Question

A town wants to replace its water pumping system they have to options gravity plan and pumping plan. Both options have a 40-year life, at a 10% interest rate no salvage value. OS Gravity Pumping DO $1,400,000 MAAR Initial Investment $2,800,000 Investment @ none year 10 TO o SO 1 TA $200,000 O ) Ne Operation and 10.000$/yr Maintenance AT Power Cost None 50,000$/yr > Average first 10 years (A value) Average next 30 N years (A Value) OOT OO A- 100,000 $/yr CD Using annual cash flow analysis which is the better option answer in numerical format.

Answer 1

ANSWER:

i = 10% and n = 40 years

1) Gravity:

pw = initial investment + o and m(p/a,i,n)

pw = 2,800,000 + 10,000(p/a,10%,40)

pw = 2,800,000 + 10,000 * 9.779

pw = 2,800,000 + 97,990

pw = 2,897,990

aw = pw(a/p,i,n)

aw = 2,897,990(a/p,10%,40)

aw = 2,897,990 * 0.1023

aw = 296,443.91

so the annual worth is $296,443.91

2) pumping:

pw = initial investment + investment in year 10(p/f,i,n) + o and m(p/a,i,n) + 1st 10 years power cost(p/a,i,n) + last 30 years power cost(p/a,i,n) (p/f,i,n)

pw = 1,400,000 + 200,000(p/f,10%,40) + 25,000(p/a,10%,40) + 50,000(p/a,10%,10) + 100,000(p/a,10%,30) (p/f,10%,10)

pw = 1,400,000 + 200,000 * 0.0221 + 25,000 * 9.779 + 50,000 * 6.145 + 100,000 * 9.427 * 0.3855

pw = 1,400,000 + 4,420 + 244,475 + 307,250 + 363,410.85

pw = 2,319,555.85

aw = pw(a/p,i,n)

aw = 2,319,555.85(a/p,10%,40)

aw = 2,319,555.85 * 0.1023

aw = 237,290.56

since the annual cost of gravity is more then that of pumping , therefore we will select pumping.