## Problem

[[images/f1.png]]

## Solution
Let the 2 known points be A and B.
To solve for a and b for point A:

[[images/f2.png]]

Substitution of `b` into original eqn for point B:

[[images/f3.png]]

To find `b`, substitute `a` back into the equation for `b`

### Checking
Using the following input as sample:
```
{stops=32, distance=39.081000 km, time=66.000000 min}
{stops=31, distance=36.573000 km, time=62.000000 min}
```
Answer should be:
```
a = speed = 0.66411...
b = stopTime = 0.2235337...
```

### Error Handling
If any of the 2 data points return "null", only one data point can be used. To handle this, the stopTime (b) is assumed to be a default value.

[[images/f4.png]]

## Coding
### Final Equations

[[images/f5.png]]

### Optimization
To optimize calculations, number of divisions should be decreased. This is because based on [Intel's arithmetic latencies](https://www.agner.org/optimize/instruction_tables.pdf#page=63). The following equations and corresponding pseudocode is proposed.

[[images/f6.png]]

In the pseudocode, `speed` refers to the mathematical constant `a` while `stopTime` refers to the constant `b`. `a` and `b` refer to the 2 data points.

```
sRatio = b.stops / a.stops
speed = (b.distance - sRatio*a.distance)/(b.time - sRatio*a.time)
stopTime = (a.time*speed - a.distance)/(a.stops * speed)
```