Chapter 7, Problem 1P

a)

Summary Introduction

To determine: The daily production quantities and sequences.

Introduction:

Lean systems in an organization would have greater impact on the customers and their values. The goal of lean thinking is to improve the productivity of the department and the goal for any lean system is to maximize the customer value.

Explanation of Solution

Given information:

It is given that the month has 20 production days. The monthly requirement of Model A is 5,000 units, Model B is 2,500 units, and Model C is 3,000 units.

Determine the daily production quantities and sequences:

Daily production quantity of Model A:

This can be calculated by multiplying the monthly requirement of Model A and the value of one divided by the monthly production days.

$\begin{array}{c}\text{Daily requirement of Model A}=\left(\text{Monthly requirement}\right)\times \frac{1}{\text{Production days}}\\ =5,000\times \frac{1}{20}\\ =5,000\times 0.05\\ =250\end{array}$

Daily production quantity of Model B:

It can be calculated by multiplying monthly requirement of Model B and the value of one divided by the monthly production days.

$\begin{array}{c}\text{Daily requirement of Model B}=\left(\text{Monthly requirement}\right)\times \frac{1}{\text{Production days}}\\ =2,500\times \frac{1}{20}\\ =2,500\times 0.05\\ =125\end{array}$

Daily production quantity of Model C:

It can be calculated by multiplying monthly requirement of Model C and the value of one divided by the monthly production days.

$\begin{array}{c}\text{Daily requirement of Model C}=\left(\text{Monthly requirement}\right)\times \frac{1}{\text{Production days}}\\ =3,000\times \frac{1}{20}\\ =3,000\times 0.05\\ =150\end{array}$

Hence, the daily production quantity of Model A is 250, Model B is 125, and Model C is 150. Here, the common multiple is 25.

Determine the sequence for Model A:

$\begin{array}{c}\text{Sequence}=\frac{\text{Daily production requirement}}{\text{Common multiple}}\\ =\frac{250}{25}\\ =10\end{array}$

Determine the sequence for Model B:

$\begin{array}{c}\text{Sequence}=\frac{\text{Daily production requirement}}{\text{Common multiple}}\\ =\frac{125}{25}\\ =5\end{array}$

Determine the sequence for Model C:

$\begin{array}{c}\text{Sequence}=\frac{\text{Daily production requirement}}{\text{Common multiple}}\\ =\frac{150}{25}\\ =6\end{array}$

Hence, the sequence is (10) A (5) B (6) C.

Note: The largest common multiplier serves as the number of runs required per day.

b)

Summary Introduction

To determine: The daily production quantities and sequences.

Introduction:

Lean systems in an organization would have greater impact on the customers and their values. The goal of lean thinking is to improve the productivity of the department and the goal for any lean system is to maximize the customer value.

Explanation of Solution

Given information:

It is given that the month has 20 production days. Monthly requirement of Model A is 2,000 units, Model B is 3,000 units, and Model C is 6,000 units.

Determine the daily production quantities and sequences:

Daily production quantity of Model A:

It can be calculated by multiplying monthly requirement of Model A and the value of one divided by the monthly production days.

$\begin{array}{c}\text{Daily requirement of Model A}=\left(\text{Monthly requirement}\right)\times {\frac{1}{\text{Production days}}}_{}^{}\\ =2,000\times \frac{1}{20}\\ =2,000\times 0.05\\ =100\end{array}$

Daily production quantity of Model B:

It can be calculated by multiplying monthly requirement of Model B and the value of one divided by the monthly production days.

$\begin{array}{c}\text{Daily requirement of Model B}=\left(\text{Monthly requirement}\right)\times \frac{1}{\text{Production days}}\\ =3,000\times \frac{1}{20}\\ =3,000\times 0.05\\ =150\end{array}$

Daily production quantity of Model C:

It can be calculated by multiplying monthly requirement of Model C and the value of one divided by the monthly production days.

$\begin{array}{c}\text{Daily requirement of Model C}=\left(\text{Monthly requirement}\right)\times \frac{1}{\text{Production days}}\\ =6,000\times \frac{1}{20}\\ =6,000\times 0.05\\ =300\end{array}$

Hence, the daily production quantity of Model A is 100, Model B is 150, and Model C is 300. Here, the common multiple is 50.

Determine the sequence for Model A:

$\begin{array}{c}\text{Sequence}=\frac{\text{Daily production requirement}}{\text{Common multiple}}\\ =\frac{100}{50}\\ =2\end{array}$

Determine the sequence for Model B:

$\begin{array}{c}\text{Sequence}=\frac{\text{Daily production requirement}}{\text{Common multiple}}\\ =\frac{150}{50}\\ =3\end{array}$

Determine the sequence for Model C:

$\begin{array}{c}\text{Sequence}=\frac{\text{Daily production requirement}}{\text{Common multiple}}\\ =\frac{300}{50}\\ =6\end{array}$

Hence, the sequence would be (2) A (3) B (6) C.

Note: The largest common multiplier serves as the number of runs required per day.