Question

Consider the following EOY cash flows for two mutually exclusive alternatives one must be chosen The MARR is 4% per year Lead Acid $9,000 $2,750 Lithium lon Capital investment Annual expenses Useful life Market value at end of useful life $15,000 $2,200 12 years 18 years 50 $3,000 Click the icon to view the interest and annuity table for discrete compounding when ,24% per year. (a) Determine which alternative should be selected based on the PW method. Assume repeatability and use a study period of 36 years. The PW of the Lead Acid is S-70100 (Round to the nearest hundreds ) The PW of the Lithium lon is s-66200 (Round to the nearest hundreds.) Which alternative should be selected? Choose the correct answer below. y Lithium lon O Lead Acid (b) would the selected alternative remain the same if the MARR were 8%? The PW of the Lead Acid is $ 46200 (Round to the nearest hundreds) The PW of the Lithium lon is s-45500 (Round to the nearest hundreds.) Which alternative should be selected? Choose the correct answer b elow Lithium Ion BLead Acid

I keep getting the Lithium Ion ones wrong, someone please help.

Answer 1

Case of Lead Acid

MARR=i=4%

Useful life=n=12 years

Capital investment=Co=9000

Salvage=S=0

Annual expanses=R=$2750

$(P/A,i,n)=\frac{1-\frac{1}{(1+i)^{n}}}{i}$

$(P/A,0.04,12)=\frac{1-\frac{1}{(1+0.04)^{12}}}{0.04}=9.385074$

PW=-Co-R*(P/A,4%,12)=-9000-2750*9.385074 =-34808.95

Since, project is to be repeated at the end of 12 years and 14 years with same cash flows. So

PW=-34808.95-34808.95(P/F,4%,12)- 34808.95*(P/F,4%,24)

(P/F,4%,12)=1/(1+4%)¹²=0.624597

(P/F,4%,24)=1/(1+4%)²⁴=0.390121

PW=-34808.95-34808.95*0.624597- 34808.95*0.390121=-70130.2

Case of Lithium ion

Useful life=n=18 years

Capital investment=Co=15000

Salvage=S=3000

Annual expanses=R=$2200

$(P/A,0.04,18)=\frac{1-\frac{1}{(1+0.04)^{18}}}{0.04}=12.659297$

(P/F,0.04,18)=1/(1+4%)^18=0.493628

PW=-Co-R*(P/A,4%,18)+S*(P/F,4%,18)=-15000-2200* 12.659297+3000*0.493628=-41369.57

Since, project is to be repeated at the end of 18 years with same cash flows. So

PW=-41369.57-41369.57(P/F,4%,18)

PW=-41369.57-41369.57*0.493628=-61790.7

PW of Lead acid=-$70100

PW of lithium ion=-61800

Cost of lithium ion is less, it should be selected.

Suppose MARR changes to 8%

Case of Lead Acid

MARR=i=8%

Useful life=n=12 years

Capital investment=Co=9000

Salvage=S=0

Annual expanses=R=$2750

$(P/A,0.08,12)=\frac{1-\frac{1}{(1+0.08)^{12}}}{0.08}=7.536078$

PW=-Co-R*(P/A,8%,12)=-9000-2750* 7.536078 =-29724.21

Since, project is to be repeated at the end of 12 years and 14 years with same cash flows. So

PW=-29724.21-29724.21(P/F,8%,12)- 29724.21*(P/F,4%,24)

(P/F,8%,12)=1/(1+8%)^12=0.397114

(P/F,8%,24)=1/(1+8%)^24=0.157699

PW=-29724.21-29724.21*0.397114-29724.21*0.157699=-46215.59

Case of Lithium ion

MARR=i=8%

Useful life=n=18 years

Capital investment=Co=15000

Salvage=S=3000

Annual expanses=R=$2200

$(P/A,0.08,18)=\frac{1-\frac{1}{(1+0.08)^{18}}}{0.08}=9.37188714$

(P/F,0.04,18)=1/(1+8%)^18=0.250249

PW=-Co-R*(P/A,8%,18)+S*(P/F,8%,18)=-15000-2200* 9.37188714+3000*0.250249

=-34867.40

Since, project is to be repeated at the end of 18 years with same cash flows. So

PW=-34867.40-34867.40 (P/F,8%,18)

PW=-34867.40-34867.40*0.250249=-43592.93

PW of Lead acid=-$46200

PW of lithium ion=-$43600

Cost of lithium ion is less, it should be selected.