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Question 1198377: Assume that your widget manufacturing company has a total annual demand of N widgets per year evenly distributed across the year. Each widget cost $b dollars in material and manufacturing costs to make. Every time you do a production run to make some widgets, you incur a set-up cost of P dollars. Any widgets awaiting sale must be stored and thus incur an average storage fee of c dollars per widget per year. Let x be the size of each production run (i.e. x is the number of widgets per production run).
Write a cost function C(x) and explain each term in the equation and how it was determined.
Write down any constraints on the allowable values of x.
Determine a formula for the value of x that minimizes total annual cost. Show all of your work.
Prove that your formula actually corresponds to the global minimum cost.
Write down a formula for the number of production runs per year as a function of x.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! ### **1. Cost Function \( C(x) \):**
The total cost \( C(x) \) includes three components:
1. **Material and manufacturing cost**:
- The total number of widgets demanded annually is \( N \).
- Each widget costs \( b \) dollars to make.
- The **total manufacturing cost** is:
\[
\text{Manufacturing Cost} = N \cdot b
\]
2. **Setup cost**:
- A production run is done every \( x \) widgets.
- The total number of production runs per year is \( \frac{N}{x} \) (since \( N \) widgets are needed annually, and \( x \) widgets are produced per run).
- Each production run incurs a setup cost \( P \).
- The **total setup cost** is:
\[
\text{Setup Cost} = \frac{N}{x} \cdot P
\]
3. **Storage cost**:
- Widgets are produced in batches of \( x \) and consumed uniformly over the year.
- The average number of widgets stored at any time is \( \frac{x}{2} \) (half of the production run, assuming constant consumption).
- The storage cost per widget per year is \( c \).
- The **total storage cost** is:
\[
\text{Storage Cost} = \frac{x}{2} \cdot c
\]
Combining these components, the total cost function is:
\[
C(x) = N \cdot b + \frac{N}{x} \cdot P + \frac{x}{2} \cdot c
\]
---
### **2. Constraints on \( x \):**
1. \( x > 0 \): The production run size must be positive.
2. \( x \leq N \): The production run size cannot exceed the total annual demand.
Thus, \( x \) must satisfy:
\[
0 < x \leq N
\]
---
### **3. Minimizing \( C(x) \):**
To minimize \( C(x) \), take the derivative of \( C(x) \) with respect to \( x \) and set it to zero.
\[
C(x) = N \cdot b + \frac{N}{x} \cdot P + \frac{x}{2} \cdot c
\]
\[
\frac{dC}{dx} = -\frac{N \cdot P}{x^2} + \frac{c}{2}
\]
Set \( \frac{dC}{dx} = 0 \):
\[
-\frac{N \cdot P}{x^2} + \frac{c}{2} = 0
\]
\[
\frac{N \cdot P}{x^2} = \frac{c}{2}
\]
\[
x^2 = \frac{2 \cdot N \cdot P}{c}
\]
\[
x = \sqrt{\frac{2 \cdot N \cdot P}{c}}
\]
---
### **4. Verifying Minimum:**
To confirm that this value of \( x \) gives a global minimum, examine the second derivative:
\[
\frac{d^2C}{dx^2} = \frac{2N \cdot P}{x^3}
\]
For \( x > 0 \), \( \frac{d^2C}{dx^2} > 0 \), indicating that \( C(x) \) is convex and has a global minimum at \( x = \sqrt{\frac{2 \cdot N \cdot P}{c}} \).
---
### **5. Number of Production Runs per Year:**
The number of production runs per year is:
\[
\text{Production Runs} = \frac{N}{x}
\]
Substitute \( x = \sqrt{\frac{2 \cdot N \cdot P}{c}} \):
\[
\text{Production Runs} = \frac{N}{\sqrt{\frac{2 \cdot N \cdot P}{c}}} = \sqrt{\frac{N \cdot c}{2 \cdot P}}
\]
---
### **Summary of Results:**
1. **Cost Function**:
\[
C(x) = N \cdot b + \frac{N \cdot P}{x} + \frac{x \cdot c}{2}
\]
2. **Optimal Production Run Size**:
\[
x = \sqrt{\frac{2 \cdot N \cdot P}{c}}
\]
3. **Number of Production Runs per Year**:
\[
\text{Production Runs} = \sqrt{\frac{N \cdot c}{2 \cdot P}}
\]
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