Chapter 10, Problem 25

Chapter 10, Problem 25

Incandescent Bulb (60 watt) / CFL Bulb (15watts)
Price: $0.50 / Price : $3.50
Bulb Life: 1,000 hours / Bulb Life: 12,000 hrs
Light Usage: 500hrs/year / Light Usage: 500hrs/year
R=10% / R = 10%
Project Life: 1,000/500 = 2 years / Project Life : 12,000/500 = 24 years
Initial Investment / Initial Investment
Bulb: $0.50
Total: $0.50 / Bulb: $3.50
Total: $3.50
OCF / OCF
OCF1 – 2 = $3.03 / OCF1 – 24 = $0.7575
Non OCF / Non OCF
Non OCF = 0 / Non OCF = 0
NPV= -0.50 – 3.03 (PVIFA2,10%) / NPV= -3.50 – 0.7575 (PVIFA24,10%)
CF0 -0.50
CF1-3.03F12
I10%
Cpt NPV -5.759 / CF0 -3.50
CF1-0.7575F124
I10%
Cpt NPV -10.3059
Compute EAC
N2
I10%
PV 5.759
cpt.Pmt 3.318
FV0
EAC = $3.318 / Compute EAC
N24
I10%
PV 10.3059
cpt. Pmt1.147
FV0
EAC = $1.147

Chapter 10, Problem 26

Set EACIncandescent= EACCFL

Incandescent / CFL
NPV = -0.50 – [(30)(Elec Cost)](PVIFA2,10%) / NPV = -3.50 – [(7.50)(Elec. Cost)](PFIVA24,10%)
EACIn:
-0.50-[(30)(Elec.Cost)](PVIFA2,10%) = EAC (PVIFA2,10%) / EACCFL:
-3.50 – [(7.50)(Elec. Cost)](PFIVA24,10%)= EAC(PFIVA24,10%)
PVIFA2,10% : (Could use Tables)
N2
I10%
Cpt PV1.735537
Pmt1
FV0 / PVIFA24,10%: (Could use Tables)
N 24
I 10%
Cpt PV8.9847
Pmt1
FV0

EACIn = -0.2881 – 30(Elec. Cost) /
EACCFL = -0.3895– 7.50(Elec. Cost)

EACIn= EACCFL

-0.2881 – 30 (Elec.Cost) = -0.3895 – 7.50 (Elec.Cost)

0.1015 = (22.50)(Elec. Cost)

Elec. Cost = $ 0.004509

Chapter 10 Problem 27

The solution to this problem is similar to the solution for question 26.

The difference is the incandescent bulb has a 1-year life.

PVIF1,10% = 0.90909

= -0.5500 – (30)(Elec. Cost)

EACIn=EACCFL

-0.5500 – (30)(Elec.Cost) = -0.3895 – 7.50(Elec.Cost)

-0.160451 = 22.50(Elec.Cost)

Elec Cost = -0.007131

The electricity cost is negative. This means the incandescent bulbs should not bereplaced until they burn out.

EAA Chapter10 Prob 25 26 27.DocxPage 1