==
24.98 min, or about 25 minutes.
E.14 T3 = 20,000, Cost = $40/hour, 85% learning curve,
Resolve for plane #1 N = 3, C = 0.773 for 85%
20,000 hr 25,873 hrs for plane #1
0.773 =
41
5
6
Now, 25,873 0.723:
so cost 18,706 hours $40 $748,240
25,873 0.686:
so cost 17,749 hours $40 $709,960
25,873 0.657:
so cost 16,999 hours $40 $67
T T C
T
T
=  =
=  =
=
=  =
=
= = 9,960
12 15
through Cost = 56,274 $40
$2,250,960
TT=
=
E.16 (a) Expected price at a volume of 1.4 1012 bits:
◼ From Table E.3:
Learning curve coefficient for 0.70 learning curve
factor and doubling:
C = 0.7000
Cost:
= 100 0.7000 = 70 milli-cents/bit
(b) Expected price at a volume of 89.6 1012 bits
◼ Using the general learning curve formula: Cost of
the nth unit = TN Ln where: T is the present unit
cost, L is the learning curve factor, n is the number
of times the volume will double over the present
328 BUSINESS ANALYTICS MODULE E LEARNI NG CURVES
◼ For the present problem:
Cost = 100 0.77 = 100 0.0823543 = 8.235 milli-cents/bit
3rd unit, Table E.3 = 0.702
Estimated time for turbine number 1 = 460/.702 = 655.27
Using factors from Table E.3
T4 = 655.27(0.64) = 419.37 hours = $25,162.
T5 = 655.27(0.596) = 390.54 hours = $23,432.
T6 = 655.27(0.562) = 368.26 hours = $22,096.
E.18 Time to produce eighth plane = 28,718 hours
Learning factor = 0.80
◼ Time to produce first plane:
require 6,000 4.339 = 26,034 hours, while four boats
would require 6,000 3.556 = 21,336 hours. Thus four
E.23 (a) In order to estimate the learning curve rate, we take the
ratios of the units that have doubled:
Report 2 56 0.848
Report 1 66
Report 4 48 0.857
56
Report 2
Report 6 45 0.849
53
Report 3
Report 8 41 0.854
48
Report 4
==
==
==
==
135 hours.
E.26 T1 = $448 million
T11 = $270 million
N11
270 .60
TT
where C10th unit is obtained from Table E.3:
◼ From Table E.3: Learning curve coefficient for 0.80
learning curve factor and 10th unit:
C = 0.477
330 BUSINESS ANALYTICS MODULE E LEARNI NG CURVES
0.877, or 88%.
CASE STUDY
1(b). What are the advantages and disadvantages to SMT from
this approach?
The advantage to SMT is that SMT has the business and knows
2. How does the SMT proposed learning rate compare with that
of other industries?
Refer to Table E.1. SMT is to produce a new machine to be used
in the manufacture of logic chips. This machine is a highly ad-
vanced design, and contains a network of piping of different mate-
fication, process, or personnel. Changes in any of these are likely
to adversely affect the curve and have negative consequences on
time and cost. Additionally, in this case both IBM and SMT are