ENGINEERING ECO ANALYSIS W/STUDY GUIDE
ENGINEERING ECO ANALYSIS W/STUDY GUIDE
13th Edition
ISBN: 9780190693053
Author: NEWNAN
Publisher: Oxford University Press
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Chapter 14, Problem 86P
To determine

The alternative that should be selected by the couple.

Expert Solution & Answer
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Answer to Problem 86P

Buying the house is an effective alternative.

Explanation of Solution

Calculation:

Alternative 1: Continue to rent the duplex home.

Determine the payment for year 1.

P1=(Houserent+Basicutilitiescost)×N ...... (I)

Here, N is the number of months per year, P1 is the payment.

Substitute $450 for house rent, $139 for basic utilities cost and 12 months for N.

P1=($450+$139)×12=589×12=$7,068

Calculate the equivalent year 0 payment in year 0 dollars.

P0=P1(1+f)n ...... (II)

Here, f is the inflation rate.

Substitute $7,068 for P1, 5% for f and 1 year for n in Equation (II).

P0=$7,068(1+0.05)1=$7,068(1.05)1=$7,068(0.9523)=$6,731.42

Calculate the equivalent interest rate.

i=(i'+f+i'×f) ...... (III)

Here, i' is the nominal interest rate and i is the equivalent interest rate.

Substitute 15.5% for i' and 5% for f.

i=(0.155+0.05+0.155×0.07)=(0.205+0.01085)=(0.21585)×100=21.58%

Calculate the present worth of house rent after 10 years.

PW10=A(PA,i,n) ...... (IV)

Substitute $6,731.42 for A, 21.58% for i and 10 years for n.

PW10=$6,731.42(PA,21.58%,10)=$6,731.42((1+0.2158)1010.2158(1+0.2158)10)=$6,731.42(3.9768)=$26,769.74

Thus, the present worth of the house rent after 10 years is $26,769.74.

Alternative 1: Buying a house.

Calculate the mortgage interest rate per month.

im=Mortgageinterestrate12

Substitute 0.08 for Mortgageinterestrate.

im=0.0812=0.66%

Calculate the number of payments.

n=N×12 ...... (V)

Substitute 30 years for N in Equation (V).

n=30×12=360payments

Calculate the down payment.

Downpayment=Housecost×inflationrate ...... (VI)

Substitute $75,000 for house cost and 5% for inflation rate.

Downpayment=$75,000×0.05=$3,750

Calculate the closing cost in constant dollars.

Closingcostinconstantdollars=[Housecost×interestrateofclosingcost] ...... (VII)

Substitute $75,000 for house cost and 1% for interest rate at closing cost in Equation (VII).

Closingcostinconstantdollars=$75,000×0.01=$750

Write the formula to Calculate the monthly payment.

A=(HousecostDownpayment)(AP,im,n) ...... (VIII)

Here, A is the monthly payment.

Substitute $75,000 for house cost, $3750 for down payment, 0.66% for im and 360 for n in Equation (VIII).

A=($75,000$3750)(AP,0.66%,360)=($75,000$3750)(0.00667(1+0.00667)360(1+0.00667)3601)=$71,250(0.00734)=$523

Calculate the mortgage balance after 10 year comparison period.

A'=A(PA,im,n) ...... (IX)

Substitute $523 for A, 0.66% for im and 240 for n in Equation (IX).

A'=$523(PA,0.66%,240)=$523((1+0.0067)24010.0067(1+0.0067)240)=$523(119.51)=$62,504

Calculate the total amount paid in payments.

n=A×N×12 ...... (X)

Here, N is number of years.

Substitute $523 for A and 10 years for N in Equation (X).

n=$523×10×12=$62,760

Calculate the principal repayments.

Principalrepayments=(HomecostDownpayment)A' ...... (XI)

Substitute $75,000 for home cost, $3750 for down payment and $62,504 for A' in Equation (XI).

Principalrepayments=($75,000$3750)$62,504=$8,746

Calculate the interest payments.

Interestpayments=nPrincipalrepayments ...... (XII)

Substitute $62,760 for n and $8,746 for Principalrepayments in Equation (XII).

Interestpayments=$62,760$8,746=$54,014

Calculate the monthly tax saving.

Monthlytaxsaving=(A×Averageinterestrateonloanpayment×Marginalincometax) ...... (XIII)

Substitute $523 for A, 0.8772 for Averageinterestrateonloanpayment, 0.3 for Marginal tax rate in Equation (XIII).

Monthlytaxsaving=(523×0.877×0.3)=$138

Calculate the cost of mortgage after tax.

Costofmortgageafterthetax=AMonthlytaxsaving ...... (XIV)

Substitute $523 for A and $138 for monthly tax saving in equation (XIV).

Costofmortgageafterthetax=523138=$385

Calculate the sale amount of the property after 10 years.

Saleamountofproperty=Homecost(FP,i,n) ...... (XV)

Substitute $75,000 for home cost and 6% for i in Equation (XV).

Saleamountofproperty=$75,000(FP,6%,10)=$75,000(1+0.06)10=$134,314

Calculate the net income from the sale.

Netincomefromthesale=(Saleamountoftheproperty(Commission×saleamountoftheproperty)A') ...... (XVI)

Substitute $134,314 for sale amount and 5% for Commission, $62,504 for A' in Equation (XVI).

Netincomefromthesale=($134,314(0.05×$134,314)$62,504)=$134,314671662504=$65,094

Calculate the present worth of home cost of owning house for year 1.

PW1=[(Downpayment+Closingcost)+(Costofmortgage×12)(PA,i',n)+(Constantdollarutilities×12)(PA,i,n)+(Constantdollarinsuranceandmaintenance×12)(PA,i,n)Netincomefromthesale(PF,i',n)] ...... (XVII)

Substitute $3,750 for down payment, $750 for closing costs, $385 for Costofmortgage, 0.155 for i', 10 years for n, 0.10 for i and $65,094 for Netincomefromthesale in Equation (XVII).

PW1=[(3750+750)+(385×12)((1+0.155)1010.155(1+0.155)10)+(160×12)((1+0.10)1010.10(1+0.10)10)+(50×12)((1+0.10)1010.10(1+0.10)10)65,094(1(1+0.155)10)]=[4500+4620(4.925)+(1920)(6.144)+(600)(6.144)65,094(0.2366)]=$27,335.14

Thus, the present worth of cost of owning the house is $27,335.14.

Conclusion:

Thus, buying the house is an affective alternative.

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ENGINEERING ECO ANALYSIS W/STUDY GUIDE

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