How many donkeys should the cruise line have in its herd to minimize costs, considering a normally distributed demand with a mean of 10,000 and a standard deviation of 3,000, a cos... How many donkeys should the cruise line have in its herd to minimize costs, considering a normally distributed demand with a mean of 10,000 and a standard deviation of 3,000, a cost of $15,000 per thousand tourists for maintaining a herd, and a spot market cost of $20 per tourist? A local donkey herder offers a rental price of $17.50 per tourist. Read more

Steps to Solve

1. Define the cost functions

Let $x$ be the number of tourists.

Cost of maintaining a donkey herd: $C_1(x) = 15000 \cdot \frac{x}{1000} = 15x$ if $x \le \text{herd size}$, and $15\cdot \text{herd size} + 17.5(x - \text{herd size})$ if $x > \text{herd size}$. Cost of using the spot market: $C_2(x) = 20x$. Cost of renting from the local herder: $C_3(x) = 17.5x$. We need to find the optimal herd size that minimizes the expected total cost.

2. Incorporate the rental option

Since the local donkey herder offers a better price than the spot market ($17.50 < $20), we will always choose the rental option if the herd size is insufficient. Therefore, if the demand $x$ exceeds the herd size, the remaining tourists will be served by the local herder at a cost of $17.50 per tourist.

3. Define the decision variable and formulate the cost function

Let $Q$ be the size of the donkey herd. The cost function will now be: $C(x, Q) = 15x$, if $x \le Q$ $C(x, Q) = 15Q + 17.5(x - Q)$, if $x > Q$

4. Simplify the cost function

$C(x, Q) = 15x$, if $x \le Q$ $C(x, Q) = 15Q + 17.5x - 17.5Q = 17.5x - 2.5Q$, if $x > Q$

5. Find the expected cost

The demand $x$ is normally distributed with mean $\mu = 10000$ and standard deviation $\sigma = 3000$. We want to find the optimal $Q$ that minimizes the expected cost $E[C(x, Q)]$.

$E[C(Q)] = \int_{-\infty}^{Q} 15x f(x) dx + \int_{Q}^{\infty} (17.5x - 2.5Q) f(x) dx$, where $f(x)$ is the probability density function of the normal distribution.

6. Differentiate the expected cost function with respect to Q and set to zero

To minimize the expected cost, we differentiate $E[C(Q)]$ with respect to $Q$ and set the derivative equal to zero. $\frac{dE[C(Q)]}{dQ} = \frac{d}{dQ} \left[ \int_{-\infty}^{Q} 15x f(x) dx + \int_{Q}^{\infty} (17.5x - 2.5Q) f(x) dx \right] = 0$

Using Leibniz rule: $15Q f(Q) + (17.5Q - 2.5Q)(-f(Q)) + \int_{Q}^{\infty} -2.5 f(x) dx = 0$ $15Q f(Q) - 15Q f(Q) - 2.5 \int_{Q}^{\infty} f(x) dx = 0$ $- 2.5 \int_{Q}^{\infty} f(x) dx = 0$

$\int_{Q}^{\infty} f(x) dx$ is not zero. There appears to be an error in the formulation. Let's check the differentiation step.

$\frac{dE[C(Q)]}{dQ} = \frac{d}{dQ} \left[ \int_{-\infty}^{Q} 15x f(x) dx + \int_{Q}^{\infty} (17.5x - 2.5Q) f(x) dx \right] = 0$ $15Q f(Q) - (17.5Q - 2.5Q) f(Q) + \int_{Q}^{\infty} \frac{d}{dQ}(-2.5Q) f(x) dx = 0$ $15Q f(Q) - 15Q f(Q) - 2.5 \int_{Q}^{\infty} f(x) dx = 0$ $-2.5 \int_{Q}^{\infty} f(x) dx = 0$. This is incorrect. Must be an easier method.

7. Marginal Analysis Let $Q$ be the herd size. Consider adding one more donkey. The cost associated with adding one more donkey is $15. The benefit of adding one donkey is that when the demand exceeds the current herd size, we save $17.50 - 15 = 2.50$ per tourist. The probability that the demand exceeds Q is $P(x > Q)$.

We want to find $Q$ such that the cost of adding one more donkey is equal to the expected benefit. $15 = 2.5 P(x > Q)$ $P(x > Q) = \frac{15}{2.5} = 6$

This result doesn't make sense as $P(x>Q)$ must be less than 1. Let's analyze it again Instead of focusing on one tourist, consider increments of 1000 tourists. The cost of owning a donkey to transport 1000 tourists is $15000$. The benefit of owning versus renting one donkey is $17.50-15=2.50$ per tourist The equivalent analysis suggests that adding one donkey saves $2.50 \cdot 1000 = $2500 So, $15000 = 2500 P(x>Q)$ $P(x > Q) = \frac{15000}{2500} = 6$

Let’s think about cost savings. Let $q$ be the number of thousand tourists. $15q = 15000$ $17.5q = 17500$ per thousand tourists If we decide to increase the size of donkey herd by 1 unit (thousand tourists) the cost is $15000$. The benefit will be in terms of decreased renting. The rental savings is $2500 per donkey/thousand tourists So $P(\text{demand} > Q) = \frac{15000}{17500-15000} = \frac{15000}{2500} = 6$ - does not make sense. This cant be larger than 1!!

8. Correct Marginal Analysis

Cost of donkey = 15 per tourist Rental price is 17.50 per tourist.

If, after servicing "Q" Tourists using the donkey and any additional tourists are served by renting, then we have to consider two costs: renting or servicing everyone usign the donkeys.

$P = \frac{17.5-15}{17.5} = \frac{2.5}{17.5} = 0.14286$ $P(x>Q) = 0.14286 -> P(x<Q) = 1 - 0.14286 = 8.85714$ $z = \frac{Q - 10000}{3000}$

Using $Z$ table, for $0.85714 -> z = +1.07$ So $Q = 3000(1.07) + 10000 = 3210 + 10000 = 13210$

Therefore, the number of donkeys should support 13210 tourists.