Save Money Like an Engineer

A pile of cold, hard, cash.

I recently wrote a post about how the difference between engineers and scientists is money.  An engineer's job it to make their employer money, either through designing a more efficient system or designing a more productive system.  They use science regularly, as it is the job of the engineer to understand the mechanical, electrical, chemical, and/or nuclear aspects of a system and how these aspects can be affected by a change to that system.  Scientists' primary focus, meanwhile, is the science itself.  They are tasked with gaining knowledge about a particular subject, and developing, refining, or debunking previously held theories.  Scientists are always worried about money because their research depends on it, but making a profit off the research is rarely their goal.  Both disciplines rely on each other to perform their work, and neither could exist without the other.

Saving Money by Evaluating Cost

How an engineer evaluates cost, however, was not discussed.  It is one thing to say an engineer makes a system more efficient, but what method does the engineer use to determine the best way to suck more money out of a manufacturing plant, power plant, or consumer product?  There are always many paths to making a plant operate more efficiently, but as I said above, the engineer has to account for all factors, be it mechanical, electrical, chemical, or nuclear.  There could be three options, each with their own pluses and minuses, but you may not have the option to implement all three.  Option A may not allow Option B from being installed, or Option C may change the operation of the plant in a way that makes Option A less economical.  In most complex systems, a small change can have wide-ranging implications.  And we haven't even mentioned the final factor that can always throw a wrench into any good plan, finances.  The engineer has to be able to evaluate all of these options, see the cascade of effects that will occur from each one, and determine the best route for the company.  How does the engineer do that?

As it turns out, engineers use the same financial tools that accountants, banks, and investors use to evaluate cost.  There are tried and true methods and equations that allow anyone, if they understand all the factors, to perform a cost analysis on anything.  These methods are not only valuable to engineers, financiers, and bankers. They are also valuable to the average person in the home.  I am going to walk through a couple fairly simple examples to help people understand one part of the engineers job, but these methods can be used anywhere, as long as the three basic tenants are followed.

Include All Cost Factors

One step I have noticed is regularly missed when I read an analysis of just about anything is including all the cost factors.  People often latch onto one aspect of cost and treat that as the only cost.  This can not only be misleading, but is often wildly inaccurate.  Most often people will look at the up-front cast of a product, but ignore the downstream costs, even if they add up to be far more than a product that costs bit more upfront but has lower or no downstream costs.  This mentality is what powers the freemium software market, charging nothing to use it, but instead charging small amounts for add-ons.  Even if a person ends up paying double for the same service, they do not weigh the cost the same because the cost is spread over multiple small transactions.  This is not only poor judgement, but a huge waste of money.

Initial, Recurring, and Final Costs

Whenever performing a cost analysis, the engineer must look at all costs.  These generally include up-front costs (construction), recurring cost (maintenance, fuel, etc.), and end-of-life costs (deconstruction).  Generally all of these come into play in one form or another for every cost analysis.  Nothing is truly maintenance free, nor does anything last forever.  If you develop a system that is missing one of these three cost factors, either go back and check your analysis or apply for a patent because you have just struck gold.

Interest Rate Adjustments

One last cost factor to be sure to include is interest rate.  This isn't engineering specific, but it is vitally important to any cost analysis.  If the company if borrowing money to pay for an upgrade, that interest rate has to be included in the recurring maintenance costs, otherwise your cost analysis will be skewed.  The inflation rate would also factor into this analysis, though as I discuss below, this often gets cancelled out because the comparison needs to be made in constant dollars in order to weigh all costs appropriately.

Return on Investment

The last cost you always have to consider is actually not a cost at all, but the reward.  What does each option provide in return?  If all options provide the same return on investment, it can be cancelled out, but if they are different at all, like a different percent savings per year or a different delay on return, then the return on investment must be added to the calculation as well.

Compare Apples to Apples

Remember back in elementary school when you learned about fractions and common denominators?  This is exactly what I am talking about.  There is no way to properly compare two fractions without finding a common denominator so that both values are on the same footing.  Likewise in the real world, systems must be placed on equal footing.

Lifespan

It is one thing to say Option A costs less up-front and costs less per year in maintenance, but it Option A only lasts one third as long as Option B, it may not be the right choice.  Both scenarios need to be placed on equal footing.  In that example, three Option As would need to be used back to back to back in order to be appropriately compared to Option B.  Then you have three times the up-front cost divided over three separate years spread evenly across the entire lifespan of both Options.  Is Option A still the best option now?  That would depend on the numbers.  

Inflation

The other factor that needs to be considered when trying to normalize two or more options is inflation.  A dollar today is not worth the same as a dollar tomorrow, or a dollar ten, twenty, or fifty years from now.  A recurring cost of $10,000 per year would actually increase slightly (around 2% for 2013).  $10,000 in year one would be $10,200 in year two, $10,404 in year three, and so on.

In the above scenario, inflation actually gets cancelled out, because to set the recurring costs in terms of constant dollars, you would have to divide by the inflation rate to get into 2013 (or whatever year you like) constant dollars.  The $10,000 per year, adjusted for inflation and readjusted back to today's dollars would be $10,000.  In that example, and in most simplified examples, the recurring costs don't need to be inflation adjusted, so $10,000 per year is just $10,000 per year.

In the real world, inflation adjustment does have to be considered for certain scenarios.  I will keep my below examples simple for clarity's sake, but unfortunately the real world is not always simple.  If a company were to outsource for maintenance, the contract would probably set a fixed yearly price for a certain number of years.  That fixed price would need to be adjusted for inflation, because $10,000 five years from now has less value than $10,000 now.  Basically, the cost for maintenance would go down every year as that $10,000 per year paid out holds less and less value.

Risk Can Affect Your Decision

Risk is an easily overlooked aspect of a cost analysis, especially when it can be hard to quantify, but it can also be the most determining factor in your decision.  The reason companies many times choose projects that are more costly in the long-term, but cheaper in the short-term is risk.  In my field, nuclear power, the primary reason why no nuclear power plants have been built in the last twenty years is risk.  Not risk of meltdown, but risk of losing their investment.  Every project runs the risk of failure for a number of reasons from delays and cost overruns to political action.  If this happens, the company loses out on at least a portion of their investment, with no hope of recovering it.

A project with small up-front costs will cause less damage to the company's bottom line if cancelled, but a project with a huge initial investment and cheaper recurring costs can cause some real financial damage.  In the case of nuclear power, a mothballed reactor can put a company out billions of dollars.  Even though the benefits of extremely cheap recurring costs might make the nuclear project better for the company over the next forty to sixty years, the fear of bankrupting a company over one single project is enough to scare away the most confident investor.

For the sake of simplicity I am not going to even touch risk analysis in this post, but the idea should at least be in the back of your head.

Engineering Cost Analysis Examples

The next few examples are going to be fairly simple versions of cost analyses, though I will try to make them increasingly complex to try and introduce all of the basic concepts.  As you will quickly figure out, not all of these examples are engineering related, and that is because this method is a very standard method of performing a cost analysis.  This method can be used to perform a cost analysis on just about anything, and I highly recommend you use it whenever you can.

Because a good infographic can tell you so much more than just numbers can, I would always suggest you draw out a cash flow diagram whenever you perform a cost analysis.  For each example, I have provided at least one diagram to represent the problem graphically.  This should provide some perspective to the problem and allow you to visually assess the answer.

And as I mentioned in an earlier post, I will provide a warning to those who dislike or are not gifted when it comes to math, even though most of what is below is graphical in nature.

WARNING: MATH AHEAD

Example 1 - The Simple Interest Problem

This is pretty much the simplest example I can come up with, and this should familiarize you with the cash flow diagram if you aren't already.  I put together a few examples of varying investment scenarios, but to start we will look at the first one.  Let's say you want to invest $10,000 into the stock market, and your stock broker guarantees you an interest rate of 8% for ten years.  How much money would you have in your portfolio at the end of that ten year period?

This is a pretty simple problem, one you could probably do without a diagram, but none-the-less, here is what that would look like:

Example #1: Stock Market Investment

So in 2013 you put in $10,000 into the stock market (negative because you have spent that money).  Over the next ten years that money is accruing interest, meanwhile you do nothing to it.  You don't add any money in, and for simplicity's sake, you don't collect any dividends.  After ten years, you sell your shares and collect your money.  Sans broker fees, taxes, etc, the amount you would have, assuming 8% interest every year, is:

where F is your final value, P is your present value, i is your interest rate, and n is your time period.  This is known as the F/P equation (solve for F, knowing P).  For those who use Microsoft Excel, this equation is built-in as FV.  For this example, the equation looks like:

So after ten years, your $10,000 investment is now worth $21,589.25.  Not too bad.  Now, let's move onto the second scenario above.  This time, maybe you can't afford $10,000 up-front, but you can afford $1,000 per year.  You still are investing the same $10,000, but now it is spread over all ten years.  How would that change the equation, and more importantly to the investor, how would that change the value of your investment?  The scenario is slightly different, and the equation then becomes:

Where A is the recurring investment.  This is known as the F/A equation (solving for F, knowing A).  Using this equation, or using the same FV Excel function, the answer becomes:

As you can see, investing $1,000 per year for ten years is not nearly as lucrative as investing $10,000 all at once, though this is intuitively obvious since the $1,000 per year has less time to accrue interest.

Example 2 - Some Simple Engineering Scenarios

Now that the basic framework has been established, I will briefly discuss a few examples of cost analyses an engineer might do.  One piece of information that the engineer needs to perform a proper analysis is to know what the company uses as a minimum acceptable rate of return (MARR).  This is the rate each investment must meet in order to be considered as a viable option.  This rate will depend on the company, but it is set somewhere higher than the rate of return (RoR) if the money were to sit in a bank or other safe investment.

For this example, we will assume a MARR of 5%.  To perform a proper apples-to-apples comparison, each scenario will be put in terms of present value.  Whichever option costs less will be the best option.  For this example, I assumed each  option had a RoR equal to the MARR.

Example #2: Simple Engineering Scenario 1

As you can see from this scenario, Option B is the cheapest, therefore should be selected over Option A and C.  To calculate the present value of each option, the following equation was used:

To determine present value, you need to include the initial investment (Ii), and add that to the P/A equation (similar to the F/A equation above, but solving for P, knowing A).  Depending on what you use at the MARR, though, the results can be very different.  Let's assume for a moment that the MARR is now 15% instead of 5%.  Here is what the three options from above look like now:

Example #2: Simple Engineering Scenario 2

Assuming a MARR of 15%, Option A is not the cheapest by a couple thousand dollars.  This is why these types of problems are never just about the engineering, but also about the status of the company.  How much interest the company can make on other investments and how much risk the company is willing to take affect the MARR and therefore the best choice for investment.

Example 3 - Less Simple Scenarios

There are many factors that go into making a cost analysis decision as an engineer or as anyone looking to take a more logical approach to their personal choices.  For the above examples, I made many assumptions and simplifications, including the same rate of return for each option, the same time period for each option, constant recurring fees, and no salvage value.  It is impossible to cover every scenario here and not become completely overwhelming, so I will just cover a few different scenarios to show you how these factors can affect the decision.

To start, let's look at the previous example, but we will make a couple changes.  First, to try and make the scenarios a bit more realistic, I have changed the constant recurring costs to a gradient.  Like maintenance on anything, such as a car, the costs tend to be less near the beginning of life, and get higher as the item ages.  In addition, there is generally something salvageable left after the product's lifespan, which helps to recoup the cost.  For the next scenario, I assume the salvage value is 25% of the original cost.  Implementing those two changes, the cost comparison changes to this:

Example #3: Less Simple Scenario 1

Like the first scenario in the first example, Option B wins out, this time by a much larger margin.  This calculation is slightly more complicated, though it is mostly adding together equations already discussed above, with the only difference being the equation for calculating the present value for a gradient, which is

where G is the value of the gradient, assuming the year one cost is zero and the gradient is constant until year n.  The entire equation for calculating PV for this scenario is

where FV is the final value, in this case, the salvage value.  I realize it is starting to look complicated at this point, but as you can probably see, each portion of the equation is normalizing for the interest rate to the present time, so that you can do an apples to apples comparison.  Also, if it is not apparent, the value for year ten does not follow the gradient because it is -$4,500 for maintenance plus $2,500 for the salvage value.

The other factor that often makes a cost analysis more complicated is lifespan.  For the above examples, the lifespan for each option was assumed to be the same, however this is rarely the case in real life.  In order to do a proper comparison, the lifespans need to be the same, like finding the lowest common denominator in fractions, or in this case the lowest common multiple (LCM).  With varying lifespans, this could mean multiplying an option three or four times, but I will still keep it relatively simple.

To keep a similar scenario, we will use the options from above, but change the lifespan of each to five years, ten years, and twenty years, respectively.  This means for Option A, we have to buy the machine, run it for its five year lifespan, sell what's left, and buy another machine until our total lifespan is equal to Option C's lifespan.  This also goes for Option B, though we only need two lifespans to match Option C's lifespan in that case.  The results are as follows:

Example #3: Less Simple Scenario 2

So these look downright complicated, but if you understand your goal and break the problem into its constituent parts, it becomes much easier.  Keep in mind that our objective it to compare apples to apples, so we are trying to normalize all costs to the same year, in this case the present value.

The first part is easy, it is just the cost of purchasing the machine at year zero.  Since the purchase occurs in the same year as we are trying to normalize to, no work needs to be done.

The second part is the cost to purchase the machine again to replace the worn out one.  In the case of Option A, we do this three times, making it four total machines purchased over the timeframe we are looking at, while Option B only requires one more machine purchased (two total), and Option C only requires the first machine purchased.  Each machine costs the same, but the costs occur at a date in the future, so must be normalized to today's costs.  Using the F/P equation way up at the top and adjusting it to solve for P knowing F (P/F equation), we get

So for Option A, the second machine purchased costs $10,000 (F) at year five (n), and we use the same MARR (i) and get

This step is then repeated for each time the machine is purchased, with only the n changing to reflect the number of years between now and when the purchase actually happens.

Similarly, the cost for salvaging the old machinery is treated the same way as the cost to purchase a new machine.  The assumption was 25% return when the old parts were sold off, so for Option A that is $2,500 in years five, ten, fifteen, and twenty.  Those values then have to be normalized to present value, which would use the same F/P equation above.

The last portion of this problem is the gradient costs.  In the first scenario in this example, we just used the gradient equation, which normalizes the total costs of a gradient to year zero.  This works again for the first gradient, as year zero for the entire problem corresponds to year zero for the gradient.  For the other gradient costs, however, the gradient equation would only normalize the costs to a point in the future since they start years after the present.  That equation does give you a single value, however, which can be treated as a future value that occurs at the start of the gradient.  This value can then be normalized to present value using the F/P equation.

Once all these separate parts of the problem are calculated, all the present values can be added together to get the entire present value of the option.  As you can see from the above diagram, Option C is by far the better choice since the $25,000 cost of purchasing the machine only happens once.  I have provided two diagrams of the expenditures and incomes which for Option A can be broken down into twelve separate parts, all normalized for present value:

1. Initial machine cost at year 0 (-$10,000)
2. Maintenance costs for the first machine over years 1-5 (-$4,118.46)
3. Salvage cost for machine one at year 5 (+$1,958.82)
4. Second machine cost at year 5 (-$7,835.26)
5. Maintenance costs for the second machine over years 6-10 (-$3,226.92)
6. Salvage cost for machine two at year 10 (+$1,534.78)
7. Third machine cost at year 10 (-$6,139.13)
8. Maintenance costs for the third machine over years 11-15 (-$2,528.38)
9. Salvage cost for machine three at year 15 (+$1,202.54)
10. Fourth machine cost at year 15 (-$4,810.17)
11. Maintenance costs for the fourth machine over years 16-20 (-$1,981.05)
12. Salvage cost for fourth machine at year 20 (+$942.22)

Example #3: Scenario 2 - Expenditures

Example #3: Scenario 2 - Incomes

Cash Flow Diagrams and Cost Analysis Uses

This was an attempt to provide a little insight into one aspect of the job of an engineer.  For those of you who are not engineers or have no desire to become one, what it the value of all this?  The answer is a ton.  You may not ever be in the situation of performing a cost analysis at work, but you already do this in an informal way in your daily life.  Formalizing the process will only make it less affected by your emotions and more by the actual costs.

Think about buying a car.  You have a prime example of a situation with multiple alternatives, initial costs (buying new car), recurring, fluctuating costs (maintenance, fuel), and salvage costs (selling old car).  Based on a car's initial cost, fuel economy, expected lifespan, salvage value, and the interest rate on a car loan, normalized for present value, you could determine the best value for you and make the best economical choice.  It really is an excellent tool any time you have multiple options regarding a personal purchase.

There are really limitless opportunities to use this method to make a rational decision on a purchase that would otherwise be left up to emotion.  I highly recommend going this route on as many purchase decisions as is possible.  Now in reality, I doubt anyone is going to perform this work to determine which coffee they should buy every day, but for big purchases like a car, a new phone, or a home, this can save you a huge headache and a feeling of buyers remorse later.

Engineering Economy
6th Edition

References

For this post I referenced Engineering Economy, 6th Edition by Leland Blank and Anthony Tarquin.  This is a good, straightforward look at economics from the engineer's perspective, and also covers more advanced topics not mentioned in this post.  If you are looking to learn more about what was discussed here, I recommend you pick up a copy.