q in percent learningin 2 for an 80 percent learning curve q 0223106931 03219 so when x 3 p 100

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q = ln(percent learning)/ln 2 For an 80 percent learning curve: q = 0.2231/0.6931 = 0.3219 So, when X = 3, p = 100, and q = 0.3219, Y = 100 - (3 - 0.3219) = 100 - 2.6781 = 97.3219 (first two units total hours) and an average of 70.3 to produce (210.6/3 units). Labor hours (rounded to the nearest tenth). Thus, it takes 100 hours to produce the first unit, an average of 70.3 hours to produce the first three units (a total of 210.6 hours for producing three units). The third unit, therefore, took 140.3 hours. Applying the Concepts Medcom Company installs computerized patient record systems in hospitals and medical centers. Medcom has noticed that each general type of system is subject to an 80 percent learning curve. The installation takes a team of professionals to set up and completely test the system. Required: 1. Assume that the first installation takes 800 hours, and the team of professionals is paid an average of $50 per hour. Complete the following table for Medcom Company (round answers to one decimal place): Cumulative Number of Systems Cumulative Average Time per System (in Hours) Cumulative Total Labor Hours 1 800 800 2 800 * 0.8 = 640 800 + 640 = 1440 4 800 * 0.8^2 = 512 1440 + 512 = 1952 8 800 * 0.8^3 = 409.6 1952 + 409.6 = 2361.6 16 800 * 0.8^4 = 327.7 2361.6 + 327.7 = 2689.3 32 800 * 0.8^5 = 262.1 2689.3 + 262.1 = 2951.4 2. Total Labor cost (round to the nearest dollar): $2951 3. Using the data table above, calculate the following for Medcom to produce a third system: a. Cumulative average time to produce the third system (round to the nearest tenth): 512 hours b. Cumulative total labor hours to produce the third system (round to the nearest tenth): 1952 hours c. Time to produce the third system (round to the nearest tenth): 512 - 1440 = -928 hours (negative value indicates that the third system was produced faster than the second system)

          q = ln(percent learning)/ln 2
For an 80 percent learning curve:
q = 0.2231/0.6931 = 0.3219
So, when X = 3, p = 100, and q = 0.3219, Y = 100 - (3 - 0.3219) = 100 - 2.6781 = 97.3219 (first two units total hours) and an average of 70.3 to produce (210.6/3 units).
Labor hours (rounded to the nearest tenth). Thus, it takes 100 hours to produce the first unit, an average of 70.3 hours to produce the first three units (a total of 210.6 hours for producing three units). The third unit, therefore, took 140.3 hours.
Applying the Concepts
Medcom Company installs computerized patient record systems in hospitals and medical centers. Medcom has noticed that each general type of system is subject to an 80 percent learning curve. The installation takes a team of professionals to set up and completely test the system.

Required:
1. Assume that the first installation takes 800 hours, and the team of professionals is paid an average of $50 per hour. Complete the following table for Medcom Company (round answers to one decimal place):

Cumulative Number of Systems Cumulative Average Time per System (in Hours) Cumulative Total Labor Hours
1 800 800
2 800 * 0.8 = 640 800 + 640 = 1440
4 800 * 0.8^2 = 512 1440 + 512 = 1952
8 800 * 0.8^3 = 409.6 1952 + 409.6 = 2361.6
16 800 * 0.8^4 = 327.7 2361.6 + 327.7 = 2689.3
32 800 * 0.8^5 = 262.1 2689.3 + 262.1 = 2951.4

2. Total Labor cost (round to the nearest dollar): $2951

3. Using the data table above, calculate the following for Medcom to produce a third system:
a. Cumulative average time to produce the third system (round to the nearest tenth): 512 hours
b. Cumulative total labor hours to produce the third system (round to the nearest tenth): 1952 hours
c. Time to produce the third system (round to the nearest tenth): 512 - 1440 = -928 hours (negative value indicates that the third system was produced faster than the second system)

q in percent learningin 2 for an 80 percent learning curve q 0223106931 03219 so when x 3 p 100 and q 03219 y 100 3 03219 first two units total hours and an average of to produce 2106 160 ia 39093

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