Question

A new highway is to be constructed.

Design A calls for a concrete pavement costing ​$85 per foot with a 16​-year ​life;three paved ditches costing ​$4 per foot​each; and four box culverts every​ mile, each costing ​$8,000 and having a 16​-year life. Annual maintenance will cost ​$11,800 per​mile; the culverts must be cleaned every four years at a cost of​$300 each per mile.


Design B calls for a bituminous pavement costing ​$40 per foot with a 8​-year ​life; two sodded ditches costing ​$1.40 per foot​ each; and two pipe culverts every​ mile, each costing ​$2,200 and having a 8​-year life. The replacement culverts will cost ​$2,400 each. Annual maintenance will cost ​$2,900 per​ mile; the culverts must be cleaned yearly at a cost of ​$230 each per​ mile; and the annual ditch maintenance will cost ​$1.45 per foot per ditch.


Compare the two designs on the basis of equivalent worth per mile for a 16​-year period. Find the most economical design on the basis of AW and PW if the MARR is 6​% per year.

Answer 1

Desing B

Total Present worth $ 502.485,35‬

Annual worth: $ 49,722.003

Option 2:

Total Present worth $ 666.441,33‬

Annual worth: $ 53,845.798

Step-by-step explanation:

Desing A

$85 x 5280 = 448,800

$4 x 3 x 5280 = 63,360

$8000 x 4 = 32,000

total cost: 544,160‬

Annual cost:

11,800 + 300 = 12,100

PV of the annual maintenance:

`C * (1-(1+r)^(-time) )/(rate) = PV\\`

C 12,100.00

time 16

rate 0.06

`12100 * (1-(1+0.06)^(-16) )/(0.06) = PV\\`

PV $122,281.3328

Present worth:

total cost to construct 544,160‬ + maintenance $122,281.33 = 666.441,33

Annual worth:

`PV / (1-(1+r)^(-time) )/(rate) = C\\`

PV 544,160

time 16

rate 0.06

`544160 / (1-(1+0.06)^(-16) )/(0.06) = C\\`

C $ 53,845.798

Desing B

a mile is equivalent to 5,280 foot

paviment $40 x 5,280 = $ 211,200

sood ditched 2 per foot x 5,280 foot per mile x $1.40 = $ 12,038.4

pipe culvert 2,200 x 2 = 4,400

Total value to construct: 227.638,4‬

PV of maintenance:

replacement 2,400 x 2 = 4,800 (in 8 years)

`(Maturity)/((1 + rate)^(time) ) = PV`

Maturity $4,800.0000

time 8.00

rate 0.06000

`(4800)/((1 + 0.06)^(8) ) = PV`

PV 3,011.5794

maintenance $2,900

culverts 2 x $ 230 = $ 460

ditch 1.45 x 5,280 x 2 = $ 15,312

Total yearly cost: 18.672‬

PV of this annuity over 16 years:

`C * (1-(1+r)^(-time) )/(rate) = PV\\`

C 18,672.00

time 16

rate 0.06

`18672 * (1-(1+0.06)^(-16) )/(0.06) = PV\\`

PV $188,697.2765

PV of the replacement bituminous concrete

`(Maturity)/((1 + rate)^(time) ) = PV`

Maturity $211,200.0000

time 16.00

rate 0.06000

`(211200)/((1 + 0.06)^(16) ) = PV`

PV 83,138.0951

Present worth:

Total value to construct: $ 227.638,4‬ +

yearly cost PB $ 188,697.28 +

concrete replacement $ 83,138.0951

culvert replacement: $ 3,011.58

Total Present worth 502.485,35‬

Annual worth:

`PV / (1-(1+r)^(-time) )/(rate) = C\\`

`502485.35 / (1-(1+0.06)^(-16) )/(0.06) = C\\`

C $ 49,722.003