Question

A new highway is to be constructed with two alternative designs. Design A calls for a concrete pavement costing $90 per foot with a 20-year life; two paved ditches costing $3 per foot each; and three box culverts every mile, each costing $9,000 and having a 20-year life. Annual maintenance will cost $1,800 per mile; the culverts must be cleaned every five years at a cost of $450 each per mile. Design B calls for a bituminous pavement costing $45 per foot with a 10-year life, two sodded ditches costing $1.50 per foot each; and three pipe culverts every mile, each costing $2,250 and having a 10-year life. The replacement culverts will cost $2,400 each. Annual maintenance will cost $2,700 per mile; the culverts must be cleaned yearly at a cost of $225 each per mile; and the annual ditch maintenance will cost $1.50 per foot per ditch. Compare the two designs using equivalent worth per mile for a 20-year period. Find the most economical design if the MARR is 6% per year.

Answer 1

Answer:

Given that:

A new highway is to constructed with two alternative designs .

Start design A in which all components have a life of 20 years . Now the capital investment for this design will be

Concrete pavement = ($90 / ft)(5,280 ft / min)

= $475,200 / mile

Paved ditches = 2($3 / ft)(5,280 ft / min)

= $31,680 / mile

Box culverts = (3 culverts / mile)($9,000 / min)

= $27,000 / mile

Therefore the total capital investment will become,

Total capital investment = $475,200 +$31,680+$27,000

= $533,880 / mile

The design maintenance cost is given as $1,800.

With this periodic cleaning of culverts will become :

(3 culverts / mile)($450 / culverts) = $1,350 / mile

Now apply the AW formula:

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

The annual worth in this case will become,

AW(6%)=-$533,880(A /P ,6%,20)-$1,800-$1,350(A / F ,6%, 5)

= -$48,594 / mile

Now calculate the present worth of design A

PW(6%)=-$533,880-[$1,800+$1,350(A /F ,6%,5)](P / A ,6%, 20)

= -$557,273 / mile

Consider design B in which all components have life of 0 years.Now the capital investment for this design in zero year will be

Bituminous pavement = ($45 / ft)(5,280 ft / min)

=$237,600

Sodded ditches = 2($150 / ft)(5,280 ft / min)

=$15,840 / mile

Pipe culverts = (3 culverts / mile)($2,250 / culverts)

=$6,750 / mile

Therefore the total capital investment will become

Total capital investment = $237,600+$15,840+$6.750

=$260,190 / mile

Consider design B in which all components have end of  years 10 years.Now the capital investment for this design in zero year will be

Bituminous pavement = ($45 / ft)(5,280 ft / min)

=$237,600

Sodded ditches = 2($150 / ft)(5,280 ft / min)

=$15,840 / mile

Pipe culverts = (3 culverts / mile)($5,280 / culverts)

=$7,200 / mile

Therefore the total capital investment will become

Total capital investment = $237,600+$15,840+$7.200

=$260,640 / mile

Calculate maintenance cost for design B

Annual cleaning of culvert = (3 culverts / mile)($225 / culvert)

=$675 / mile

Annual ditch maintenance = 2($150 / ft)($5,280 ft / min)

=$15,840 / mile

The annual pavement maintenance is given as $2,700 / mile.

Now calculate annual worth and its relation as shown below:

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

The annual worth in this case will become,

AW(6%)=-[$260,190+$260,640(P /F ,6%,10)](A/P , 6% ,20)-$9,215

= -$54,595 / mile

Now present worth will become

PW(6%)=-$260,190-$260,640(P /F ,6%,10)]-$19,215(P / A ,6%, 20)

= -$626,126 / mile