Chapter 13, Problem 14P

A jewelry firm buys semiprecious stones to make bracelets and rings. The supplier quotes a price of $8 per stone for quantities of 600 stones or more, $9 per stone for orders of 400 to 599 stones, and $10 per stone for lesser quantities. The jewelry firm operates 200 days per year. Usage rate is 25 stones per day, and ordering costs are $48.

a. If Carrying costs are $2 per year for each stone, find the order quantity that will minimize total annual cost.

b. If annual carrying costs are 30 percent of unit cost, what is the optimal order size?

c. If lead time is six working days, at what point should the company reorder?

Summary Introduction

To determine: The Optimal order quantity that minimizes the total cost.

Introduction: Inventory is a stock or store of goods. Every company store lots of goods as inventory which will be used during replenishment periods. Management of inventory is so much essential to manage cost and also to reduce cost.

Answer to Problem 14P

The Optimal order quantity that minimizes the total cost is 600 stones.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=25\text{stones/day}\times \text{200}\text{days/year}\\ \text{=5,000}\text{stones/year}\\ \text{S}=\text{\$48}\end{array}$

Number of Stones	Price per Stone (P)
1-399	$10
400-599	$9
600+	$8

Formula:

$\text{Q}=\sqrt{\frac{\text{2DS}}{\text{H}}}$

$\text{TC}=\left(\frac{\text{Q}}{\text{2}}\right)\text{H}+\left(\frac{\text{D}}{\text{Q}}\right)\text{S}+\text{PD}$

Calculation of Economic Order Quantity:

H=$2

$\begin{array}{c}\text{Q}=\sqrt{\frac{\text{2}\times \text{5000}\times 48}{2}}\\ =490\text{stones}\end{array}$

The economic order quantity is calculated by dividing the product of 2 and 5,000 and 48 with 2 and taking square root which gives the resultant value as 490 stones.

Calculation of total cost:

$\begin{array}{c}{\text{TC}}_{490}=\left(\frac{490}{\text{2}}\right)2+\left(\frac{5,000}{490}\right)48+9\left(5,000\right)\\ =\$45,979.80\end{array}$

$\begin{array}{c}{\text{TC}}_{\text{600}}=\left(\frac{600}{\text{2}}\right)2+\left(\frac{5,000}{600}\right)48+8\left(5,000\right)\\ =\$41,000\end{array}$

The optimal order quantity is 600 stones which is the minimal total cost.

Hence, the optimal order quantity is 600 stones.

Summary Introduction

To determine: The Optimal order size.

Introduction: Inventory is a stock or store of goods. Every company store lots of goods as inventory which will be used during replenishment periods. Management of inventory is so much essential to manage cost and also to reduce cost.

Answer to Problem 14P

The Optimal order size is 600 stones.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=25\text{stones/day}\times \text{200}\text{days/year}\\ \text{=5,000}\text{stones/year}\\ \text{S}=\text{\$48}\end{array}$

Number of Stones	Price per Stone (P)
1-399	$10
400-599	$9
600+	$8

Formula:

$\text{Q}=\sqrt{\frac{\text{2DS}}{\text{H}}}$

$\text{TC}=\left(\frac{\text{Q}}{\text{2}}\right)\text{H}+\left(\frac{\text{D}}{\text{Q}}\right)\text{S}+\text{PD}$

Calculation of optimum order size:

• P=$8

$\begin{array}{c}\text{Q}=\sqrt{\frac{\text{2}\times 5,0\text{00}\times 48}{0.30\times 8}}\\ =447\text{stones}\end{array}$

The economic order quantity is calculated by dividing the product of 2 and 5,000 and 48 with the product of 0.30 and 8, square root is taken which gives the resultant value as 447 stones which is not feasible.

• P=$9

$\begin{array}{c}\text{Q}=\sqrt{\frac{\text{2}\times 5,0\text{00}\times 48}{0.30\times 9}}\\ =422\text{stones}\end{array}$

The economic order quantity is calculated by dividing the product of 2 and 5,000 and 48 with the product of 0.30 and 9, square root is taken which gives the resultant value as 422 stones which is feasible.

Calculation of total cost:

$\begin{array}{c}{\text{TC}}_{422}=\left(\frac{422}{\text{2}}\right)\left(0.30\times 9\right)+\left(\frac{5,000}{422}\right)48+9\left(5,000\right)\\ =\$46,138.42\end{array}$

$\begin{array}{c}{\text{TC}}_{\text{600}}=\left(\frac{600}{\text{2}}\right)\left(0.30\times 8\right)+\left(\frac{5,000}{600}\right)48+8\left(5,000\right)\\ =\$41,120\end{array}$

The optimal order quantity is 600 stones which is the minimal total cost.

Hence, the optimal order quantity is 600 stones.

Summary Introduction

To determine: The reorder point.

Introduction: Inventory is a stock or store of goods. Every company store lots of goods as inventory which will be used during replenishment periods. Management of inventory is so much essential to manage cost and also to reduce cost.

Answer to Problem 14P

The reorder point is 150 stones.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=25\text{stones/day}\times \text{200}\text{days/year}\\ \text{=5,000}\text{stones/year}\\ \text{S}=\text{\$48}\end{array}$

Number of Stones	Price per Stone (P)
1-399	$10
400-599	$9
600+	$8

Calculation of reorder point:

$\begin{array}{c}\text{ROP}=25\text{stones/day}\times 6\text{days}\\ =150\text{stones}\end{array}$

The reorder point is calculated by multiplying 25 stones/day with 6 days which gives 150 stones.

Hence, the reorder point is 150 stones.