Solved Problems

Solved Problem M3-1 Terry Wagner is considering self-publishing a book on yoga. She has been teaching yoga for more than 20 years. She believes that the fixed costs of publishing the book will be about $10,000. The variable costs are $5.50, and the price of the yoga book to bookstores is expected to be $12.50. What is the break-even point for Terry?

Solution

This problem can be solved using the break-even formula in the module:

$\begin{array}{l} Break-even point in units & = & \frac{$ 10, 000}{$ 12.50 - $ 5.50} \\ = & \frac{$ 10, 000}{$ 7} \\ = & 1, 429 units \end{array}$ $\begin{array}{l} Break-even point in units & = & \frac{$ 10, 000}{$ 12.50 - $ 5.50} \\ = & \frac{$ 10, 000}{$ 7} \\ = & 1, 429 units \end{array}$
Solved Problem M3-2 The annual demand for a new electric product is expected to be normally distributed with a mean of 16,000 and a standard deviation of 2,000. The break-even point is 14,000 units. For each unit less than 14,000, the company will lose $24. Find the expected opportunity loss.

Solution

The expected opportunity loss (EOL) is

$EOL = K σ N (D)$ $EOL = K σ N (D)$

We are given the following:

$\begin{array}{l} K & = & loss per unit = $24 \\ μ & = & 16, 000 \\ σ & = & 2, 000 \end{array}$ $\begin{array}{l} K & = & loss per unit = $24 \\ μ & = & 16, 000 \\ σ & = & 2, 000 \end{array}$

Using Equation M3-6, we find

$\begin{array}{l} D & = & | \frac{μ - Break-even point}{σ} | & = & | \frac{16, 000 - 14, 000}{2, 000} | & = & 1 \\ N (D) & = & N (1) & = & 0.08332 from Appendix M3 .2 \\ EOL & = & K σ N (1) & = & 24 (2, 000) (0.08332) & = & $ 3, 999.36 \end{array}$ $\begin{array}{l} D & = & | \frac{μ - Break-even point}{σ} | & = & | \frac{16, 000 - 14, 000}{2, 000} | & = & 1 \\ N (D) & = & N (1) & = & 0.08332 from Appendix M3 .2 \\ EOL & = & K σ N (1) & = & 24 (2, 000) (0.08332) & = & $ 3, 999.36 \end{array}$

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