Question

The Florida Department of Agriculture & Consumer Services is investigating two proposals for increasing the capacity of the drainage canal in an agricultural region. (15 points) i) Proposal A requires removing weeds and sediment that have accumulated during previous years. The dredging equipment costs $650,000 to purchase. The equipment is expected to have a 10-year life with a $17,000 salvage value. The annual operating costs are estimated to be $50,000. To control weeds in the canal, an additional $120,000 is required per year for the weed control program. ii) Proposal B is to line the canal with concrete at an initial cost of $4 million. The lining is assumed to be permanent, but maintenance will be required at a cost of $5,000 each year. In addition, lining repairs will be made every 5 years at a cost of $30,000. Compare the alternative using the EUAW with an interest rate of 5% per year. State which is the better alternative and provide one sentence explaining why.

Answer 1

we will compare present value of costs of both of the projects in order compare them. whichever has lower present value of the cost, we will choose that proposal.

let v = 1/(1+i)

v = 1/(1+0.05)

v = 0.95

Proposal A

Annual operating costs + Cost for Weed control program = 50000 + 120000 = 170000

we are assuming that these annual costs occur at the end of each year.

Present Value of Costs = 650000 + 170000*( v^1 + V^2 +v^3 +v^4 +v^5 +v^6 +v^7 +v^8 +v^9 +v^10) - 17000*v^10

we will use formula of sum of GP series in order to find sum of v^1 to v^10

PV = 650000 + 170000*[1 - (0.95)^10]/[1 - 0.95] - 17000*[0.95]^10

= 650000 + 170000*[1 - 0.5987]/[0.05] - 10178.528

= 650000 + 1364420 - 10178.528

= 2,004,241.47

this present value of the costs is for 10 years. but at 11th year this investment will take place again, so at the 11th year (costs for next 10 year) its present value will be the same as 2004241.41. its is also same at 21st year (cost for the next 10 years). this will go on forever. so we have to find out the present value at time 0 for all the investments.

PV = 2004241.47 + 2004241.47*[v^10 +v^20 +v^30 +v^40 +v^50...]

PV = 2004241.47 + 2004241.47*v^10*[v^1 +v^2 +v^3 +v^4 +v^5...]

we will use the formula for finding the sum of infinite number in a GP series.

PV = 2004241.47 + 2004241.47*(0.95)^10*[1/(1-0.95)]

= 2004241.47 + 2004241.47*(0.5987)*[20]

= 2004241.47 + 23998787.4

= 26,003,023.2

Proposal B

PV = 4000000 + 5000*[v^1 + v^2 + v^3 + v^4 + v^5...] + 30000*[v^5 + v^10 + v^15 + v^20 + v^25...]

PV = 4000000 + 5000*[v^1 + v^2 + v^3 + v^4 + v^5...] + 30000*v^5*[v^1 + v^2 + v^3 + v^4 + v^5...]

= 4000000 + 5000*[(0.95)^1 +(0.95)^2 + (0.95)^3 + (0.95)^4 + (0.95)^5...] + 30000*(0.95)^5*[(0.95)^1 +(0.95)^2 + (0.95)^3 + (0.95)^4 + (0.95)^5...]

= 4000000 + 5000*[1/(1-0.95)] + 30000*(0.7738)*[1/(1-0.95)]

= 4000000 + 5000*[20] + 30000*(0.7738)*(20)

= 4000000 + 100000 + 464280

= 4,564,280

Present Value of Proposal B is way less than Proposal A, so we will choose Proposal B.

This is because for Proposal A we have to purchase dredging equipment every 10 year. so if we consider lifetime, then proposal A requires huge amount of investment over the spread of many years.