Economics

The Time Value of Money and Inflation

To make a point, I’ll start this blog entry in an unusual way, that is, by talking about vectors. A vector is basically an ordered row of numbers. Consider this expression for example:

(12, 3, 5)

This vector could represent a lot of things. For example a point in a three dimensional coordinate system, with the vector components being the x-, y- and z-values respectively. Or for a company offering three products, it could stand for the sales of these products in a certain year.

Why this talk about vectors? You were probably very surprised when you heard grandma say that she paid only 150 $ for her first car. It seems so amazingly cheap. But it is not. Your dear grandma is talking about 1950’s money, while you are thinking of today’s money. These two have a very different value.

If you want to specify the costs of a good precisely, merely giving an amount of money will not be sufficient. The value of money changes over time and thus to be absolutely precise, you should always couple this amount with a certain year. For example, this is what grandma’s car really cost:

(150 $, 1950)

This is far from (150 $, 2012), which is what you were thinking of when grandma shared the story with you. Using an online inflation calculator, we can conclude that this is actually what the car would cost in today’s money:

(1410 $, 2012)

Not an expensive car, but certainly more than 150 $ in today’s money. Now you can see why I started this chapter using vectors. They allow us to easily and clearly couple an amount with a year. A true pedant would even ask for one more component since we are still missing the respective months. But let’s not get too pedantic.

How can we justify saying that 150 $ in 1950’s money is the same as 1410 $ in today’s money? We can look at how much of a certain good these amounts would buy in the given year. With 150 $ in 1950 you could fill your basket with about as many apples as you can with 1410 $ today. The same goes for most other common goods: oranges, potatoes, water, cinema tickets, and so on.

This is inflation, goods get more expensive each year. At a later point we will take a look at what reasons there are for inflation to occur. But before that, let’s define the rate of inflation and see how it is measured …

This was an excerpt from the ebook “Business Math Basics – Practical and Simple”, available for Kindle here: http://www.amazon.com/dp/B00FXB8QSO.

Physics (And The Formula That Got Me Hooked)

A long time ago, in my teen years, this was the formula that got me hooked on physics. Why? I can’t say for sure. I guess I was very surprised that you could calculate something like this so easily. So with some nostalgia, I present another great formula from the field of physics. It will be a continuation of and a last section on energy.

To heat something, you need a certain amount of energy E (in J). How much exactly? To compute this we require three inputs: the mass m (in kg) of the object we want to heat, the temperature difference T (in °C) between initial and final state and the so called specific heat c (in J per kg °C) of the material that is heated. The relationship is quite simple:

E = c · m · T

If you double any of the input quantities, the energy required for heating will double as well. A very helpful addition to problems involving heating is this formula:

E = P · t

with P (in watt = W = J/s) being the power of the device that delivers heat and t (in s) the duration of the heat delivery.

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The specific heat of water is c = 4200 J per kg °C. How much energy do you need to heat m = 1 kg of water from room temperature (20 °C) to its boiling point (100 °C)? Note that the temperature difference between initial and final state is T = 80 °C. So we have all the quantities we need.

E = 4200 · 1 · 80 = 336,000 J

Additional question: How long will it take a water heater with an output of 2000 W to accomplish this? Let’s set up an equation for this using the second formula:

336,000 = 2000 · t

t ≈ 168 s ≈ 3 minutes

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We put m = 1 kg of water (c = 4200 J per kg °C) in one container and m = 1 kg of sand (c = 290 J per kg °C) in another next to it. This will serve as an artificial beach. Using a heater we add 10,000 J of heat to each container. By what temperature will the water and the sand be raised?

Let’s turn to the water. From the given data and the great formula we can set up this equation:

10,000 = 4200 · 1 · T

T ≈ 2.4 °C

So the water temperature will be raised by 2.4 °C. What about the sand? It also receives 10,000 J.

10,000 = 290 · 1 · T

T ≈ 34.5 °C

So sand (or any ground in general) will heat up much stronger than water. In other words: the temperature of ground reacts quite strongly to changes in energy input while water is rather sluggish. This explains why the climate near oceans is milder than inland, that is, why the summers are less hot and the winters less cold. The water efficiently dampens the changes in temperature.

It also explains the land-sea-breeze phenomenon (seen in the image below). During the day, the sun’s energy will cause the ground to be hotter than the water. The air above the ground rises, leading to cooler air flowing from the ocean to the land. At night, due to the lack of the sun’s power, the situation reverses. The ground cools off quickly and now it’s the air above the water that rises.

Image
———————-

I hope this formula got you hooked as well. It’s simple, useful and can explain quite a lot of physics at the same time. It doesn’t get any better than this. Now it’s time to leave the concept of energy and turn to other topics.

This was an excerpt from my Kindle ebook: Great Formulas Explained – Physics, Mathematics, Economics. For another interesting physics quicky, check out: Intensity (or: How Much Power Will Burst Your Eardrums?).

From Simple to Compound Interest

Imagine you loan a bank the principal P = 10000 $ at an interest rate of i = 5 %. This is the amount of interest you would receive with simple interest, given the duration t of the loan:

t = 1 year
→ I = 10000 $ * 0.05 * 1 = 500 $

t = 2 years
→ I = 10000 $ * 0.05 * 2 = 1000 $

t = 3 years
→ I = 10000 $ * 0.05 * 3 = 1500 $

As you can see, the interest grows linearly with the duration of the loan. For each additional year, you get an additional 500 $, which is just 5 % of the principal 10000 $. In other words: each year the interest rate is applied to the principal. How could that be any different?

Consider this: At the end of the first year, you’ll receive an interest payment in the amount of 500 $. This means that your bank statement will now read 10000 $ + 500 $  = 10500 $. So why not apply the interest rate to this updated value? This would lead to an interest payment of 10500 $ * 0.05 = 525 $ for the second year instead of just 500 $.

Continuing this train of thought, at the end of the second year your bank statement would read 10000 $ + 500 $ + 525 $ = 11025 $. Again we would rather have the interest rate applied to this updated value instead of the unchanging principal. This would result in an interest payment of 11025 $ * 0.05 = 551.25 $ for the third year.

For comparison, here’s what the final pay out would be for the simple interest plan:

10000 $ + 500 $ + 500 $ + 500 $ = 11500 $

And this is what we would get with the “not simple” interest plan, where we apply the interest rate to the updated amounts instead of the principal:

10000 $ + 500 $ + 525 $ + 551.25 $ = 11576.25 $

The latter is called compound interest. It means that we include already paid interests in the calculation of next year’s interest, which leads to the amount received growing exponentially instead of linearly.

(This was an excerpt from “Business Math Basics – Practical and Simple”. You can get it here: http://www.amazon.com/Business-Math-Basics-Practical-Simple-ebook/dp/B00FXB8QSO/)