Chapter 13, Problem 11P

A company is about to begin production of a new product. The manager of the department that will produce one of the components for the product wants to know how often the machine used to produce the item will be available for other work. The machine will produce the item at a rate of 200 units a day. Eighty units will be used daily in assembling the final product. Assembly will take place five days a week, 50 weeks a year. The manager estimates that it will take almost a full day to get the machine ready for a production nm, at a cost of $300. Inventory holding costs will be $10 a year.

a. What runs quantity should be used to minimize total annual costs?

b. How many days does it take to produce the optimal run quantity?

c. What is the average amount of inventory?

d. If the manager wants to run another job between runs of this item, and needs a minimum of 10 days per cycle for the other work, will there be enough time?

e. Given your answer to part d, the manager wants to explore options that will allow this other job to be performed using this equipment. Name three options the manager can consider.

f. Suppose the manager decides to increase the new size the new product. How many additional units would be needed to just accommodate the other job? How much will that increase the total annual cost?

Summary Introduction

To determine: The run quantity to minimize 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 11P

The run quantity to minimize the total cost is 1,414 units.

Explanation of Solution

Given information:

$\begin{array}{c}D=250\text{days}\times 80\text{/day}=20,000\\ \text{u}=80\text{/day}\\ \text{p}=200\text{/day}\end{array}$

$\begin{array}{c}\text{H}=\$10\\ \text{S}=\$300\end{array}$

Formula:

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

Calculation of run quantity to minimize the total cost:

$\begin{array}{c}{\text{Q}}_{\text{o}}=\sqrt{\frac{\text{2}\times 20,000\times 300}{10}}\sqrt{\frac{200}{200-80}}\\ =1,414\text{units}\end{array}$

The optimal run size is calculated by dividing the product of 2 and 200,000 and 300 with 10 and taking square root which gives the resultant value as 1095 units and 200 is divided with the difference of 200 and 80 and square root is taken which gives 1.290, both values are multiplies which gives the optimal run size as 1414 units.

Hence, the run quantity to minimize the total cost is 1,414 units.

Summary Introduction

To determine: The number of days to produce optimal run quantities.

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 11P

The number of days to produce optimal run quantities are 7.70 days.

Explanation of Solution

Given information:

$\begin{array}{c}D=250\text{days}\times 80\text{/day}=20,000\\ \text{u}=80\text{/day}\\ \text{p}=200\text{/day}\end{array}$

$\begin{array}{c}\text{H}=\$10\\ \text{S}=\$300\end{array}$

Formula:

$\text{Days}\text{to}\text{produce}\text{optimal}\text{run}\text{quantity}=\frac{{\text{Q}}_{\text{p}}}{\text{p}}$

Calculation of days to produce optimal run quantities:

$\begin{array}{l}=\frac{1,414}{200}\\ =7.07\text{days}\end{array}$

The number of days to produce optimal run quantity is calculated by dividing 1,414 with 200 which gives 7.07 days.

Hence, the number of days to produce optimal run quantities is 7.70 days.

Summary Introduction

To determine: The average inventory.

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 11P

The average inventory is 625 units.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=250\text{days}\times 80\text{/day}=20,000\\ \text{u}=80\text{/day}\\ \text{p}=200\text{/day}\end{array}$

$\begin{array}{c}\text{H}=\$10\\ \text{S}=\$300\end{array}$

Formula:

$\text{Average}\text{inventory}=\frac{{\text{I}}_{\text{max}}}{2}$

Calculation of average inventory:

$\begin{array}{c}{\text{I}}_{\text{max}}=\frac{{\text{Q}}_{\text{p}}}{\text{p}}\left(\text{p}-\text{u}\right)\\ =\frac{1,414}{200}\left(200-80\right)\\ =848.4\text{units}\end{array}$

$\begin{array}{c}\text{Average}\text{inventory}=\frac{848.4}{2}\\ =424.2\text{units}\end{array}$

The average inventory is calculated by dividing 848.4 by 2 which yields 424.2 units.

Hence, the average inventory is 424.2 units.

Summary Introduction

To explain: Whether there is adequate time for another job between the run of items.

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 11P

There will not be adequate time for the other job.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=250\text{days}\times 80\text{/day}=20,000\\ \text{u}=80\text{/day}\\ \text{p}=200\text{/day}\end{array}$

$\begin{array}{c}\text{H}=\$10\\ \text{S}=\$300\end{array}$

Formula:

$\text{Cycle}\text{time}=\text{Run}\text{time}+\text{Consumption}\text{time}$

Calculation to check whether there is adequate time to do the job between the production batches:

$\begin{array}{c}\text{Cycle}\text{time}=\frac{{\text{Q}}_{\text{p}}}{\text{u}}\\ =\frac{1,414}{80}\\ =17.68\text{days}\end{array}$

The cycle time is calculated by dividing 1,414 with 80 which yields the resultant as 17.68 days.

$\begin{array}{c}\text{Run}\text{time}=\frac{{\text{Q}}_{\text{p}}}{\text{p}}+\text{Setup}\text{time}\\ =\frac{1,414}{\text{200}}+1\\ =8.07\text{days}\end{array}$

The run time is calculated by dividing 1,414 by 200 and adding the resultant with setup time which yields 8.07 days.

$\begin{array}{c}\text{Cycle}\text{time}=\text{Run}\text{time}+\text{consumption}\text{time}\\ \text{Consumption}\text{time}=17.68-8.07\\ =9.61\text{days}\end{array}$

Consumption time is calculated by subtracting cycle time of 17.68 with the consumption time of 8.07 which yields the resultant as 9.61 days which is not sufficient since other job requires 10 days.

Hence, there will not be adequate time for the other job.

Summary Introduction

To explore: Three options for managers that will allow him to do other job.

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.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=250\text{days}\times 80\text{/day}=20,000\\ \text{u}=80\text{/day}\\ \text{p}=200\text{/day}\end{array}$

$\begin{array}{c}\text{H}=\$10\\ \text{S}=\$300\end{array}$

Three options for managers that will allow him to do other job:

The three options for manager to explore the ways to do other jobs are,

• The setup time for other job can be shortened.
• Increase the run time of the new product which allows longer run time between runs which is there less need to run the component.
• Reducing the run time for other jobs.

Summary Introduction

To determine: The addition units produced from the new product and the total annual 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 11P

Additional units produced and increase in cost is 53 units per run and $5.7 per year.

Explanation of Solution

Given information:

$\begin{array}{c}\text{D}=250\text{days}\times 80\text{/day}=20,000\\ \text{u}=80\text{/day}\\ \text{p}=200\text{/day}\end{array}$

$\begin{array}{c}\text{H}=\$10\\ \text{S}=\$300\end{array}$

Formula:

$\text{Cycle}\text{time}=\text{Run}\text{time}+\text{Consumption}\text{time}$

Calculation of addition units produced from the new product and the total annual cost:

The additional units produced is computed by finding the value of Q_p

$\begin{array}{c}\text{Cycle}\text{time}=\text{Run}\text{time}+\text{Consumption}\text{time}\\ \frac{{\text{Q}}_{\text{p}}}{80}=\left[\frac{{\text{Q}}_{\text{p}}}{200}+1\right]+10\end{array}$

Solving for Q_p, we get the values as,

${\text{Q}}_{\text{p}}=1467\text{units}$

The additional units per run = 1,467 – 1,414 = 53 units per run.

The additional unit per run is calculated by subtracting 1,467 and 1,414 which gives 53 units per run.

Increase in total cost:

$\begin{array}{c}{\text{I}}_{\text{max}}=\frac{{\text{Q}}_{\text{p}}}{\text{p}}\left(\text{p}-\text{u}\right)\\ =\frac{1,467}{200}\left(200-80\right)\\ =880.2\text{units}\end{array}$ $$

$\begin{array}{c}{\text{TC}}_{1,467}=\left(\frac{{\text{I}}_{\text{max}}}{\text{2}}\right)\text{H}+\left(\frac{\text{D}}{\text{Q}}\right)\text{S}\\ =\left(\frac{880.2}{\text{2}}\right)10+\left(\frac{20,000}{1,467}\right)300\\ =\$4,401+\$4,089.98\\ =\$8,485.28\end{array}$

$\begin{array}{c}{\text{TC}}_{1,414}=\left(\frac{{\text{I}}_{\text{max}}}{\text{2}}\right)\text{H}+\left(\frac{\text{D}}{\text{Q}}\right)\text{S}\\ =\left(\frac{848.2}{\text{2}}\right)10+\left(\frac{20,000}{1,414}\right)300\\ =\$4,242+\$4,243.28\\ =\$5.70/\text{year}\end{array}$

Difference in total cost:

$\begin{array}{c}\text{Increase}=\$8,490.98-\$8,485.28\\ =\$5.70/\text{year}\end{array}$

The increase in total cost is calculated by calculating the difference between $8,490.98 and $8,485.28 which is $5.70 per year.

Hence, additional units produced and increase in cost is 53 units per run and $ 5.7 per year.