Two options for a new bridge are being considered. The first bridge costs $2,000

Question

Two options for a new bridge are being considered. The first bridge costs $2,000,000 to build and lasts for 40 years. The bridge costs $200,000 a year to maintain in year 1, and this cost increases by 0.5% each year. The second bridge costs $1,5000,000 to build and lasts for 30 years. The maintenance cost of the 2nd bridge is $195,000 a year, increasing by $1,110 per year. The annual interest rate is 5%. If a bridge is going to be needed for the indefinite future, Present Worth Analysis to determine whether the 1st or 2nd bridge should be built.

Explanation / Answer

Option-A

The initial cost of the first bridge is $2,000,000 lasting for 40 years.The maintenance cost of the first bridge is $200,000 which increases by 0.5% making the annual increase=(0.005*$200,000)=$1000

The interest rate or i is 5% or 0.05 and the timeline of the project is 40 years which can be divided annually or n=1

Hence,the Present Worth of Project A or PW(A) can be written as:-

PW(A)=-$2,000,000-$200,000/i-$1000*{i/((1+i)^n)-1}

PW(A)=-$2,000,000-$200,000/0.05-$1000{0.05/((1+0.05)^1)-1}

PW(A)=-$2,000,000-$40,00,000-$1000(0.05/0.05)

PW(A)=-$2,000,000-$40,00,000-$1000

PW(A)=-$60,01,000

Option-B

The initial cost of the bridge is $15,000,000 and the maintenance cost of first year is $195,000 increasing $1110 per year.The project timeline is 30 years

Hence,the Present Worth of Project B or PW(B) can be written as:-

PW(B)=-$15,000,000-$195,000/i-$1110*{i/((1+i)^n)-1}

PW(B)=-$15,000,000-$195,000/0.05-$1110{0.05/((1+0.05)^1)-1}

PW(B)=-$15,000,000-$39,00,000-$1110(0.05/0.05)

PW(B)=-$15,000,000-$39,00,000-$1110

PW(B)=-$189,01,110

Notice from the above calculations that option-A yields a higher present worth PW(A) compared to the present worth of option-B or PW(B).Therefore,it would be reasonable to choose option-A or the first bridge considering the indefinite time period in the future.

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