Chapter 7, Problem 6P

a)

Summary Introduction

To determine: The graph illustrating the three total cost lines on the same chart.

Answer to Problem 6P

Graph illustrating the three cost lines:

Explanation of Solution

Given information:

Refurbishing cost = $800

Cost of making modifications  = $1,100

Purchase of new equipment  = $1,800

Refurbishing labor and material costs = $1.10 / board

Modifications labor and material costs = $0.70 / board

New equipment variable costs = $0.40 / board

Formation of cost lines denoted in the graph:

The cost line is formed by multiplying the labor cost and number of units and summing the value with the process cost. Let ‘Y’ denote the cost line. Let ‘x’ denote the number of units.

Refurbishing cost line (A):

The cost line (A) mentioned in the graph is derived as follows:

$\begin{array}{c}\text{Y}=\text{Refurbishing cost}+\left(\text{Refurbishing labor cost}\times \text{Number of units}\right)\\ =\text{800}+\left(\text{1}\text{.10}\times \text{x}\right)\\ =\text{800}+\text{1}\text{.10x}\end{array}$ (A)

Modifications cost line (B):

The cost line (B) mentioned in the graph is derived as follows:

$\begin{array}{c}\text{Y}=\text{Modifications cost}+\left(\text{Modifications labor cost}\times \text{Number of units}\right)\\ =\text{1,100}+\left(\text{0}\text{.70}\times \text{x}\right)\\ =\text{1,100}+\text{0}\text{.70x}\end{array}$ (B)

New equipment cost line (C):

The cost line (C) mentioned in the graph is derived as follows:

$\begin{array}{c}\text{Y}=\text{New equipment cost}+\left(\text{New equipment labor cost}\times \text{Number of units}\right)\\ =\text{1,800}+\left(\text{0}\text{.40}\times \text{x}\right)\\ =\text{1,800}+\text{0}\text{.40x}\end{array}$ (C)

Calculation of cost at 0 units and arbitrary units of 4,000 using the cost lines (A), (B) and (C):

Refurbishing:

At 0 units:

$\begin{array}{c}\text{Y}=\text{800}+\text{1}\text{.10x}\\ =\text{800}+\left(\text{1}\text{.10}\times \text{0}\right)\\ =\text{800}\end{array}$

At 4,000 units:

$\begin{array}{c}\text{Y}=\text{800}+\text{1}\text{.10x}\\ =\text{800}+\left(\text{1}\text{.10}\times \text{4,000}\right)\\ =\text{800}+\text{4,400}\\ =\text{5,200}\end{array}$

Modification:

At 0 units:

$\begin{array}{c}\text{Y}=\text{1,100}+\text{0}\text{.70x}\\ =\text{1,100}+\left(\text{0}\text{.70}\times \text{0}\right)\\ =\text{1,100}\end{array}$

At 4,000 units:

$\begin{array}{c}\text{Y}=\text{1,100}+\text{0}\text{.70x}\\ =\text{1,100}+\left(\text{0}\text{.70}\times \text{4,000}\right)\\ =\text{1,100}+\text{2,800}\\ =\text{3,900}\end{array}$

New equipment:

At 0 units:

$\begin{array}{c}\text{Y}=\text{1,800}+\text{0}\text{.40x}\\ =\text{1,800}+\left(\text{0}\text{.40}\times \text{0}\right)\\ =\text{1,800}\end{array}$

At 4,000 units:

$\begin{array}{c}\text{Y}=\text{1,800}+\text{0}\text{.40x}\\ =\text{1,800}+\left(\text{0}\text{.40}\times \text{4,000}\right)\\ =\text{1,800}+\text{1,600}\\ =\text{3,400}\end{array}$

The cost values at 0 and 400 units are plotted on the graph.

b)

Summary Introduction

To determine: The alternative to choose when SB company thinks it can sell more than 3,000 boards.

Answer to Problem 6P

The best alternative when more than 3,000 boards can be sold is alternative 1.

Explanation of Solution

Given information:

Refurbishing cost = $800

Cost of making modifications  = $1,100

Purchase of new equipment  = $1,800

Refurbishing labor and material costs = $1.10 / board

Modifications labor and material costs = $0.70 / board

New equipment variable costs = $0.40 / board

Formation of cost lines:

The cost line is formed by multiplying the labor cost and number of units and summing the value with the process cost. Let ‘Y’ denote the cost line. Let ‘x’ denote the number of units.

Refurbishing cost line:

$\begin{array}{c}\text{Y}=\text{Refurbishing cost}+\left(\text{Refurbishing labor cost}\times \text{Number of units}\right)\\ =\text{800}+\left(\text{1}\text{.10}\times \text{x}\right)\\ =\text{800}+\text{1}\text{.10x}\end{array}$ (A)

Modifications cost line:

$\begin{array}{c}\text{Y}=\text{Modifications cost}+\left(\text{Modifications labor cost}\times \text{Number of units}\right)\\ =\text{1,100}+\left(\text{0}\text{.70}\times \text{x}\right)\\ =\text{1,100}+\text{0}\text{.70x}\end{array}$ (B)

New equipment cost line:

$\begin{array}{c}\text{Y}=\text{New equipment cost}+\left(\text{New equipment labor cost}\times \text{Number of units}\right)\\ =\text{1,800}+\left(\text{0}\text{.40}\times \text{x}\right)\\ =\text{1,800}+\text{0}\text{.40x}\end{array}$ (C)

Calculation of cost at 3,000 boards:

Alternative 1:

Alternative 1 is refurbishing cost calculated by using equation (A)

$\begin{array}{c}\text{Y}=\text{800}+\text{1}\text{.10x}\\ =\text{800}+\left(\text{1}\text{.10}\times 3\text{,000}\right)\\ =\text{800}+3\text{,300}\\ =4\text{,100}\end{array}$

Alternative 2:

Alternative 2 is modifications cost calculated by using equation (B)

$\begin{array}{c}\text{Y}=\text{1,100}+\text{0}\text{.70x}\\ =\text{1,100}+\left(\text{0}\text{.70}\times 3\text{,000}\right)\\ =\text{1,100}+\text{2,100}\\ =\text{3,200}\end{array}$

Alternative 3:

Alternative 3 is new equipment cost calculated by using equation (C)

$\begin{array}{c}\text{Y}=\text{1,800}+\text{0}\text{.40x}\\ =\text{1,800}+\left(\text{0}\text{.40}\times 3\text{,000}\right)\\ =\text{1,800}+\text{1,200}\\ =\text{3,000}\end{array}$

The total cost of alternative 1 is less than alternative 2 and 3 (3000 < 4100, 3200).

Hence, the best alternative for making 3,000 boards is alternative 3.

c)

Summary Introduction

To determine: The alternative to choose when SB company thinks that the market for boards will be between 1,000 and 2,000 boards.

Answer to Problem 6P

The best alternative when the market for boards will be between 1,000 and 2,000 boards is making modifications.

Explanation of Solution

Given information:

Refurbishing cost = $800

Cost of making modifications  = $1,100

Purchase of new equipment  = $1,800

Refurbishing labor and material costs = $1.10 / board

Modifications labor and material costs = $0.70 / board

New equipment variable costs = $0.40 / board

Formation of cost lines:

The cost line is formed by multiplying the labor cost and number of units and summing the value with the process cost. Let ‘Y’ denote the cost line. Let ‘x’ denote the number of units.

Refurbishing cost line:

$\begin{array}{c}\text{Y}=\text{Refurbishing cost}+\left(\text{Refurbishing labor cost}\times \text{Number of units}\right)\\ =\text{800}+\left(\text{1}\text{.10}\times \text{x}\right)\\ =\text{800}+\text{1}\text{.10x}\end{array}$ (A)

Modifications cost line:

$\begin{array}{c}\text{Y}=\text{Modifications cost}+\left(\text{Modifications labor cost}\times \text{Number of units}\right)\\ =\text{1,100}+\left(\text{0}\text{.70}\times \text{x}\right)\\ =\text{1,100}+\text{0}\text{.70x}\end{array}$ (B)

New equipment cost line:

$\begin{array}{c}\text{Y}=\text{New equipment cost}+\left(\text{New equipment labor cost}\times \text{Number of units}\right)\\ =\text{1,800}+\left(\text{0}\text{.40}\times \text{x}\right)\\ =\text{1,800}+\text{0}\text{.40x}\end{array}$ (C)

Calculation of best alternative:

The best alternative is calculated by equating (A) with (B) and (B) with (C).

Calculation of lower limit value:

Lower limit value is calculated by Equating (A) with (B):

$\begin{array}{c}\text{800}+\text{1}\text{.10x}=\text{1,100}+\text{0}\text{.70x}\\ \text{1}\text{.10x}-\text{0}\text{.70x}=\text{1,100}-\text{800}\\ \text{0}\text{.4x}=\text{300}\end{array}$

$\begin{array}{c}\text{x}=\frac{\text{300}}{\text{0}\text{.4}}\\ \text{x}=\text{750 units}\end{array}$

Calculation of Upper limit value:

Upper limit value is calculated by Equating (B) with (C):

$\begin{array}{c}\text{1,100}+0.\text{70x}=\text{1,800}+\text{0}\text{.40x}\\ \text{0}\text{.70x}-\text{0}\text{.40x}=\text{1,800}-1,1\text{00}\\ \text{0}\text{.3x}=7\text{00}\end{array}$

$\begin{array}{c}\text{x}=\frac{\text{700}}{\text{0}\text{.3}}\\ \text{x}=2,333\text{ units}\end{array}$

The boards must be in the range of 1,000 to 2,000. Hence, the best alternative is making major modifications.