How Can Calculus of Variations Optimize Stock Sales in a Bear Market?

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In summary: C(dN'/dt) = -b - 2CN''So the optimal N(t) and N'(t) for maximizing profit are N(t) = N0/2 + (b/2C)t^2 and N'(t) = (b/C)t + N0/2C. If B is too large, the price per share will decrease too quickly over time, making it difficult to sell the shares at a high price and therefore reducing potential profit.
  • #1
Liquidxlax
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This is for Classical mechanics 2

I'm not sure how to put partials in and the N with the dot beside it was supposed to be the derivative of N with respect to time

Suppose that you have N0 shares of stock and you want to make as money as you can by selling all of them in a single day. If N0 is large, and you sell your shares in small batches, the money you make can be written approximately as an integral

∫dt N'(t)P(N,N';t) t1 to t2

where t1 and t2 are the opening and closing times for the stock exchange, N(t) is a smooth function that is approximately the number of shares you have sold at time t (which satisfies N(t1) = 0 and N(t2) = N0, N' is the rate at which you sell the stock (in shares per hour, say) and P(N, N';t) is the price per share as a function of time. The interesting thing is that the price depends on how you sell the shares. For example, if you sell them too fast, the price will drop. That is why P depends on N and N'.

a) Suppose that P(N,N';t) = P0 - Bt -CN' for P0, B, C all positive (This is a "bear market" because of the -Bt term as the stock price is going down with time). Find N(t) and N' that allow you to make the most money.

b) Discuss briefly what happens if B is too large.


2. Homework Equations

dP/dN−d/dt(dP/dN')=0



3. The Attempt at a Solution

I've really got nothing for this because I'm unsure of how to start minus checking the relevant equation

dP/dN=0=d/dt(dP/dN).

so

dP/dN.=−C

and i think there has to be an auxillary equation to do with the fact that N(t1) = 0 and N(t2) = N0
 
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  • #2
Liquidxlax said:
This is for Classical mechanics 2

I'm not sure how to put partials in and the N with the dot beside it was supposed to be the derivative of N with respect to time

Suppose that you have N0 shares of stock and you want to make as money as you can by selling all of them in a single day. If N0 is large, and you sell your shares in small batches, the money you make can be written approximately as an integral

∫dt N'(t)P(N,N';t) t1 to t2

where t1 and t2 are the opening and closing times for the stock exchange, N(t) is a smooth function that is approximately the number of shares you have sold at time t (which satisfies N(t1) = 0 and N(t2) = N0, N' is the rate at which you sell the stock (in shares per hour, say) and P(N, N';t) is the price per share as a function of time. The interesting thing is that the price depends on how you sell the shares. For example, if you sell them too fast, the price will drop. That is why P depends on N and N'.

a) Suppose that P(N,N';t) = P0 - Bt -CN' for P0, B, C all positive (This is a "bear market" because of the -Bt term as the stock price is going down with time). Find N(t) and N' that allow you to make the most money.

b) Discuss briefly what happens if B is too large.


2. Homework Equations

dP/dN−d/dt(dP/dN')=0



3. The Attempt at a Solution

I've really got nothing for this because I'm unsure of how to start minus checking the relevant equation

dP/dN=0=d/dt(dP/dN).

so

dP/dN.=−C

and i think there has to be an auxillary equation to do with the fact that N(t1) = 0 and N(t2) = N0

The optimality (Euler) equation is dL/dN = (d/dt) (dL/dN'), where L is the integrand. For L = N'*(P0-bt-cN'), what do you get?

RGV
 
  • #3
Ray Vickson said:
The optimality (Euler) equation is dL/dN = (d/dt) (dL/dN'), where L is the integrand. For L = N'*(P0-bt-cN'), what do you get?

RGV

dL/dN = d/dt(dL/dN') = (d/dt)(P0 - Bt -2CN') = -b
 

Related to How Can Calculus of Variations Optimize Stock Sales in a Bear Market?

1. What is calculus by variations?

Calculus by variations is a branch of mathematics that deals with finding optimal values or solutions for mathematical functions by considering small variations or changes in the function. It is used to solve problems in physics, engineering, economics, and other fields.

2. How is calculus by variations different from traditional calculus?

In traditional calculus, we find the maximum or minimum values of a function by taking derivatives and setting them equal to zero. In calculus by variations, we consider small changes in the function and find the optimal value by minimizing or maximizing these changes.

3. What are some real-world applications of calculus by variations?

Calculus by variations is used in various fields such as physics, where it is used to find the path of least resistance between two points, and in economics, where it is used to optimize production and minimize costs. It is also used in engineering for structural analysis and optimization.

4. What are the essential concepts in calculus by variations?

The essential concepts in calculus by variations include functionals, variations, Euler-Lagrange equation, and the fundamental lemma. Functionals are functions of functions, and variations are small changes in the function. The Euler-Lagrange equation is used to find the optimal value, and the fundamental lemma is used to prove the equation.

5. What are some challenges in solving problems using calculus by variations?

One of the main challenges in solving problems using calculus by variations is finding the appropriate functional to optimize. This requires a deep understanding of the problem and the underlying mathematical concepts. Another challenge is solving the resulting equations, which can be complex and require advanced mathematical techniques.

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