Question

In a store, the daily demand for milk (in litres) is a random variable having Exp(λ) distribution, where λ > 0. At the beginning of the day, the store purchases c (> 0) litres of milk at a fixed price b (> 0) per litre. The milk is then sold to the customers at a fixed price s (> b) per litre. At the end of the day, the unsold milk is discarded. Then the value of c that maximizes the expected net profit for the store equals

• A. \(-\frac{1}{λ}\ln(\frac{b}{s})\)
• B. \(-\frac{1}{λ}\ln(\frac{b}{s+b})\)
• C. \(-\frac{1}{λ}\ln(\frac{s-b}{s})\)
• D. \(-\frac{1}{λ}\ln(\frac{s}{s+b})\)

Answer 1

The Correct Option is A

The problem involves maximizing the expected net profit for a store based on the daily demand for milk, which follows an exponential distribution. Let's break down the problem step-by-step:

1. **Understand the Variables and Distribution:** - Let \( X \) be the daily demand for milk in liters, which is an exponentially distributed random variable with parameter \(\lambda\). The probability density function (pdf) of an exponential distribution is given by: \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\).

2. **Profit Calculation:** - The store purchases \( c \) liters of milk at a cost \( b \) per liter, so the cost is \( bc \). - If the store sells \( y \) liters, the revenue from selling milk is \( sy \), where \( s \) is the selling price per liter. - The profit from selling \( y \) liters is \( sy - bc \). - However, \( y \) depends on whether the demand \( X \) is less than or greater than the supply \( c \): - If \( X \leq c \), then \( y = X \). - If \( X > c \), then \( y = c \).

3. **Expected Profit Calculation:** - The expected profit is given by: \(E[\text{Profit}] = \int_{0}^{c} (sx - bc) \lambda e^{-\lambda x} \, dx + \int_{c}^{\infty} (sc - bc) \lambda e^{-\lambda x} \, dx\)

4. **Simplifying the Expected Profit:** - Compute each integral separately: \(E[\text{Profit}] = \int_{0}^{c} (sx - bc) \lambda e^{-\lambda x} \, dx + (sc - bc) e^{-\lambda c}\)- Solve this integral. The detailed calculation and solving will show that:

5. **Maximization:** - Differentiate the expected profit with respect to \( c \) and set it to zero to find the optimal \( c \). - After solving, the value of \( c \) that maximizes the expected net profit is found to be: \(c^* = -\frac{1}{\lambda}\ln\left(\frac{b}{s}\right)\)

6. **Conclusion:** - Thus, the correct value of \( c \) that maximizes the expected net profit is: \(-\frac{1}{\lambda}\ln\left(\frac{b}{s}\right)\)

This matches with the provided correct answer option: \(-\frac{1}{\lambda}\ln(\frac{b}{s})\).