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# Application of Exponential Modeling in Forecasting

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Researchers have been interested in modeling life circumstances to allow data users to understand the underlying trend. There is the derivation of a straight line given the variables under investigation varies at a constant rate. For example, various economists believe that the demand schedule follows a straight line. However, there are situations in which the variables do not vary at a constant rate (Strickland, Lile, Rush & Stoops, 2016). That is, they may increase rapidly at first and later increase slowly. On the contrary, they might increase slowly at first before increasing rapidly. The phenomena call for the application of exponential modeling. The model uses the parameter as an exponent. Therefore, unlike the linear model, its equation can assume more than one degree (Strickland et al., 2016). The model does not only assist in modeling but also finding solutions to real-world circumstances.

Exponential Modeling and Compound Interest

There exist two forms of interests that banks charge. That is the compound and the simple interest. The two methods illustrate a constant percentage growth and constant growth rate respectively (Podmanik, 2016). Therefore, simple interest allows money to multiply in a straight line. On the contrary, compound interest represents exponential growth of the investment. The application of exponential modeling is possible using the following illustration.

An investor deposits \$1,000 in a bank for 3 years. The bank offers him a 10% compound interest. The interests will be;

1st Year = \$ 1,000 * 10/100 = \$100

2nd Year = (\$ 1,000 + \$100) * 10/100 = \$110

3rd Year = (\$ 1,000 + \$100 + \$110) * 10/100 = \$121

It is evident that we are multiplying the previous balance with the growth factor (Podmanik, 2016). The growth factor is 1.1 in the scenario. Hence, the compound interest for 3 years works as follows:

\$1,000 * 1.1 * 1.1 * 1.1 = \$1,000 * 1.13= \$1,000 * 1.331 = \$1,331

The amount at the end of three years using the first method was \$1,000 + \$100 + \$110 + + \$121 = \$1,331

Therefore, the second formula is of the form

Amount = Principal (1 + growth rate)3 = A = P (1 + r)n, which is an exponential function, where n represents the number of periods.

Therefore, a loan of \$34,800 lent at 12% compound interest after ten years would amount to:

A = P (1 + r) n =\$ 34, 800 (1 + 0.12)10 = \$108,083.52

Applications of Exponential Decay

There exist various scenarios, which display exponential decay. The first one is the radioactive decay. The model can assist in determining the initial number of atoms left (Github, n.d.). The model assumes that the original number of atoms is N0, the formula will be as follows: N0p = N0e-ʎt. The exponent t represents the time after the start of the decaying process starts. It can also assist in determining the half-life of the atom since it shows how the atoms reduce or survive during the timespan. Compaction of sediment offers another illustration of applying the exponential model. The reasoning is that the compaction increases rapidly at first and later slowly as one move deeper into the earth's crust. The formula applied is ʃ0 L1, 0 φ1

dz = ʃ ᶓ- L1ᶓ φ1 dz.

Applications of Exponential Growth

Some circumstances display an exponential growth pattern. For example, population growth is one such good illustration. Previous studies claim that population in some continents like Africa may be following that pattern (Passy, 2013). A hypothetical formula of the growth is (2generation)2. The formula tries to show that the population increases rapidly as the generations increases. However, during the initial stages of growth, the population was growing rather slowly. Concurrently, bacteria grow at a prolonged rate in a refrigerator (Passy, 2013). However, after the food stays in the open space, they grow exponentially.

Consumption of drugs is another illustration of exponential growth. The factors affecting the growth are price, intensity, and elasticity. The equation in logarithmic notation is as follows: Log10Q = log10 (Q0) + k (e-α∗Q0∗C - 1). The original quantity is Q0 while the new one is Q. Moreover, there is a C representing commodity price and lambda representing demand elasticity (Strickland et al., 2016). The model assists in determining the relationship between demand and prices of addictive drugs.

Exponential Modeling and Forecasting

The exponential modeling can assist policymakers to formulate specific forecasts. However, there have to be smoothing since the data always have some levels of uncertainty (OTEXTS, n.d). The uncertainty exists due to seasonal and trend factors. Thus, the smoothing assists in increasing the precision of the forecasts. It is an example of a weighted moving average framework. The researcher can use an observed value or previous experience in the smoothing process. The disparity between the observed values and predicted ones gives the residuals. Eventually, the sum of squared differences yields the sum of square errors. After that, the researcher derives the error term which smoothens the data (OTEXTS, n.d). The model can forecast oil production, which will be available in ten years. As a result, the smoothing process will help reduce the impacts of external elements such as government regulations. The researcher only requires the formula and after that use time to predict future observations.

References

Podmanik, M. (2016). Compound interest and exponential growth. In By the Numbers (pp. 166-187). The USA. Lumen Learning Center.

Strickland, J., Lile, J., Rush, C., & Stoops, W. (2016). Comparing exponential and exponentiated models of drug demand in cocaine users. PubMed Central Journal, 24(6). 447-455.

September 18, 2023
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