Question

Software Learning Curve

Corporate executives for a national company are considering implementing new software that will make their business run more efficiently, thereby saving the company a substantial amount of money in the long run. Two competing systems are being reviewed. Software A is the latest version of the software the company currently uses and many of the functions operate similarly to the current version. The software company claims the latest version will increase productivity by 20%. Software B is a completely different system, so it will take longer to learn, but its manufacturers claim it performs at 150% of its leading competitor (the software currently in use by this company).

Before committing to either product, the executives tested the competing software for four weeks in two of their offices. The first office installed software A, which proved easy for the employees to learn as expected. After one week, productivity was already 107.59% compared to previous levels; after two weeks it had jumped to 117.80%. A learning curve model A\left(t\right)=M-Ce^{-kt}A(t)=M−Ce−ktcan be used to represent the productivity at this office t weeks after the new software is installed, where M is the maximum level of productivity with the software. Use the given data to construct the specific learning curve for software A. Based on this model, how productive were the employees of this office when using this new software without any training (i.e., when the new software was first installed)?

The second office installed software B, which required lengthy training for most of the employees. Prior to training the employees could not use this software at all, so the initial productivity was zero in this trial. After four weeks of training and practice using the new software, many of the employees were still struggling and productivity was only 57.18%, of the previous output for this office. Use the given data, including the manufacturer’s performance claim, to construct a learning curve function B\left(t\right)B(t) for software B. How many weeks will it take for this office to reach its previous level of productivity?

Construct a table of values for both A\left(t\right)A(t) and B\left(t\right)B(t) for the first 12 weeks. Solve algebraically to obtain a precise estimate of the number of weeks after installing new software until productivity at the two offices would be equal.

Question#3

Use the given data to construct the specific learning curve function A\left(t\right)=M-Ce^{-kt}A(t)=M−Ce −kt for software A. Identify the values for each parameter below, giving M and C as integers and rounding k to the nearest hundredth.

M = Question Blank type your answer... ; C = Question Blank type your answer... ; k = Question Blank type your answer...

Answer 1

for software A
claim is 20% increase in productivity
test :
after one week, increase in productivity is 7.59%
after two weeks, increase in productivity is 117.8 %
learning curve model
A(t) = M - C*e^(-kt)*A(t)
A(t) = M/(1 + C*e^(-kt))

now, M = 120
from the data
107.59 = 120/(1 + C*e^(-k))
117.8 = 120/(1 + C*e^(-2k))

Ce^(-k) = 0.11534529231341202714006877962636
C*e^(-2k) = 0.01867572156196943972835314091681
e^(k) = 6.1762161065999712714091373818115
k = 1.8207058034639142047
C = 0.712397452206577223

so without training
at t = 0,
A(0) = M/(1 + C) = 120/(1 + 0.712397452206577223 ) = 70.0771890 %

for software B
M = 150%
after 0 week, productivity = 0
after 4 weeks, 57.18

also, the model is B(t) = M(1 - e^(-mt))
We assume
then for t = 0
0 = M(1 - 1)/(1 + 1)

for t = 4
e^(-4m) = 1 - 57.18/150 = 0.6188
m = 0.1199932900
hence
model is B(t) = 150(1 - e^(-0.11999*t))
to reach 100% productivity
100 = 150(1 - e^(-0.119999*t))
t = 9.155 weeks

so it will take 9.155 weeks for the people at office B to use software B and reach previous productivity

below is the table for many weeks of productivity of A and B

	A	B
0	70.07719	0
1	107.59	16.96104
2	117.8	32.00424
3	119.6382	45.34644
4	119.9413	57.18
5	119.9905	67.67549
6	119.9985	76.98422
7	119.9998	85.24038
8	120	92.56298
9	120	99.05759
10	120	104.8178
11	120	109.9267
12	120	114.458
13	120	118.4768
14	120	122.0413
15	120	125.2027
16	120	128.0066
17	120	130.4935
18	120	132.6991
19	120	134.6554
20	120	136.3905
21	120	137.9294
22	120	139.2942
23	120	140.5048
24	120	141.5784
25	120	142.5307
26	120	143.3753
27	120	144.1244
28	120	144.7887
29	120	145.378
30	120	145.9006
31	120	146.3641
32	120	146.7753
33	120	147.1399
34	120	147.4633
35	120	147.7501
36	120	148.0045
37	120	148.2302
38	120	148.4303
39	120	148.6078

,
we see B overtakes A at time t
120/(1 + 0.712397*e^(-1.82t)) = 150(1 - e^(-0.11999t))
0 = 0.2 - e^(-0.11999t) + 0.712397*e^(-1.82t) - 0.712397*e^(-1.93999t)
solving we get
t = 13.413 weeks

horizontal asymptote of graph A(t) is gievn by x = 120%