Engineering EconomyPractice Problems

for Exam 5

ByDouglas Rittmann, Ph.D., P.E

1. The cost of a machine for producing a certain part is $40,000. The machine is expected to have a maintenance cost of $14,000 and an $8,000 salvage value after its 5-year economic life. If the variable cost for producing the part is $1.50 per unit and the part can be sold for $4.00 per unit, how many parts per year must the company sell in order to breakeven at an interest rate of 12% per year?

Solution - Let x represent the number of parts per year required for breakeven. The annual worth equation is:

        0 = -40,000 (A/P, 12%, 5) - 14,000 + 8000 (A/F, 12%, 5) - 1.50x + 4.00x        0 = -40,000 (0.27741) - 14,000 +8000 (0.15741) + 2.50x  -2.50x = -23,837        x = 9,534 parts/yr

Thus, if the company expects to sell more than 9,534 parts per year, it should produce the part. At any sales level below 9,534 parts per year, the company would lose money and, therefore, should not invest in the machine.

2. A company can purchase a certain machine or rent one. If purchased, the machine will cost $15,000 and will have a 5-year life with a 10% salvage value. It’s operating cost will be $8000 per year. If the machine is rented, it will cost $400 per day. At an interest rate of 10% per year, the minimum number of days the machine must be needed to justify its purchase is:

Solution: -15,000 (A/P, 10%, 5) - 8,000 + 1,500 (A/F, 10%, 5) = -400x                   -15,000 ( 0.2638 ) - 8,000 + 1,500 (0.1638)  = -400x                                                                                         x = 29.2

Present Worth Computations with Inflation

Inflation is an increase in the amount of money necessary to obtain the same amount of goods. This occurs when the value of the currency goes down from one time period to the next. Different-valued currencies (in different time periods) must be accounted-for, because economic calculations span many time periods. That is to say, we must account for not only for money's time value, but also its actual value. Up to now, we assumed that the value (i.e. worth) of the currency was the same from one time period to the next.

There are two ways to make meaningful economic calculations when the value of the currency is changing: (1) Convert the amounts that occur in the different time periods into amounts that make the currencies have the same value before time value calculations are made, or (2) change the interest rate that is used in the economic equations in such a way that it accounts for the different valued currencies as well as the time value of money.

The first case is called making calculations in constant-value dollars (also known as today's dollars). Money in one period of time, t1, will have the same value as money in another period of time, t2, when the t2 amount is divided by the inflation that occurred between the two time periods. The general equation is

       [Eq 14.1]  Dollars in period  t1 =     dollars in period t2                                                                 Inflation between t1 and t2

Let dollars in period t1 be called today's dollars and dollars in period t2 be called future dollars or then-current dollars. If f represents the inflation rate per period and n is the number of periods between t1 and t2, Equation [14.1] becomes        Today's dollars =   then-current dollars                                             (1 + f)n

Once this conversion has been made, then the interest rate that must be used in the economic equations is the real interest rate. The real interest rate, i, is the rate that represents time-value-of-money-only considerations when money is moved from one time period to another. This is the rate that was used in all calculations up to this point. The market interest rate (or inflated interest rate) is a combination of the real interest rate, i, and the inflation rate, f. In equation form, the inflated interest rate, if, is    if = i + f + ( i * f ) ,  where i = real interest rate; f = inflation rate; if = inflated interest rate

Cost IndexesA cost index is a ratio of the cost of something today to its cost at some time in the past. As such, it is a tool which can be used to estimate the cost of things today based on their cost some time ago. This endeavor is especially important to engineers who are involved in design because cost is probably the single most important factor in the design of anything.

Many of the indexes that are frequently used by engineers are updated monthly and published in professional journals like Engineering News Record (ENR) and Chemical Engineering. Table 15-1 shows the values for three of the most common indexes for the years 1985 thru 1999.

News Record (ENR) and Chemical Engineering. Table 15-1 shows the values for three of the most common indexes for the years 1985 thru 1999.                                                                                                                                     

The general equation for updating costs through the use of a cost index is:                        Ct = Co (It)   /   Io                                             [1]               where Ct = estimated cost at present time t                        Co = cost at previous time to

                        It = Index value at time t                        Io = Index value at time to

The next example illustrates its use:Example 15.1 – An engineer involved in a major construction project discovered that a similar project had been completed in 1990.  If the previous project had a construction cost of $1.545 million, what would the estimated construction cost be in 1999 based on the ENR construction cost index?

Solution: From Table 15.1, the ENR construction cost index rose from a value of 4770.03 in 1990 to 6059.47 in 1999.  Therefore, the estimated construction cost in 1999 would be                        C1999 = 1.545 (6059.47)  / 4770.03                                = $1,963,000

A 20 HP centrifugal pump (with motor) can be purchased for $1500. For an exponent value of 0.46 in the cost-capacity equation, the cost of a 100 HP pump would be estimated to be

(A) $1,653      (B) $2,295      (C) $2,835     (D) $3,145

Solution: C100 = 1500 (100/20)0.46 = $3,145Answer is (D)

The ENR Construction Cost Index values for 1990 and 1997 are 4770.03 and 5851.80, respectively. If the index is changed so that 1990 is to have a base value of 100, the value in 1997 would be closest to

(A) 115.21      (B) 122.68       (C) 1250.03     (D) over 126

Solution: Divide 1997 value by the 1990 value and multiply by 100:1997 value = (5851.80) (100) / 4770.03

= 122.68Answer is (B)

Depreciation and DepletionOne definition of the word depreciation is to lessen in estimated value:  lower the worth of. In general, the value of an asset decreases with time because of age, wear, or obsolescence. A number of methods for systematically expressing the decreasing value of assets with time have been developed over the years. These so-called depreciation models result in values which (1) affect income taxes, and (2) provide information to investors about the worth of the assets of public companies.Depreciation affects income taxes because it is one of the deductions (from income) that businesses can take before calculating the amount of taxes they owe per the following equation:

Taxes = (income – deductions) (tax rate)

Since depreciation is a deduction (just as labor costs, rent, and other expenses are for businesses), the taxes owed are reduced by an amount equal to the depreciation times the tax rate (assuming income stays the same).  For example, a business that has a depreciation deduction of $5000 in a year when its tax rate is 40% would have its tax bill reduced by $2000 that year (i.e. $5000 * 0.40). The depreciation calculated for this purpose is called tax depreciation and must be determined using only IRS-approved models. Book depreciation refers to the depreciation procedures used by corporations to more accurately reflect to shareholders the value of their assets.  Two of the models used for depreciating assets are discussed below.The modified accelerated cost recovery system (MACRS) is the only currently-approved depreciation model allowed for tax depreciation in the United States.  According to this model, the depreciation charge for a given year is calculated by multiplying the asset’s depreciable amount (known as its basis, B, which is usually its first cost) by a depreciation rate. In equation form, MACRS depreciation is

Dt = dtB where: dt = depreciation rate, %

                       B = Asset’s basis, $

Thus, the depreciation charge in year 3 for a $10,000 asset which has a 5-year recovery would be $1,920 (i.e. $10,000 * 0.1920)

The book value of an asset refers to its undepreciated amount and is represented as the difference between the first cost, B, and the sum of the depreciation that has been charged up to that time. In equation form,

BVt = B - D           (B - Sums of D's )

For example, an asset that costs $10,000 and has a 5-year recovery period would have a book value at the end of year 3 equal to:

BV3 = 10,000 – 10,000 (0.20) – 10,000 (0.32) – 10,000 (0.1920)           = 10,000 – 10,000 (0.20 + 0.32 + 0.1920)                   = $2,880

While the MACRS model is the only one approved for tax depreciation, a model frequently used by corporations for book depreciation is the straight line model. Under this model, the depreciation charge is the same each year per the following equation:

D = (B – SV) / n where: B = Asset first cost

                               SV= Asset salvage value                                  n = Asset life

Thus, an asset which has a first cost of $10,000 with an expected salvage value of $2000 after a 5-year useful life would have a depreciation charge of $1,600 each year i.e.(10,000–2000)/5.

The book value of an asset depreciated by the straight line method would beBVt = B – tD where t = no. of years asset has been depreciated

The depreciation models discussed up to this point apply to assets that can be replaced. For assets that cannot be replaced, like natural resources, different procedures are used for tax accounting purposes. There are two methods which can be used to account for this so-called depletion: cost depletion and percentage depletion.Cost depletion involves the multiplication of a cost factor, pt, by the amount of resource removed in a given year. The factor, Pt, is equal to the first cost of the resource divided by the total amount of recoverable resource:           Pt = (first cost)  / resource capacity

The annual depletion deduction is Pt times the amount of resource harvested in that year. For example, if a gold mine which cost $1,000,000 had an estimated 4000 ounces of gold, the depletion factor would be $250 per ounce (i.e. 1,000,000/4000). The depletion charge in a year when 1,000 ounces is removed would be $250,000 (i.e. 250 * 1000).Percentage depletion is a procedure wherein a certain percentage of the income from harvesting the resource is taken as the deduction. The percentage that is taken is a function of the type of resource involved as shown in the table below:

Thus, if 10,000 ounces of gold are harvested in a year when gold is selling for $270 per ounce, the depletion allowance deduction would be (0.15) (10,000) (270) = $405,000 (subject to certain tax law restrictions).

Questions?