# Maximum payoff with scenarios

Suppose I have Items $$H$$ and $$K$$, which cost me $$5$$ and $$2$$ dollars respectively. If sold, I will gain $$10$$ dollars for item $$H$$ and $$5$$ dollars for item $$K$$ Suppose also I have scenarios A and B and the likelihood of each happening is 50%. Next, in scenario A, people will buy all at most 500 items $$H$$ for $$10$$ dollars and $$0$$ item $$K$$, and in scenario B, people will buy at most $$100$$ item $$H$$ and $$1000$$ item $$K$$. I have $$2500$$ dollars, How can I combo my items of $$H$$ and $$K$$ so that I can maximize my gain (minimize loss) no matter which scenario happened?

I only know that we should follow the function $$5H+2K \leq 2500$$ since it sets the number of items that I can buy to sell. However, I don't have any idea how to relate this inequality to the gain/loss and also the two scenarios. I suppose I have to find one or more functions to represent these two quantities? Any help is appreicated.

• I have corrected my answer. I assumed that any items not sold were considered lost. Jan 21, 2022 at 22:13

Let $$r$$ denote the number of items of type H that you purchase.

Let $$s$$ denote the number of items of type K that you purchase.

First, establish a baseline proposed purchase of

$$r = 100, s = 1000,$$ which results in an expenditure of $$2500$$.

Under this baseline proposal, your profits are

• Under A : $$5 \times r = 500.$$
• Under B : $$(5 \times r) + (3 \times s) = 3500.$$

So, the disparity is $$3500 - 500 = 3000.$$

This $$3000$$ must be bridged, by increasing the value of $$r$$, and decreasing the value of $$s$$.

Suppose that you increase the value of $$r$$ by $$2k$$ and decrease the value of $$s$$ by $$5k$$, where $$k \in \Bbb{Z^+}$$ and $$k \leq 200$$.

Since $$(5 \times 2k) - (2 \times 5k) = 0$$, you will continue to spend exactly $$2500$$.

Further, you will gain $$10k$$ under A, and lose $$15k$$ under B. So, the net effect willl be to bring the results of A and B, $$25k$$ closer.

Since the amount to be bridged is $$25 \times 120 = 3000$$

you should set $$k = 120$$.

This implies that the final computation is

$$r = 100 + 2(120) = 340, ~s = 1000 - 5(120) = 400.$$
Then

• Under A, profit is: $$340 \times 5 = 1700.$$
• Under B, profit is: $$(100 \times 5) + (400 \times 3) = 1700.$$

Edit
Well, this is embarrassing. I went off the rails, overlooking that the actual profit is gross sales minus expenditure, where the expenditure is fixed at 2500.

That is, I should have assumed that unsold inventory is trashed.

So, the method is right, but the math was wrong.

For example, if $$r = 340, s = 400$$, then your gross sales under A are $$[3400]$$, and your gross sales under B are $$[(1000) + (2000) = 3000 \neq 3400].$$

The corrected math is:

Under the baseline, your gross sales are

• Under A: $$100 \times 10 = 1000.$$
• Under B: $$(100 \times 10) + (5 \times 1000) = 6000.$$

So, the disparity in gross sales to be bridged is $$(5000)$$.

When $$r$$ is increased by $$(2k)$$, and $$s$$ is decreased by $$(5k)$$

• Under A: gross sales increase by $$(20k)$$.
• Under B: gross sales decrease by $$(25k)$$.

So the net effect is to bring the gross sales of A $$(45k)$$ closer to the gross sales of B.

Since the amount to be bridged is $$5000 \approx 45 \times 111$$, the natural inclination is to set $$k = 111$$.

This implies that the approximate final computation, which will need fine tuning is:

$$r = 100 + 2(111) = 322, ~s = 1000 - 5(111) = 445.$$
Then

• Under A, gross sales are: $$322 \times 10 = 3220.$$
• Under B, gross sales are: $$(100 \times 10) + (445 \times 5) = 3225.$$

This can be improved by holding $$r$$ at $$322$$, and reducing $$s$$ to $$444$$. Under A, since your expenditure is decreased by $$2$$, you have increased your profit by $$2$$. Under B, since you have purchased and sold one less of item K, you have lost a profit of 3.

Under this scenario, $$r = 322, s = 444$$ (instead of $$445$$), the profit (gross sales minus expenditures) under A and B match.