Question

Two contractors submitted the following cost estimates for a proposed bridge:

Bridge A               Bridge B

First Cost             P20M     P27M

Life        20 years               30 years

Salvage Value     P500,000             P750,000

Annual Upkeep P160,000             P50,000

Repairs P1.5M every 4 years       P1M every 6 years

        If interest is 7.5% compounded annually, which bridge is cheaper in terms of its capitalized cost? and by how much?

Answer 1

Bridge A:

X.First Cost=20,000,000

ANNUAL UPKEEP COST

Uniform Series Present Worth Factor(USPWF)=(P/A,i,N)=(((1+i)^N)-1)/(i*((1+i)^N))

i=Interest Rate=7.5%=0.075, N=Number of Years=20

USPWF=(P/A,7.5%,20)=(((1+0.075)^20)-1)/(0.075*((1+0.075)^20))=10.19449

Y.PRESENT VALUE OF ANNUAL UPKEEP=160000*USPWF=160000*10.19449==1,631,119

REPAIRS COST

Interest rate in 4 years=((1+0.075)^4)-1=0.335469

i=33.55%

N=Number of 4 year Period=20/4=5

USPWF=(P/A,33.5%,5)=(((1+0.3355)^5)-1)/(0.3355*((1+0.3355)^5))=2.2792

Z. PRESENT VALUE OF REPAIR COST=1500000*2.2792=3,418,736

SALVAGE VALUE

S. PRESENT VALUE OF SALVAGE=500000/(1.075^20)=117,707

CAPITALIZED COST=X+Y+Z-S=20000000+1631119+3418736-117707=24,932,148

Bridge B:

X.First Cost=27,000,000

ANNUAL UPKEEP COST

Uniform Series Present Worth Factor(USPWF)=(P/A,i,N)=(((1+i)^N)-1)/(i*((1+i)^N))

i=Interest Rate=7.5%=0.075, N=Number of Years=30

USPWF=(P/A,7.5%,30)=(((1+0.075)^30)-1)/(0.075*((1+0.075)^30))=11.81039

Y.PRESENT VALUE OF ANNUAL UPKEEP=50000*USPWF=50000*11.81039=590,519

REPAIRS COST

Interest rate in 6 years=((1+0.075)^6)-1=0.5433

i=54.33%

N=Number of 6year Period=30/6=5

USPWF=(P/A,54.33%,5)=(((1+0.5433)^5)-1)/(0.5433*((1+0.5433)^5))=1.630364

Z. PRESENT VALUE OF REPAIR COST=1000000*1.630363=1,630,363

SALVAGE VALUE

S. PRESENT VALUE OF SALVAGE=750000/(1.075^30)=85,666

CAPITALIZED COST=X+Y+Z-S=27000000+590519+1630363-85666=29,135,216

BRIDGE A has lower Capitalized Cost

Lower by: 29,135,216-24,932,148=4,203,068