Spatial arbitrage refers to a trading strategy that involves exploiting price differences of the same asset or security in different geographic locations. It involves buying the asset or security in a lower-priced market and simultaneously selling it in a higher-priced market, with the aim of profiting from the price differential.
For example, if gold is trading at $1,800 per ounce in New York and $1,820 per ounce in London, a spatial arbitrageur could buy gold in New York and sell it in London for a profit of $20 per ounce. This type of trading is common in global financial markets where prices of assets can vary depending on factors such as exchange rates, taxes, and transportation costs.
Spatial arbitrage can be executed using various financial instruments, such as futures contracts, options, and exchange-traded funds (ETFs). However, it requires a high level of skill, knowledge, and access to information and technology to be successful.
The intersection of hyperbolas could be used to identify potential arbitrage opportunities by providing information about the relationship between prices of a given asset in two different locations.
For example, let's consider a scenario where there are two different markets for a commodity, Market A and Market B. The prices of the commodity in both markets are influenced by various factors such as supply, demand, transportation costs, and storage costs.
Assuming that the prices of the commodity in both markets are represented by hyperbolic functions, the intersection of these hyperbolas can provide information about the equilibrium price of the commodity where the prices in both markets are equal.
If the actual prices in the two markets deviate from this equilibrium price, there may be an opportunity for spatial arbitrage. Traders can buy the commodity in the market where the price is lower and sell it in the market where the price is higher, earning a profit from the price difference.
Overall, the intersection of hyperbolas can be used as a tool to identify potential arbitrage opportunities and make informed trading decisions.
The code is solving for the intersection point of two hyperbolas using algebraic methods. The hyperbolas are defined by the equations:
```
(a1 * x) / (b1 + x) = y and (a2 * x) / (b2 + x) = y
```
where a1, b1, a2, and b2 are constants, and x and y are variables.
To solve for the intersection point, the code first computes intermediate values for a, b, and c, which are coefficients of a quadratic equation of the form:
```
a * x^2 + b * x + c = 0
```
The quadratic formula is then used to find the roots of this equation, which correspond to the x-coordinates of the intersection points.
The code then checks if the roots are valid, i.e., if they lie between 0 and min(b1, b2). If a valid root is found, it is returned. Otherwise, the function returns 0.
Here is a brief explanation of the math involved in the code:
1. Compute intermediate values a, b, and c:
```
a = a1 * b1 - a2 * b2 b = 2 * b1 * b2 * (a1 + a2) c = b1 * b2 * (a1 * b2 - a2 * b1)
```
2. Compute the discriminant m = b^2 - 4ac.
3. Check if the discriminant is positive (i.e., real roots exist) and compute the roots:
```
if m > 0 { let m_sqrt = m.sqrt().unwrap(); let x1 = (-b + m_sqrt) / (2 * a); let x2 = (-b - m_sqrt) / (2 * a);
```
4. Check if the roots are valid:
```
if (x1 > 0 && x1 < b1 && x1 < b2) || (x2 > 0 && x2 < b1 && x2 < b2) { // wrong order, return nothing zero } else if x1 > 0 && x1 < b1 && x1 < b2 { x1 } else { x2 }
```
5. Return the valid root, or 0 if no valid root was found.
```rust
pub fn calculate_spatial_arbitrage(pair0: Pair, pair1: Pair) -> Decimal {
let a1 = pair0.a;
let b1 = pair0.b;
let a2 = pair1.a;
let b2 = pair1.b;
let zero = Decimal::zero();
let one = Decimal::one();
let two = one + one;
let four = two * two;
let a = a1 * b1 - a2 * b2;
let b = two * b1 * b2 * (a1 + a2);
let c = b1 * b2 * (a1 * b2 - a2 * b1);
let m = b.sqrt().unwrap() - four * a * c;
if m > zero {
let m_sqrt = m.sqrt().unwrap();
let x1 = (-b + m_sqrt) / (two * a);
let x2 = (-b - m_sqrt) / (two * a);
// 0 < x < b1 and 0 < x < b2
if (x1 > zero && x1 < b1 && x1 < b2) || (x2 > zero && x2 < b1 && x2 < b2) {
// wrong order, return nothing
zero
} else if x1 > zero && x1 < b1 && x1 < b2 {
x1
} else {
x2
}
} else {
// complex number, return nothing
zero
}
}
```