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Pop-Donuts located in Los Angeles of California makes two products, donuts and cakes. Pop-Donuts has bottlenecks in its capital used to purchase flour that is required in making both donuts and cakes, and also in labor hours used to make these two products. These two products are very popular in the local market and all products are sold quickly. The goal of Pop-Donuts is to make as much profits as possible within its capacities. Currently contribution margins for a dozen donuts and a cake are $\\\$ 3$ and $\\\$ 9$, respectively. Pop-Donuts will adjust sales prices when variable costs change to keep the contribution margins constant. Required direct labor hours are $15 \mathrm{~min}(0.25$ h) per dozen donuts and $1 \mathrm{~h}$ per cake. Pop-Donuts is expecting uncertainties in its budgets for direct materials as well as usage of flour per unit. The budget to purchase flour has a distribution with $f(x)=\frac{60000}{x^2}, \quad 15000 \leq x \leq 20000$, the cost of flour for a dozen donuts has a distribution with $f(x)=-750 x^2+900 x-262.5, \quad x \in(0.5,0.7)$ and the cost of flour for a cake has a fixed value of $\\\$ 1.3$.

Solution: We now build a continuous SLP problem and convert it to a discrete model by dividing these ranges into sub-intervals. Let us assume the following intervals in this discrete model. We use the first interval to demonstrate calculations and other intervals follow. Given the interval of $(15000,17500)$ and pdf $f(x)=\frac{60000}{x^2}, 15000 \leq x \leq 20000$, then the cumulative distribution function value $(\mathrm{CDF})=\int_{15000}^{17500} \frac{60000}{x^2} d x=0.5714$ and its mean value is $\int_{15000}^{17500} \frac{60000}{x} d x=16,186$ As these distributions are independent, hence this discrete model can have four combinations in this case. Note that given large computational complexities of the SAA method, we uses equal interval in sampling rather than equal probability in sampling.

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As indicated before, if more sub-intervals are selected and hence more combinations are used, the obtained optimal solution will get closer to the global optimal solution to the original continuous SLP problem.


Q: I couldn't understand how they construct the four scenarios from the solution, especially the two constraints. Like I guess the 1st constraint is cost of flour but what is the 2nd constraint? And what they mean by, we uses equal interval in sampling rather than equal probability in sampling.?

Q: In the 1st constraint $\bbox[4px, border: 2px solid red]{0.5625} x_1+1.3 x_2 \leq \bbox[4px, border: 2px solid red]{16186}$, is the coefficients come from $\int x\left(-750 x^2+900 x-262.5\right) d x$ and $\int_{15000}^{17500} \frac{60000}{x} d x$ ? because then for $(0.5,0.7)$ interval isn't it should be $\int_{0.5}^{0.7} x\left(-750 x^2+900 x-262.5\right) d x=0.6$ ? but it was given $0.5625$ Another question is, they mention probability as $0.2857$, how?

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2nd constraint is related to direct labor hours. It seems 15 min per dozen donuts, $x_1$ could imply a dozen donuts. But I am not sure how 0.25 is turning into 2.5 while per cake labor remains at 1. It should be
$0.25x1 + x2 \le L$ where L is Labor hours available.
As for Sample Average Approximate, its a continuous distribution over given interval like (15000,20000) or (0.5,0.7). So if you are to take a random variable of x from these intervals, you can choose a random value with a uniform probability of 1/N where N=2000-15000 or as for the range, 0.7-0.5. This will have have its complexities of large number of values possible. So the paper is suggesting to consider equal intervals with may be of size=100 and then take the average which you have correctly calculated as $\int\ xf(x)$ over the intervals.

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  • $\begingroup$ Thanks for your response @Sutanu. In the 1st constraint $\bbox[5px, border: 2px solid red]{0.5625}x_1+1.3x_2\leq 16186$, is the coefficient mean value come from $\int x\left(-750\:x^2+900\:x-262.5\right)dx$? because then for $(0.5,0.7)$ interval isn't it should be$\int _{0.5}^{0.7}\:x\left(-750\:x^2+900\:x-262.5\right)dx=0.6$? but it was given $0.5625$ Another question is, they mention probability as $0.2857$, how? $\endgroup$
    – falamiw
    Jan 8, 2023 at 4:18

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