What is the Kovat's Index used for?
Why do we use the Kováts index? 


The Kováts index is used to normalize GC data. This Wikipedia page(though not very long), sums up why you would want to do so.

Retention times of the same compound on even two different versions of identical instrument with the same temperature and pressure program using the same column from the same manufacturer might not be the same! Retention time is dependent on so many things:


The identity of the analyte
The identity of the stationary phase
The length of the column (did you know that columns sometimes get trimmed as they age to remove deposits?)
The diameter of the column
The thickness of the stationary phase film
The flow rate of the carrier gas
The temperature or temperature program of the oven 


With so many variables, gas chromatography (indeed all chromatography) can seem like a forbidding black art in which one has no hope of being able to compare results from one instrument to the next or one lab to the next.

Fortunately, we're smarter than that. We can always compare our analyte's retention time to some retention times of well behaved standards. If we know the elution order (will not change) of a series of well-behaved standards on several instruments, and we know when our analyte elutes in relation to those standards on one instrument, we have a good starting point for predicting the retention time of that analyte on another instrument.

Ervin Kováts took this idea a step farther and created a quantitative index relating the retention time of an unknown analyte to the retention times of known standards in a way that produced system-independent constants describing the elution behavior of compounds. The system-independent part is important, given all of the factors listed above that can influence retention time. Every analyte should have (nearly) the same Kováts index value on any GC system under any set of parameters. Thus, if we determine the Kováts index of analyte $\ce{A}$ on one GC, we can know something about how $\ce{A}$ will behave on other instruments or with a different temperature program, etc.


  Why the linear alkanes?


The choice of the n-alkanes was probably not arbitrary, but I cannot say for certain. Of the two links from the Wikipedia article that might contain a rationale, one is dead and the other is in German behind a paywall. However, any well-understood series of standards would have worked. My guess would be that Kováts chose the linear alkanes for the following reasons:


They form a homologous series increasing one carbon at a time. 
A great many of them are (were - Kováts developed this index in the 1950s) available commercially in high purity (high enough purity that they can be easily purified further). At Sigma Aldrich, you can buy all of the n-alkanes at least up to eicosane $\ce{C20H42}$.
The first 20 linear alkanes cover a wide range of boiling points (and thus retention times): Methane $\text{BP}=-161.5\ ^\circ \text{C}$, pentane (the first liquid at room temperature) $\text{BP}=35.9\ ^\circ \text{C}$, decane $\text{BP}=173.8\ ^\circ \text{C}$, and eicosane $\text{BP}=343.1\ ^\circ \text{C}$.
There is a relatively reliable linear relationship between $\log(t_r)$ and the number of carbon atoms in the chain.



  How do you determine and use the Kováts index?


Lets say we have an analyte $\ce{A}$ that has a retention time of 193 seconds on some system. First, we identify the linear alkanes that have retention times on either side of the analyte. Lets say that heptane has a retention time of 170 seconds and octane has a retention time of 200 seconds in this system.

The formula for the index looks like this (for non-isothermal programs):

$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$


$I$ is the Kováts index
$n$ is the number of carbon atoms in the alkane with the lower retention time
$N$ is the number of carbon atoms in the alkane with the higher retention time
$a$ is the analyte
$t_r$ is retention time, so $t_{r,a}$ is the retention time of the analyte


For our example:

$$I=100\times \left[7+\left(8-7\right)\dfrac{193-170}{200-170} \right]$$
$$I=100\times \left[7+\dfrac{23}{30}\right]=777$$

Now, we want to analyze $\ce{A}$ on a different instrument, where the retention times of heptane and octane are 217 and 289 seconds, respectively. We can plug values back into the equation for the index to predict the retention time of $\ce{A}$ on this system:

$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$
$$777=100\times \left[7+\left(8-7\right)\dfrac{t_{r,a}-217}{289-217} \right]$$
$$7.77=\left[7+\dfrac{t_{r,a}-217}{72} \right]$$
$$0.77=\dfrac{t_{r,a}-217}{72}$$
$$55.2={t_{r,a}-217}$$
$$t_{r,a}=272$$