Applications of Money- Time Relationships MARR by PW Method r r All cash inflows and outflows are...

Applications of Applications of Money-Time Money-Time

RelationshipsRelationships

MARR by PW MethodMARR by PW Method

All cash inflows and outflows are discounted to the present point in time at an interest rate that is generally the MARR.

PW(i) = 0; i = MARR If PW(i) > 0 => The investment is “good”. If the return rate of an investment is greater

than MARR then the investment is “profitable”.

MARRMARR

MARR(hurdle rate) - maximize the economic MARR(hurdle rate) - maximize the economic well-being of a company.well-being of a company.

Example 1Example 1 An investment of An investment of $10,000$10,000 can be made in a can be made in a

project that will produce a uniform annual project that will produce a uniform annual revenue of revenue of $5,310$5,310 for five years and then have a for five years and then have a market (salvage) value of market (salvage) value of $2,000$2,000. Annual . Annual expenses will be expenses will be $3,000$3,000 each year. The company each year. The company is willing to accept any project that will earn is willing to accept any project that will earn 10%10% per year or more, before income taxes, on per year or more, before income taxes, on all invested capital. Show whether this is a all invested capital. Show whether this is a desirable investment by using the PW method. desirable investment by using the PW method.

PW(10%) = -10,000 + 5,310(P/A, 0.1, 5) +PW(10%) = -10,000 + 5,310(P/A, 0.1, 5) +2,000 (P/F, 0.1, 5) - 3,000(P/A, 0.1, 5) = 02,000 (P/F, 0.1, 5) - 3,000(P/A, 0.1, 5) = 0

=> This investment is marginally acceptable.=> This investment is marginally acceptable.

Example 2Example 2 A piece of new equipment has been proposed by engineers to

increase the productivity of a certain manual welding operation. The investment cost is $25,000. and the equipment will have a market value of $5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm's MARR is 20% per year, is this proposal a sound one?

PW(20%) = $8,000(P/A,20%,5) +$5,000(P/F,20%,5)

- $25,000 = $934.29

Because PW(20%) > 0 => “Investment justifiable”

Bond Value DeterminationBond Value Determination Z = face, or par, value Z = face, or par, value C = redemption or disposal price (usually equal C = redemption or disposal price (usually equal

to Z) to Z) r = r = bond rate (nominal interest) per interest bond rate (nominal interest) per interest

period period N = N = number of periods before redemption number of periods before redemption i = bond i = bond yield yield rate per period rate per period VVN N = = value (price) of the bond value (price) of the bond N N interest periods interest periods

prior to redemption prior to redemption VVNN = C(P/F, i %, N) + rZ(P/A, i %, N) = C(P/F, i %, N) + rZ(P/A, i %, N)

Example 1Example 1 Find the current price (PW) of a 10-year bond

paying 6% per year (payable semiannually) that is redeemable at par value, if bought by a purchaser to yield 10% per year. The face value of the bond is $1,000.

Sol: i6mon = (1.1)1/2 - 1 = 0.049

VVNN = 1,000 (P/F, 4.9%, 20) + = 1,000 (P/F, 4.9%, 20) +

1,000 (0.03) (P/A, 4.9%, 20) 1,000 (0.03) (P/A, 4.9%, 20)

= 384.10 + 377.06 = 761.16

Example 2Example 2A bond with a face value of $5,000 pays interest of A bond with a face value of $5,000 pays interest of 8% per year. This bond will be redeemed at par 8% per year. This bond will be redeemed at par value at the end of its 20-year life, and the first value at the end of its 20-year life, and the first interest payment is due one year from now.interest payment is due one year from now.

(a) How much should be paid now for this bond in (a) How much should be paid now for this bond in order to receive a yield of 10% per year on the order to receive a yield of 10% per year on the investment?investment?

(b) If this bond is purchased now for $4,600, what (b) If this bond is purchased now for $4,600, what annual yield would the buyer receive?annual yield would the buyer receive?

(a) (a) VVNN= 5,000= 5,000(P/F,10%,20) + 5,000(0.08)(P/A,10%,20) (P/F,10%,20) + 5,000(0.08)(P/A,10%,20)

= 743 + 3,405.44 = $4,148.44= 743 + 3,405.44 = $4,148.44

(b) 4,600 = 5,000(P/F, i, 20) + 5,000(0.08)(P/A, i, 20)(b) 4,600 = 5,000(P/F, i, 20) + 5,000(0.08)(P/A, i, 20)

i = 8.9% per year.i = 8.9% per year.

Example 3Example 3 A bond that matures in A bond that matures in 88 years has a face value of years has a face value of $10,000$10,000. .

The bond stipulates a fixed nominal interest rate of The bond stipulates a fixed nominal interest rate of 8%8% per per year, but interest payments are made to the bondholder every year, but interest payments are made to the bondholder every three monthsthree months. A prospective buyer of this bond would like to . A prospective buyer of this bond would like to earn earn 10%10% nominal interest per year on his or her investment nominal interest per year on his or her investment because interest rates in the economy have risen since the because interest rates in the economy have risen since the bond was issued. How much should this buyer be willing to bond was issued. How much should this buyer be willing to pay for the bond?pay for the bond?

i3mon = (1.1)1/4 - 1 = 0.024 = 2.4%

VVNN = 10,000 = 10,000(0.02) (P/A, 2.4%, 32) + 10,000(P/F, 2.4%, 32) (0.02) (P/A, 2.4%, 32) + 10,000(P/F, 2.4%, 32)

= 4,369.84 + 4,537.71 = 8,907.55 = 4,369.84 + 4,537.71 = 8,907.55

Thus, the buyer should pay no more than $8,907.55 Thus, the buyer should pay no more than $8,907.55

when 10% nominal interest per year is desiredwhen 10% nominal interest per year is desired

MARR by FW MethodMARR by FW Method

All cash inflows and outflows are calculated to All cash inflows and outflows are calculated to the future point in time at an interest rate that is the future point in time at an interest rate that is generally the MARR. generally the MARR.

FW(i) = 0; i = MARRFW(i) = 0; i = MARR If FW(i) > 0 => The investment is “good”.If FW(i) > 0 => The investment is “good”.

ExampleExample A piece of new equipment has been proposed by engineers

to increase the productivity of a certain manual welding operation. The investment cost is $25,000. and the equipment will have a market value of $5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm's MARR is 20% per year, is this proposal a sound one?

FW(20%) = $8,000(F/A,20%,5) + $5,000

- $25,000(F/P,0.2,5) = $2324.8 > 0

Because FW(20%) > 0, i.e., “Investment justifiable”

The Annual Worth Method (AW)

Capital Recovery (CR) - the minimum annual profit to Capital Recovery (CR) - the minimum annual profit to recover the initial investmentrecover the initial investment

The capital recovery (CR) amount for a project is the The capital recovery (CR) amount for a project is the equivalent uniform annual equivalent uniform annual cost cost of the capital invested. of the capital invested. It is an annual amount that covers the following two It is an annual amount that covers the following two items:items:1. Loss in value of the asset 1. Loss in value of the asset 2. Interest on invested capital (i.e., at the MARR)2. Interest on invested capital (i.e., at the MARR)

CR(i) = I(A/P, i, n) - S(A/F, i, n)CR(i) = I(A/P, i, n) - S(A/F, i, n)I = initial investmentI = initial investment

S = market (salvage) value at the end of projectS = market (salvage) value at the end of project n = project durationn = project duration

The Annual Worth Method (AW)

AW(i) = R - E - CR(i)AW(i) = R - E - CR(i)

R - Annual RevenueR - Annual Revenue

E - Annual ExpenseE - Annual Expense

If AW(i) > 0 => The project is “good”If AW(i) > 0 => The project is “good” If i = MARR thenIf i = MARR then

PW = FW = AW = 0PW = FW = AW = 0 Consistency among PW, FW, and AW methodsConsistency among PW, FW, and AW methods

Example 1Example 1 A piece of new equipment has been proposed by engineers

to increase the productivity of a certain manual welding operation. The investment cost is $25,000. and the equipment will have a market value of $5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm's MARR is 20% per year, is this proposal a sound one?

AW = R - E - CR(i), R - E = 8,000AW = R - E - CR(i), R - E = 8,000

AW(20%) = 8,000 - [25,000 (A/P,20%,5) - 5,000 AW(20%) = 8,000 - [25,000 (A/P,20%,5) - 5,000 (A/F,20%,5)] = 312.4 > 0 => The project is good.(A/F,20%,5)] = 312.4 > 0 => The project is good.

ExampleExample An investment company is considering building a 25-unit An investment company is considering building a 25-unit

apartment complex in a growing town. Because of the apartment complex in a growing town. Because of the long-term growth potential of the town, it is felt that the long-term growth potential of the town, it is felt that the company could average 90% of full occupancy for the company could average 90% of full occupancy for the complex each year. If the following items are reasonably complex each year. If the following items are reasonably accurate estimates, what is the minimum monthly rent that accurate estimates, what is the minimum monthly rent that should be charged if a 12% MARR (per year) is desired?should be charged if a 12% MARR (per year) is desired? Land investment costLand investment cost $ 50,000 $ 50,000 Building investment cost Building investment cost $225,000 $225,000 Study period, NStudy period, N 20 years 20 years Rent per unit per monthRent per unit per month RR Upkeep expense per unit per month Upkeep expense per unit per month $35 $35 Property taxes and insurance per year 10% of Property taxes and insurance per year 10% of total total initial initial investmentinvestment

Initial investment cost = 50,000 + 225,000 = 275,000Initial investment cost = 50,000 + 225,000 = 275,000

Taxes and insurance/year = 0.1 (275,000) = 27,500Taxes and insurance/year = 0.1 (275,000) = 27,500

Upkeep/year = 35(12 x 25)(0.9) = 9,450Upkeep/year = 35(12 x 25)(0.9) = 9,450

CR cost/year = 275,000(A/P,12%,20) - 50,000(A/F,12%,20) CR cost/year = 275,000(A/P,12%,20) - 50,000(A/F,12%,20) = $36,123= $36,123Revenue = Rx12x25x0.9 = 270RRevenue = Rx12x25x0.9 = 270R

Equivalent AW (of costs) = R - E - CR(i)Equivalent AW (of costs) = R - E - CR(i)

= 270R - (9450 +27,500) - 36,123 = 0 => R = 270.64= 270R - (9450 +27,500) - 36,123 = 0 => R = 270.64

this is the minimum annual rental required equals 73,073 this is the minimum annual rental required equals 73,073 and with annual compounding (M = 1) the monthly rental and with annual compounding (M = 1) the monthly rental amountamount

If M = 12, 73,073 = 0.9R(F/A, 0.01, 12)(25) = 260If M = 12, 73,073 = 0.9R(F/A, 0.01, 12)(25) = 260

Internal Rate of Internal Rate of ReturnReturn(IRR)(IRR)

IRR CalculationIRR Calculation

PW = Rk(P/F, i'%, k) - Ek(P/F, i'%, k) = 0

wherewhere RRkk = net revenues or savings for the = net revenues or savings for the kkthth

yearyear EEkk = net expenditures including any = net expenditures including any

investment costs for the kth yearinvestment costs for the kth year

N = project life (or study period)N = project life (or study period)

OR

FW = Rk(P/F, i'%, N-k) - Ek(P/F, i'%, N-k) = 0k = 0

N

0k = 0

N

Negative IRRNegative IRR

The IRR will be positive positive if satisfies two conditions:(1) both receipts and expenses are present in (1) both receipts and expenses are present in

the cash flow pattern and the cash flow pattern and

(2) the sum of receipts exceeds the sum of all (2) the sum of receipts exceeds the sum of all

cash outflows. cash outflows. Be sure to check both of these conditions in

order to avoid the unnecessary work involved with finding that the IRR is negativenegative

ExampleExample

A capital investment of $10,000$10,000 can be made in a project that will produce a uniform annual revenue of $5,310$5,310 for five years and then have a salvage value of $2,000$2,000. Annual expenses will be $3,000$3,000. The company is willing to accept any project that will earn at least 10%10% per year, before income taxes, on all invested capital. Determine whether it is acceptable by using the IRR method.

SOLUTIONSOLUTIONSum of positive cash flows ($13,550)Sum of positive cash flows ($13,550) exceeds the exceeds the

sum of negative cash flows ($10,000) => IRR >0.sum of negative cash flows ($10,000) => IRR >0.

At i' = 15%: PW = -10,000 + 2,310(3.3522)At i' = 15%: PW = -10,000 + 2,310(3.3522)

+ 2,000(0.4972) = - 1,262+ 2,000(0.4972) = - 1,262

PW= 0 = - 10,000 + (5,310 -3,000)(P/A,i'%,5) PW= 0 = - 10,000 + (5,310 -3,000)(P/A,i'%,5) + 2,000(P/F, i'%, 5)+ 2,000(P/F, i'%, 5)

At i' = 5%: PW = - 10,000 + 2,310(4.3295)At i' = 5%: PW = - 10,000 + 2,310(4.3295)

+ 2,000(0.7835) = 1,568+ 2,000(0.7835) = 1,568

1568 - 0 = 0.05 - i’ => i’ = 10.5%1568 - 0 = 0.05 - i’ => i’ = 10.5%

1568 - (-1262) 0.05 - 0.15 1568 - (-1262) 0.05 - 0.15

Due to approximation error, the true i’ = 10%Due to approximation error, the true i’ = 10%

For more accuracy: PW(10.5%) < 0 ,PW (8.5%) >0 => For more accuracy: PW(10.5%) < 0 ,PW (8.5%) >0 => Linear Interpolate 10.5% & 8.5%Linear Interpolate 10.5% & 8.5%

ExampleExample A piece of new equipment has been proposed by A piece of new equipment has been proposed by

engineers to increase the productivity of a certain engineers to increase the productivity of a certain manual welding operation. The investment cost is manual welding operation. The investment cost is $25,000, and the equipment will have a salvage $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five value of $5,000 at the end of its expected life of five years. Increased productivity attributable to the years. Increased productivity attributable to the equipment will amount to $8,000 per year after equipment will amount to $8,000 per year after extra operating extra operating costs costs have been subtracted from the have been subtracted from the value of the additional production. Evaluate the IRR value of the additional production. Evaluate the IRR of the proposed equipment. Is the investment a good of the proposed equipment. Is the investment a good one? Recall that the MARR is 20%.one? Recall that the MARR is 20%.

SolutionSolutionPW = 8,000 (P/A, i',5) + 5,000 PW = 8,000 (P/A, i',5) + 5,000 (P/F, i', (P/F, i', 5) - 25,0005) - 25,000

PW(MARR) = PW(0.2) = 934.3 > 0PW(MARR) = PW(0.2) = 934.3 > 0

=> IRR > MARR => This investment is a good one.=> IRR > MARR => This investment is a good one.

To calculate IRRTo calculate IRR

PW(0.25) = -1847.1PW(0.25) = -1847.1

Linear InterpolationLinear Interpolation

943 - 0 = 0.2 - i’ => i’ = 21.58%943 - 0 = 0.2 - i’ => i’ = 21.58%

943 - (-18147.1) 0.2 - 0.25 943 - (-18147.1) 0.2 - 0.25

i’ > MARR = 20% => This equipment is economically attractivei’ > MARR = 20% => This equipment is economically attractive

ExampleExample

In 1915 Albert Epstein allegedly borrowed $7,000 from a In 1915 Albert Epstein allegedly borrowed $7,000 from a

bank on the condition that he would repay 7% of the bank on the condition that he would repay 7% of the

loan every three months, until a total of 50 payments loan every three months, until a total of 50 payments

had been made. At the time of the 50th payment, the had been made. At the time of the 50th payment, the

$7,000 loan would be completely repaid. Albert $7,000 loan would be completely repaid. Albert

computed his interest rate to be computed his interest rate to be

[0.07(7,000) x 4] / 7,000 = 0.28 (28%).[0.07(7,000) x 4] / 7,000 = 0.28 (28%).

(a)What true (a)What true effective effective annual interest rate did Albert pay? annual interest rate did Albert pay?

(b) What, if anything, was wrong with his calculation?(b) What, if anything, was wrong with his calculation?

ExampleExampleIn 1915 Albert Epstein allegedly borrowed $7,000 from a In 1915 Albert Epstein allegedly borrowed $7,000 from a

bank on the condition that he would repay 7% of the loan bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments had been every three months, until a total of 50 payments had been made. At the time of the 50th payment, the $7,000 loan made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert computed his interest would be completely repaid. Albert computed his interest rate to be rate to be

[0.07(7,000) x 4] / 7,000 = 0.28 (28%).[0.07(7,000) x 4] / 7,000 = 0.28 (28%).



PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarterPW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter

0.07(P/A, i'%, 0.07(P/A, i'%, 50) = 150) = 1


bank on the condition that he would repay 7% of the bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments loan every three months, until a total of 50 payments had been made. At the time of the 50th payment, the had been made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert $7,000 loan would be completely repaid. Albert computed his interest rate to be computed his interest rate to be

[0.07(7,000) x 4] / 7,000 = 0.28 (28%).[0.07(7,000) x 4] / 7,000 = 0.28 (28%).




0.07(P/A, i'%, 0.07(P/A, i'%, 50) = 150) = 1 => (P/A, i’%, 50) = 14.28 => (P/A, i’%, 50) = 14.28

(P/A, 6%, (P/A, 6%, 50) = 15.77 (P/A,7%,50) = 13.850) = 15.77 (P/A,7%,50) = 13.8

15.77 - 14.28 = 0.06 - i’ => i’ = 6.73%15.77 - 14.28 = 0.06 - i’ => i’ = 6.73%

15.77 - (13.8) 0.06 - 0.0715.77 - (13.8) 0.06 - 0.07


bank on the condition that he would repay 7% of the bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments loan every three months, until a total of 50 payments had been made. At the time of the 50th payment, the had been made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert $7,000 loan would be completely repaid. Albert computed his interest rate to be computed his interest rate to be

[0.07(7,000) x 4] / 7,000 = 0.28 (28%).[0.07(7,000) x 4] / 7,000 = 0.28 (28%).




0.07(P/A, i'%, 0.07(P/A, i'%, 50) = 1 => (P/A, i’%, 50) = 14.2850) = 1 => (P/A, i’%, 50) = 14.28

(P/A, 6%, (P/A, 6%, 50) = 15.77 (P/A,7%,50) = 13.850) = 15.77 (P/A,7%,50) = 13.8

15.77 - 14.28 = 0.06 - i’ => i’ = 6.73%15.77 - 14.28 = 0.06 - i’ => i’ = 6.73%

15.77 - (13.8) 0.06 - 0.0715.77 - (13.8) 0.06 - 0.07

Effective RateEffective Rate = (1 + 0.673) = (1 + 0.673)44 - 1 = 30% - 1 = 30%

ExampleExample A finance company advertises a "bargain 6% A finance company advertises a "bargain 6%

plan" for financing the purchase of automobiles. plan" for financing the purchase of automobiles. To the amount of the loan being financed, 6% is To the amount of the loan being financed, 6% is added for each year money is owed. This total is added for each year money is owed. This total is then divided by the number of months over which then divided by the number of months over which the payments are to be made, and the result is the the payments are to be made, and the result is the amount amount of the monthly payments. For example, a woman purchases a $10,000 automobile under this plan and makes an initial cash payment of $2,500. She wishes to pay the balance in 24 monthly payments

ExampleExample A finance company advertises a "bargain 6% plan" for financing A finance company advertises a "bargain 6% plan" for financing

the purchase of automobiles. To the amount of the loan being the purchase of automobiles. To the amount of the loan being financed, 6% is added for each year money is owed. This total is financed, 6% is added for each year money is owed. This total is then divided by the number of months over which the payments then divided by the number of months over which the payments are to be made, and the result is the amount of the monthly are to be made, and the result is the amount of the monthly payments. For example, a woman purchases a $10,000 payments. For example, a woman purchases a $10,000 automobile under this plan and makes an initial cash payment of automobile under this plan and makes an initial cash payment of $2,500. She wishes to pay the balance in 24 monthly payments$2,500. She wishes to pay the balance in 24 monthly payments

Purchase price = $10,000 - 2,500 = 7,500 = PPurchase price = $10,000 - 2,500 = 7,500 = P00

6% finance charge = 0.06 x 2 years x $7,500 = 9006% finance charge = 0.06 x 2 years x $7,500 = 900 =>Total to be paid = 8,400 =>Total to be paid = 8,400 Monthly payments = $8,400/24 = $ 350Monthly payments = $8,400/24 = $ 350What effective annual rate of interest does she actually pay?What effective annual rate of interest does she actually pay?

In Reality …..In Reality …..

PP00 = A(P/A, i'%, = A(P/A, i'%, N) => N) => 7,500 = 350(P/A, i'%,24)7,500 = 350(P/A, i'%,24)(P/A, i'%, (P/A, i'%, 24) = 7,500 / 350 = 21.4324) = 7,500 / 350 = 21.43(P/A, 0.75%, 24) = 21.89 and (P/A, 0.75%, 24) = 21.89 and (P/A, 1%, 24) = 21.24(P/A, 1%, 24) = 21.24

Linear InterpolationLinear Interpolation

21.43 - 21.24 = i’ - 1% => i’ = 0.93%21.43 - 21.24 = i’ - 1% => i’ = 0.93%

21.89 - 21.24 0.75% - 1%21.89 - 21.24 0.75% - 1%

Effective RateEffective Rate = (1 + 0.0093) = (1 + 0.0093)1212 -1 = 12% -1 = 12%

Difficulties with IRR MethodDifficulties with IRR Method

The PW, AW, and FW methods assume that net The PW, AW, and FW methods assume that net receipts less expenses (positive recovered funds) each receipts less expenses (positive recovered funds) each time period are reinvested at the time period are reinvested at the MARRMARR during the during the study period, N. study period, N.

IRR method assuming that recovered funds, if not IRR method assuming that recovered funds, if not consumed in each time period, are reinvested at consumed in each time period, are reinvested at IRRIRR (i'%) rather than at the MARR. This assumption may (i'%) rather than at the MARR. This assumption may not mirror reality in some problems, thus making IRR not mirror reality in some problems, thus making IRR an unacceptable method for analyzing engineering an unacceptable method for analyzing engineering alternatives.alternatives.

Multiple IRRMultiple IRR

External Rate of Return (ERR)External Rate of Return (ERR)

The reinvestment assumption of the IRR method noted previously may not be valid in an engineering economy study. For instance, if a firm's MARR is 20% per year and the IRR for a project is 42.4%, it may not be possible for the firm to reinvest net cash proceeds from the project at much more than 20%.

EEkk(P/F, (P/F, %, k)(F/P, i'%, N) = %, k)(F/P, i'%, N) =

RRkk(F/P, (F/P, %, N - k) %, N - k)

where where RRkk = = excessexcess of receipts over expenses in of receipts over expenses in

period period k k

EEkk = = excessexcess of expenditures over receipts in of expenditures over receipts in

period period k k

N = N = project life or number of periods for the study project life or number of periods for the study

= = external reinvestment rate per periodexternal reinvestment rate per period

k = 0

N

k = 0

N

ExampleExample

The equipment investment cost is The equipment investment cost is $25,000$25,000, and the , and the equipment will have a salvage value of equipment will have a salvage value of $5,000$5,000 at the at the end of its expected life of five years. The equipment end of its expected life of five years. The equipment will amount to will amount to $8,000$8,000 per year after extra operating per year after extra operating costs costs have been subtracted from the value of the have been subtracted from the value of the additional production. Suppose that additional production. Suppose that

= MARR = 20%= MARR = 20% What is the alternative's external rate of return, and is What is the alternative's external rate of return, and is

the alternative acceptable?the alternative acceptable?

ExampleExample The equipment investment cost is The equipment investment cost is $25,000$25,000, and the , and the

equipment will have a salvage value of equipment will have a salvage value of $5,000$5,000 at the at the end of its expected life of five years. The equipment end of its expected life of five years. The equipment will amount to will amount to $8,000$8,000 per year after extra operating per year after extra operating costs costs have been subtracted from the value of the have been subtracted from the value of the additional production. Suppose that additional production. Suppose that

= MARR = 20%= MARR = 20% What is the alternative's external rate of return, and is What is the alternative's external rate of return, and is

the alternative acceptable?the alternative acceptable?

25,000(F/P, i'%,5) = 8,000(F/A,20%,5) + 5,00025,000(F/P, i'%,5) = 8,000(F/A,20%,5) + 5,000

(F/P,i'%, (F/P,i'%, 5) 5) = = 2.5813, i' = 20.88% 2.5813, i' = 20.88%

Because i' > MARR, the alternative is barely justified.Because i' > MARR, the alternative is barely justified.

When When = 15% and MARR = 20%, determine whether the = 15% and MARR = 20%, determine whether the project whose total cash flow diagram appear below is project whose total cash flow diagram appear below is acceptable. Notice in this example that the use of an acceptable. Notice in this example that the use of an is is different from MARRdifferent from MARR.. This might occur if, for some This might occur if, for some reason, part or all of the funds related to a project are reason, part or all of the funds related to a project are "handled" outside the firm's normal capital structure."handled" outside the firm's normal capital structure.

ExampleExample

Eo = $10,000 Eo = $10,000 (k = 0), (k = 0), E1 = $5,000 E1 = $5,000 (k = 1)(k = 1)

Rk = $5,000 Rk = $5,000 for k = 2,3, . . .,6for k = 2,3, . . .,6

| - 10,000 - 5,000(P/F,15%, 1)|(F/P, i'%, 6)| - 10,000 - 5,000(P/F,15%, 1)|(F/P, i'%, 6)

= 5,000(F/A,15%,5); i'% = 15.3%= 5,000(F/A,15%,5); i'% = 15.3%

Advantages of ERR over IRRAdvantages of ERR over IRR

It can usually be It can usually be solved directlysolved directly rather than rather than by trial and error.by trial and error.

ERR is unique.unique.

The Payback Period MethodThe Payback Period Method

Suppose that a project where all capital Suppose that a project where all capital investment occurs at time 0.investment occurs at time 0.


Suppose that a project where all capital investment Suppose that a project where all capital investment occurs at time 0.occurs at time 0.

Simple Payback periodSimple Payback period ( ( NN ) is the smallest ) is the smallest integer such thatinteger such that

( R( Rkk - E - Ekk ) ) I (Initial Investment) I (Initial Investment)k = 0

N


Suppose that a project where all capital investment Suppose that a project where all capital investment occurs at time 0.occurs at time 0.

Simple Payback periodSimple Payback period ( ( NN ) is the smallest ) is the smallest integer such thatinteger such that

( R( Rkk - E - Ekk ) ) I (Initial Investment) I (Initial Investment)

Discounted Payback periodDiscounted Payback period ( ( NN ) is the smallest ) is the smallest integer such thatinteger such that

( R( Rkk - E - Ek k )(P/F, i, k) )(P/F, i, k) I ; i = MARR I ; i = MARR

k = 0

N

k = 0

N

ExampleExample

The equipment investment cost is The equipment investment cost is $25,000$25,000, and the , and the equipment will have a salvage value of equipment will have a salvage value of $5,000$5,000 at the at the end of its expected life of five years. The equipment end of its expected life of five years. The equipment will amount to will amount to $8,000$8,000 per year after extra operating per year after extra operating costs costs have been subtracted from the value of the have been subtracted from the value of the additional production.additional production.

ExampleExample


(8,000) (8,000) 25,000 => 25,000 => Simple Payback Period = 4Simple Payback Period = 4

k = 0

4

ExampleExample


(8,000) (8,000) 25,000 => 25,000 => Simple Payback Period = 4Simple Payback Period = 4

8,000(P/A,20%,5) + 5,000(P/F,20%,5) > 25,000

=> Discounted Payback Period = 5

k = 0

4

k = 05

HomeworkHomework

Page 177Page 177 # 2, 3, 5, 9, 12, 13, 19, 24, 25, 30, # 2, 3, 5, 9, 12, 13, 19, 24, 25, 30,

33, 37, 41, 49, 5133, 37, 41, 49, 51 Due Date: Oct. 27, 1998Due Date: Oct. 27, 1998

Applications of Money- Time Relationships MARR by PW Method r r All cash inflows and outflows are...

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