Chapter 7.S, Problem 3P

a)

Summary Introduction

To determine: The number of days for first 10 installations.

Introduction: Learning curve is referred as the cost curve, experience curve, efficiency curve or productivity curve. The leaving curve gives information about the cost, efficiency, productivity and performance of an organization.

Answer to Problem 3P

The number of days for first 10 installations is 56.93 days.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Initial unit}{\text{T}}_1=8\text{days}\\ \text{Learning}\text{curve}=\text{85\%}\end{array}$

Formula:

${\displaystyle \sum {\text{T}}_{\text{n}}}={\text{T}}_{\text{1}}\times \text{Unit}\text{factor}$

Calculation of number of days for first 10 installations:

At 85% learning curve, and 10 unit number, the unit factor is 7.116 (using learning curve coefficient table)

$\begin{array}{c}{\text{T}}_{1-10}=8\times \text{7}\text{.116}\\ \text{=56}\text{.93}\text{days}\end{array}$ (1)

The times for all 10 installations are calculated by multiplying first unit of 8 with the unit factor 7.116 which yields 56.93 days.

Hence, the number of days for first 10 installations is 56.93 days.

b)

Summary Introduction

To determine: The number of days for second 10 installations.

Introduction: Learning curve is referred as the cost curve, experience curve, efficiency curve or productivity curve. The leaving curve gives information about the cost, efficiency, productivity and performance of an organization.

Answer to Problem 3P

The number of days for second 10 installations is 42.29 days.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Initial unit}{\text{T}}_1=8\text{days}\\ \text{Learning}\text{curve}=\text{85\%}\end{array}$

Formula:

${\displaystyle \sum {\text{T}}_{\text{n}}}={\text{T}}_{\text{1}}\times \text{Unit}\text{factor}$

Calculation of number of days for second 10 installations:

To calculate the number of days for second 10 installations which is from 11 to 20, the numbers of days of installations for 1 to 20 have to be calculated and then it has to be subtracted from number of days of installations of 1 to 10.

Number of installation of first 20 units:

At 85% learning curve, and 20 unit number, the unit factor is 12.402 (using learning curve coefficient table)

$\begin{array}{c}{\text{T}}_{1-20}=8\times \text{12}\text{.402}\\ \text{=99}\text{.22}\text{days}\end{array}$

The times for all 20 installations are calculated by multiplying first unit of 8 with the unit factor 12.402 which yields 99.22 days.

From the computed number of days, subtract the number of needed for first 10 installations,

$\begin{array}{c}\text{Time}\text{for}\text{11-20}\text{installations}=\left(\begin{array}{l}\text{Time}\text{for}\text{1-20}\text{installations}\\ \text{-Time}\text{for}\text{1-10}\text{installations}\end{array}\right)\\ =\text{99}\text{.22}-\text{56}\text{.93}\\ =\text{42}\text{.29}\text{days}\end{array}$

Hence, the number of days for second 10 installations is 42.29 days.

c)

Summary Introduction

To determine: The number of days for last 10 installations.

Introduction: Learning curve is referred as the cost curve, experience curve, efficiency curve or productivity curve. The leaving curve gives information about the cost, efficiency, productivity and performance of an organization.

Answer to Problem 3P

The number of days for last 10 installations is 37.51 days.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Initial unit}{\text{T}}_1=8\text{days}\\ \text{Learning}\text{curve}=\text{85\%}\end{array}$

Formula:

${\displaystyle \sum {\text{T}}_{\text{n}}}={\text{T}}_{\text{1}}\times \text{Unit}\text{factor}$

Calculation of number of days for last 10 installations:

To calculate the number of days for last10 installations which is from 21 to 30, the numbers of days of installations for 1 to 30 have to be calculated and then it has to be subtracted from number of days of installations of 1 to 20.

Number of installation of first 30 units:

At 85% learning curve, and 30 unit number, the unit factor is 17.091 (using learning curve coefficient table)

$\begin{array}{c}{\text{T}}_{1-30}=8\times \text{17}\text{.091}\\ =\text{136}\text{.73}\text{days}\end{array}$

The times for all 30 installations are calculated by multiplying first unit of 8 with the unit factor 17.091 which yields 136.73 days.

From the computed number of days, subtract the number of needed for first 20 installations,

$\begin{array}{c}\text{Time}\text{for}21\text{-30}\text{installations}=\left(\begin{array}{l}\text{Time}\text{for}\text{1-30}\text{installations}\\ -\text{Time}\text{for}\text{1-20}\text{installations}\end{array}\right)\\ =\text{136}\text{.73}-\text{99}\text{.22}\\ =37.51\text{days}\end{array}$

Hence, the number of days for final 10 installations is 37.51 days.