Category: Engineering Economy

Engineering economy is an important subject; how do we present alternatives? The posts include the definition of and solutions to problems.

  • 9a- Easy approach to types of assets-part-2.

    9a- Easy approach to types of assets-part-2.

    The Types of assets-part-2.

    The definitions are quoted from different sources.

    What are the accounts receivable AR?

    What are Accounts Receivable and Accounts Payable? These are business terms primarily used in accounting. The link is from Investopedia.

    What is account recievable?

    The real difference between Accounts Receivable and Accounts Payable is clear from the names. When a business buys a product and doesn’t pay for it upfront, the amount it owes is accounts payable.

    difference between Accounts Receivable vs Accounts Payable

    On the other hand, if a customer buys from you as a business but does not pay immediately, then that amount is your accounts receivable.

    What are the notes receivable?

    Notes receivable, which is the following definition, is quoted from the strategic CFO link. The balance sheet format comes in three sections: liabilities and owner’s equity. Notes receivable are an account on the balance sheet. The balance sheet is a financial statement that shows a company’s financial position at a point in time.

    Notes receivable definition.

    The balance sheet format comes in three sections: assets, liabilities, and owner’s equity.

    The next slide image explains the definition of Notes receivable and gives an example of a bank issuing Notes receivable for 90 days with a value of $100,000 to a Toy’s company with an interest rate of 5%. The interest paid is recorded as notes receivable.

    A note receivable is a written promise to receive a specific amount of cash from another party on one or more future dates. This is treated as an asset by the holder of the note and a liability by the borrower. Overdue accounts receivable are sometimes converted into notes receivable, thereby giving the debtor more time to pay, while also sometimes including a personal guarantee by the owner of the debtor entity. The guarantee provision makes the note receivable easier to collect than a standard account receivable. When a note receivable originates from an overdue receivable, the payment tends to be relatively short – typically less than one year.

    Notes receivable examples.

    Another example of Notes Receivable, quoted from Accounting Coach

    A company lends one of its important suppliers $10,000, and the supplier gives the company a written promissory note to repay the amount in six months, along with interest at 8% per year. The company will debit its current asset account Notes Receivable for the principal amount of $10,000. The credit of $10,000 will be to Cash.

    Suppose a company borrows $100,000 from its bank and signs a promissory note to pay 6% interest quarterly and the principal amount in 9 months. In that case, the bank will debit its current asset account Notes Receivable and will credit Cash or Customers’ Deposits for the principal amount of $100,000.

     What are marketable securities?

    Marketable securities are unrestricted financial instruments that can be readily sold on a stock or bond exchange. They are often classified into two groups: marketable equity securities and marketable debt securities. Marketable equity securities include shares of common stock and most preferred stock that are traded on a stock exchange and for which there are quoted market prices.


    What are the marketable securities?

    Marketable debt securities include government, and corporate bonds traded on a bond exchange and for which there are quoted market prices.

    Definition of materials.

    Material is a substance or mixture of substances that constitute an object. Materials can be pure or impure, living or non-living matter.

    Definition of material.

    Materials can be classified based on their physical and chemical properties, geological origin, or biological function. Materials science is the study of materials and their applications.

    Materials are assets until sold.

    Raw materials can be processed in different ways to influence their properties, such as purification, shaping, or the introduction of other materials. New materials can be produced from raw materials. Materials are assets until they are sold.

    Click on any image to get a slide show of all images for a clearer view.

    Engineering Economy. This link illustrates different types of economies and how to make economic decisions—the Time value of money, Applying Theory to Practice.

    this is another link from Wiki for the term account receivable.

    The next post contains the remaining types of Assets for Economy part- 3.

  • 9- Easy introduction to the different types of Assets.

    9- Easy introduction to the different types of Assets.

    Types of assets?

    This post provides a more detailed explanation of assets and discusses the Terminology of Assets for item number 1, as shown in the next slide image.

    Content of Engineering economy-5

    The general definition of types of assets.

    Based on Physical Existence: Tangible assets are types of assets that can be seen. For instance, in the construction field, we need land, cash for site expenses, machinery like excavators, and office supplies to construct buildings. Other examples include inventory, a building, rolling stock, manufacturing equipment or machinery, and office furniture. There are two types of tangible assets: inventory and fixed assets.

    Definition of an asset.

    Intangible assets are types of assets that can not be seen; for instance, Goodwill, a reputable company, can take loans from banks or gain many from patents or brands, trade secrets, etc. Please find the list of Assets in the next slide image.

    What are the differences between tangible assets and intangible assets?

    Again, assets are classified based on whether they are current or fixed. 1. Based on Convertibility.

    In the next slide image, there are two main categories of Assets. The first category is current assets that contain cash and notes receivable, which will be discussed in part 2, Marketable securities, materials, and supplies.

    What are the types of current assets and fixed assets?

    What are the Fixed assets?

    Fixed assets are fixed in that they are not readily convertible into cash. They require elaborate procedures and time for their sale and are converted into cash: land, buildings, plants, machinery, and equipment.

    Brief description of Fixed assets.

    Furniture is an example of a fixed asset. Other names for fixed assets are non-current assets, long-term assets, or hard assets. Generally, fixed asset value reduces over time (known as depreciation).

    What are the examples of the Fixed assets?

    What are the current assets?

    The definition is quoted from https://efinancemanagement.com/.

    Current assets are one type of asset on the asset side of the balance sheet, which majorly comprises cash and bank balances, inventories, and account receivables/debtors.

    The Key features of current assets are their short-lived existence, fast conversion into other assets, recurring and quick decisions, and interlinking with each other. Virtually, current asset management is almost as good as working capital management.

    Definition of current assets.

    The term current asset is formed with two words: current and asset. Current means circulating, and asset means valuables. Current assets are the assets or valuables of a business that keep circulating. The typical time frame for circulation is the financial period, which is usually one year.

    Cash or Bank Balances.

    Cash and bank balances are the balances that a company holds for its urgent needs. These balances fluctuate, and a particular balance lasts only a week or two. The balance will be high when the customer collects and again reduced when payment is made for purchasing raw materials.

    Inventories.

    Inventories form all kinds of inventories, whether raw material, work-in-progress stock, or finished goods. The time frame of their conversion is also generally between 2 to 60 days, and the rest depends on the industry in which the company operates. This is an excellent link to inventory. This is the title, What Is Inventory? Definition, Types, & Examples.Key Takeaways

    While there are many ways to count and value your inventory, the importance lies in accurately tracking, analyzing, and managing it. Insights gained from inventory evaluations are necessary for success as they help companies make more innovative and cost-efficient business decisions.

    Inventory, which describes goods ready for purchase, directly affects an organization’s financial health and prosperity.

    While there are many types of inventory, the four major ones are raw materials and components, work in progress, finished goods and maintenance, repair, and operating supplies.

    Types of current assets.

    Engineering Economy. This link illustrates different types of economies and how to make economic decisions—the time value of money, Applying Theory to Practice.

    There is another link to the definition of current assets: Current assets (definition), What are current assets? Examples of current assets, Current assets in accounting

    The next post, Easy approach to types of assets-part-2.

  • 8c-Arithmetic Gradient-part 2 in the Economy.

    8c-Arithmetic Gradient-part 2 in the Economy.

    Arithmetic Gradient-part 2 in the Economy.

    Arithmetic Gradient-part 2-Solved problem 4-8 how to find P, with given I%,n, G?

    In the Arithmetic Gradient-part 2, we will discuss a solved problem 4-8.
    A man has purchased a new automobile. He wishes to set aside enough money in a bank account to pay the maintenance on the car for the first 5 years.
    It has been estimated that the maintenance is as follows For year- 1 the maintenance cost = is $120, year- 2 the maintenance cost is $150, year- 3 the maintenance cost is $180, year- 4  the maintenance cost is $210, and for year- 5  the maintenance cost =$240.

    Assume the maintenance costs occur at the end of each year and that the bank pays 5% interest.

    Arithmetic Gradient-part 2-Solved example 4-8 How to find the P value with given arithmetic gradient G, i, n?

    How much should the car owner deposit in the bank now?

    Solution: In the Arithmetic Gradient-part 2. A-The man deposits P at year 0, to repay unequal values of maintenance at the end of each year. We draw the timeline diagram, at t=1, we have a value of $120  and at t=2, we have a value of $150 at t=3.

    We have a value of $180, at t=4, we have a value of $210, at t=5, we have a value of $240, The shape of a trapezium is created.

    B-We check whether the G has a constant value by subtracting the value at t=1 from t=2, which is 150-120=$30, which is the same value obtained by subtracting the value at t2 from t3, which is 180-150=$30.

    Solved example 4-8 How to find the P value with given arithmetic gradient G, i, n?

    Arithmetic Gradient-part 2- G value for the deposit.

    This is the same value obtained by subtracting the value at t3 from t4,  which is 210-180=$30, and the same value is obtained by subtracting the value at t3 from t4, which is 240-210=$30. This is called the G value.

    We will consider the trapezium shape as consisting of two shapes, the first shape is rectangular starts at t=1, and ends at t=5, where i=5% and n=5 years, and the second shape is the shape of a triangle that starts at t=1 and ends at t=5.

    The value at t=1 is 0, and the value at t=5=$120, the $120 represents the difference between 240-120=$120.
    For the first shape, the value of A is $120 as a constant value. Our G value is $30.

    Part b, of the Solved example 4-8 How to find the P value with given arithmetic gradient G, i, n?

    The present value P needs to be estimated, which is a sum of P1 due to uniform series with the known values of A, I%, and n, Plus P2  due to arithmetic gradient with the known values of G, I%, and n.

    Arithmetic Gradient-part 2-Estimate P/A and P/G values by using formulae.

    A/P=(I%)*(1+i%)^n/(1+i%)^n -1)(0.05)*(1+0.05)^5/(1.05)^5-1, we have i5=5%,n=5 years, substitute in the formula we can get the value of A/P which is equal to 0.2309.

    For the value of P/A, it is the reciprocal of A/P which will be =1/0.2309=4.329. For the P/G value, we have the relation, (P/G)=((1+i)^n-i*n-1))/(i^2*(1+i)^n)). If we want to estimate the values of P/G, by the relations, we find P/G we substitute in the shown equation as presented in the slide image.

    Solved problem 4.3-Arithmetic gradient-P/G.
    Solved problem 4.3-Arithmetic gradient-P/G.

    We can estimate the final P-value as the sum of P1 and P2 these values are shown in the slide image.

    P1 and p2 due to A&G values

    We get the P/A value and under the column of gradient present worth, find P given G, which is 8.237. P1=120*4.329. For the value of P2, we have P2=30*(8.237).

    To get the present value of the money to be deposited in the bank we will sum P1 and P2 as follows: P=P1+P2=519+247.11=$766.

    Arithmetic Gradient-part 2-Estimate P/A and P/G values by using table.

    We use the table of I=5%, for the compound amount factors, check the known data, which is the number of years n=5 and we need to estimate the  P/A for the first shape and  P/G for the second shape. We move horizontally to the left and under the column of present worth factor, find P with the given A.

    Part C, of the Solved example 4-8 How to find the P value with given arithmetic gradient G, i, n?

    P/G is shown by using the table to be =8.236. We will get the final value of p as a summation of the two values of P1 and P2. The final result matches the previous estimations done by substitutions.

    The pdf file used in the illustration of this post and the previous post.

    External resource in Engineering Economy. This is a great link that illustrates different types of economies, how to make economic decisions. the time value of money, Applying Theory to Practice.

    The next post will be about Types of assets.

  • 8b-A solved problem for EAW/ Equivalent Annual Worth.

    8b-A solved problem for EAW/ Equivalent Annual Worth.

    A Solved problem for EAW/ Equivalent Annual Worth.

    From our last post, we have considered a machine that costs $20,000. The salvage value is $4000 which Has revenue of $5000 yearly for the next five years, while the operation cost and maintenance are $500 annually for the next five years.

    We have converted the Initial investment and the salvage to annual Cost or CR. And then, we added the Cost of operation and maintenance.  The sum is $5120.76, and the Revenue value is $5000. We have concluded that this investment will lead to a loss.

    Review of solved problem

    What is the relation between EAB, EAC, and EAW?

    In the next slide, there are three expressions. We will introduce a definition for EAB-EAC-EAW. The first item EAB or sometimes written as EUAB stands for the equivalent uniform annual benefits.

    EUAB is the revenue you earn from operating a machine or the income you earn due to cash in or inflows, and this is represented by a positive sign(+).

    The second expression is EAC, sometimes written as EUAC, which represents the equivalent. Uniform Annual Cost. The uniform term expresses that these costs are equal in value and the term Annual means that the Cost is per year and equivalent.

    Definition of EAB-EAC-EAW.

    The second expression is EAC, sometimes written as EUAC, which represents the equivalent. Uniform Annual Cost. The uniform term expresses that these costs are equal in value and the term Annual means that the Cost is per year and equivalent.

    Definition of EAB-EAC-EAW.

    EAC we obtain from two sources; the first source is from the Capital recovery that we have expressed.
     The second source is the Cost of operation and maintenance sources. Both sources we represented by downward arrows with negative (-) signs.

    The Third expression EAW or EUAW or the equivalent uniform annual worth=EAB-EAC Or EUAB-EUAC will lead to a positive profit value.

    If the difference is positive, this will indicate that you have a good investment. If the difference is negative, this will indicate that EUAC is bigger than EUAB, this will conclude that costs are higher than benefits and the investment is not profitable.

    A solved problem for EAW or the Equivalent Annual Worth.

    We will have a solved problem for EAW or the Equivalent Annual Worth. An asset has an initial cost of $100,000 and an estimated salvage value of $40,000 after its 6-year service life.
    I draw a time scale from t=0 to t=6 years. Estimated O&M costs are $50,000 in year one, increasing by $6,000 per year after that.

    We have an item to consider, which is the uniform gradient. The start value is $50,000, G Gradient value of $6000 is added at the end of each year from year one to year 6.

    The initial value is $50,000, at the end of year two we have $56000, then $62000 at the end of year 3, followed by 68000,74000 and finally $80,000.

    The increase is the Gradient value G. G value is $6000. The assets are expected to generate annual benefits of $110,000. This is $110,000 represented as upward arrows in the time scale during the next six years from t=1 to t=6. Is this a desirable investment if MARR is 20%? how can we judge or decide on this investment? This is done by using the Annual Equivalent Cost.

    The solved problem for EAW or the Equivalent Annual Worth.

    It is important to convert the costs, in the solved problem, whether initial cost or final to annual cost. That is why we will use the conversion table relations. Using the table for the relation between A/F& A/P and G/A.

    The conversion equations in Engineering economy.

    How to estimate EAC?

    In the next slide, of the solved problem, the  EAC is composed of two items; the first one is capital recovery, and the second one is operation and maintenance. I will estimate one by one. The Capital recovery CR can be calculated from the initial cost, which is -100,000, and the salvage value S of 40,000.

    For the first estimate, we have- 100,000*(A/P, MARR=20%,n=6 years). The slide image shows both the cash-in and cash-out diagram with the years from t=0 to t=6.

    The cash-in and cash-out diagram.

    The second term is+40,000*(A/F,20%,6). If we do not have the table, we will use the conversion equations. A/P=(i)*(i+i)^n/(1+i)^n-1,  for i=20%, we can get the value of A/p as equals (20*1/100)*(1+0.20)^6/(1+0.20)^(6-1)=0.300071. For A/F, it is (A/P)*(P/F)= From the Previous relation of A/P and then multiplied by (1+i)^n. The final value of A/F=i/(1+i)^n-1. A/F=0.20/(1+0.20)^6-1=0.1007. I have completed the estimate for the capital recovery.

    The solved problem, how to get the values of A/P and A/F?

    In the next slide, for EAC for CR=-100,000(0.30071)+ 40,000(010071). EAC=-26,042. For the estimate of EAC for operation and maintenance. We have an annual gradient cost, the initial value=$50,000, and the G value=6000.

    The estimated value of A/G.

    I
    The annual gradient cost can be estimated as =-50,000, will be left unchanged drawn as downward arrows,+(- G(A/G,i=20%,6). A/G=(1/i)-((n/(1+i)^n)-1)). i=20%, then A/G=(1/0.20)-((6/(1.20)^6-1))=1.9788.

    In the next slide, The total value of EAC for operation and maintenance, EUAC, is the sum of -87.715. The last step is to calculate EUAB, which is easy to calculate since its data is given as $110,000 distributed each for the next six years. The last equation is EUAW or the equivalent annual worth =EUAB -EUAC=110,000-87,915=22,085. This investment causes profit since the difference is more than zero, and the investment is acceptable.

    How to estimate EAC for the solved problem?

    The same procedure is done for the solution.

    What is the value of EUAW?

    The pdf used for the illustration of this post can be downloaded from this link.

    The next post will give an introduction to the different types of Assets.
    For a useful external resource, Engineering Economy. Applying Theory to Practice.

  • 8A-What is Capital Recovery in the Economy?

    8A-What is Capital Recovery in the Economy?

    What is Capital Recovery in the Economy?

    18 The new Item is Capital recovery or CR; From Investopedia, Capital recovery or CR stands for Capital recovery is a term that has several related meanings in the world of business. It is, primarily, the earning back of the initial funds put into an investment. When an investment is first made in an asset or a company, the investor initially sees a negative return, until the initial investment is recouped.

    The return of that initial investment is known as capital recovery. Capital recovery must occur before a company can earn a profit on its investment.

    If you wish to review the pdf data used in the illustration, please continue reading.

    This is the Title of the Content that will be included in this post and in the next post.

    Title of the post-8a.

    These are the objectives of this lecture, but in this post, we will discuss Capital Recovery.

    Objectives of the post.

    Capital recovery is the equivalent annual cost of obtaining the Assets plus the Salvage, and we convert the Purchase price. The Capital is divided into two parts, the purchase price or the initial investment P and the salvage price. The salvage price is the selling price of the equipment at the end of its life.

    There are two values of great concern for a capital recovery: Purchase price & earning and Salvage. These values are converted to annual worth. As we remember, these annual worth are values that are equally spaced in time. I quote, Aw is comprised of two components. The capital recovery for the initial investment P at a stated interest rate (MARR) and the equivalent annual amount A.

    Capital recovery and annual worth.

    We have P as the initial investment in the next slide, the total first Cost of all assets and services. They are required to initiate the alternative. While the salvage value  (SV)-The terminal estimated value of Assets at the end of their useful lives.

    Term A is the Annual amount, the equivalent annual amount; typically, this is the annual operating Cost (AOC). We will discuss these terms in the coming solved problem, in God’s will.

    First, the initial investment then considers the operating costs in the form of an Annual amount and then compares alternatives. Compare the purchase price and maintenance cost and then evaluate the capital recovery.

    The maintenance is bigger or smaller than the profit that he can achieve from the given option at the end. There is a stated MARR that the company defines. These are the objectives of this lecture, but in this post, we will discuss Capital Recovery.

    Definitions of the different terms in Capital Recovery.

    The General equation to estimate Capital recovery.

     To convert into annual values.  We have two values; the General monetary transaction Associated with the purchase and eventual retirement of a capital asset is its initial Cost (I), and its salvage value (S) takes these sums into account.

    The capital recovery CR(I) at the stated Interest as MARR or as given in the example is equal to An investment that I(A/P, I, N), which is then converted P into annual value by the relation A/P. Where i is the stated interest rate and N is the number of years, which are the lifetime minus the salvage S, the price at the end, or the future worth, to be returned as A multiplied by (A/F) factor, I, N.

    The general equation used for CR, Capital Recovery.

    This is the other alternative equation for CR(I).In the next slide, as we can see, we have an investment I as cash out, and while there is a cash-in at the end S. This cash in and out can be converted into CR(I), the capital recovery with an equal amount and equal periods same as annual worth. The A is cash out considered a negative value.

    The sketch shows the relation between P&S. for Getting CR value.

    A Solved problem for capital recovery.

    We have a solved problem. Consider a machine that costs $20,000 and has five – year useful life. At the end of the five years, it can be sold for $4000— the annual operating and maintenance are $500. The firm can earn after tax $5,000 per year with this machine.

    In the cash flow diagram, we draw A annual worth as upward arrows as cash-in from time t=1 to t=5. Should it be purchased at an interest rate of 10%? All benefits and costs associated with the machine are accounted for in these figures. We need to estimate the capital recovery Using the equation CRI ( A/P, I, N).-S(A/F, I, N). The interest rate is given as 10%, N value=5 years.

    A Solved problem for Capital recovery.

    We can use the tables for i%=10% and N=5 years.

    We get the value of A/P; the value is 0.2638. We get the value of A/F=0.1638. The purchase price of 20,000 to be multiplied by 0.2638 will produce a downward A value. We need to estimate the capital recovery Using the equation CRI ( A/P, i, N).-S(A/F, I, N).

    The interest rate is given as 10%, N value=5 years. We get the value of A/P; the value is 0.2638. We get the value of A/F=0.1638.The purchase price of 20,000 to be multiplied by 0.2638 will produce a downward A value. We can use the tables for i%=10% and N=5 years.

    The ratio A/Pand A/F for i=10% and N=5 years.

    While the salvage value of $4000 is to be multiplied by 0.1638 and will be considered upward. We make a sum=-5276+655.2. We get =-4620.8 and include $500 as operation pointing downwards.So we get =-4620+(-500)=-5120.8.

    A3 represents the operating and maintenance annual cost.

    This is the alternative equation, which we can use to get The Value of CR(I), this equation will produce the final value of Cr exactly the same value as estimated by letting CR((I)=I(A/p,I, N)-S(A/F, I,N)

    Other Equation to find CR.

    While from revenue we have income =$5000. So finally, there is a loss, which is equal to (-5120.80+5000=-120.80. This value is the difference between the revenue and The Cost of Capital. Then the offer is to be rejected. This is the final solution done on one page using the formula for CR, Capital recovery.

    The solution is by using the general equation of CR.

    In the next post we will review the different terms EAB, EAC, and EAW, there is a solved example that demonstrates the photo use of EAC to check whether an investment is profitable or not.

    This is the link to download the pdf files.

    For a useful external resource, Engineering Economy. Applying Theory to Practice.

  • 8- Easy Illustration of the Arithmetic Gradient.

    8- Easy Illustration of the Arithmetic Gradient.

    Illustration of the Arithmetic Gradient.

    The discussion will be about the Arithmetic gradient.

    Our new subject for discussion will be the arithmetic gradient, but we will review the relationship between P and F, P is the present- value, and F is the future value. 

    Content of Engineering economy post.

    Review of the different types of investments.

    Deposit money to get the retirement amount at the end of your investment (P/F) relation.

    If you deposit money in a bank for the investment of amount=P, for which, you will receive a compound interest i% to get a future  F-value for your investment=F.

    We will find that we have different signs, P is represented by a downward arrow, or minus sign since it is a cash-out but, when you receive an F value it will be cash-in and the value is drawn by an upward arrow. The P-value in terms of F, i%, n, and F in terms of P, i%, and n are shown in the slide.

      For the P-value,  it is P=F*(P/F, i%, n). For F- value it is=F=P*(1+i)^n. P- value= F* (1+i) ^-n. The relation is based on annual compound interest i%, but, if we have changed the annual interest to become a quarterly compounded interest. The new interest rate i2=(i1/4%) and raised to the power of 4*(number of years), or 4n. That was the relation of F with a given P, I, and n.

    The relation between Future value F with present value P with known I, n.

    Deposit money to buy a car at the end of your investment (P/F) relation.

    There is another relation, for instance, if somebody wishes to buy a car, he or she takes a loan from a bank with interest i% and payment F will be at the end of time n, this is from the bank’s point of view.

    Cash in will be with a plus sign  + and the money received F is will be cash-out, with a negative sign (-).

    There is another type of relationship when someone receives money to buy goods and pays through installment or uniform series of compound interest after one year of receiving the cash-in for the agreed period. The P-value is cash in or with a positive sign.

    The cash out with equal amounts will be cash-out with negative signs, its value =A. The value for A in terms of P, i, n, A=iP(1+i)^n/((1+i)^n-1)), where i is the interest rate, and n is the time in years. The values of A are equal, the same as uniformly distributed load with A value as cash in installments in n the bank account, with an interest I and for some time.

    The relation between uniform series A with present value P with known I, n.

    Deposit money as a uniform series to get insured money at the end of an investment (A/F) relation.

    For instance, if somebody wants to receive insured money after some time, he will pay equal installments of value  A with an interest rate of i% and for n time, start after one year, the diagram shows that type, the A values are shown as upward cash in and at the end of time, for instance, time=8 years. The F- value received at the end of year 8, can be estimated based on a compound interest i%, the F- value= A(F/A, i%, n), also F=A/i((1+i)^n-1).

    How to find F with given A, i, n?

    The arithmetic gradient factors.

    We will discuss a new subject, which is the arithmetic gradient factors.
    While paying installments, these installments are not of equal value, but the difference between each consecutive payment is constant and equals  G, which is the same difference between the second and third installments also between the third and fourth installments as well as between the fourth and fifth installments.

    After one year of the agreement, somebody pays an installment with the value A. The cash-in diagram has a shape of a trapezium shape, for which the smaller value is A and the larger base=A+4G  for the sketch shown where n=5 years. But if we have n of years, the larger base value of the trapezium will be A+(n-1)G. But if we have n of years, the larger base value will be A+(n-1)*G.

    Introduction to Arithmetic gradient.
    Introduction to Arithmetic gradient.

    The previous shape consists of a trapezium, which can be solved by converting it into two shapes.

    The first shape is an equal installment diagram with a cash-in value of A for 5 years.
    The second part is the shape of a triangle with an increasing hypotenuse from 0 to 4G in the case of 5 years, as shown in the sketch, with a slope of 4G/4 =G.
    If we want to estimate the future- value in terms of A, i%, n. The relation is F=(A/i)*((1+i)^n-1)). If we want to estimate A in terms of P,  i, n, The relation is F=(i*P)*(1+i)^n/((1+i)^n-1)).

    While for the triangular shape, the value for F is in terms of i%, n, G. F=(G/I)*(1+i)^n-1)/I)-n. P/G Find P in terms of G. The( P/G, i,n) has a relation, which is shown in the slide.

    How to find F with given arithmetic gradient G, i, n?general expression.

    External resource in Engineering Economy. This link illustrates different types of economies and how to make economic decisions—the time value of money, Applying Theory to Practice.

    The two coming posts are “What is Capital Recovery in the Economy?” and “A Solved Problem for EAW/ Equivalent Annual Worth.”

    The third post: Arithmetic Gradient-part 2. How do you find P with the given I%,n, and G?