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Problem 2 [20p]: A company offers the following two options to sell a car: Option 1: you do not have to make any payments for(F/P,i,n) = (1 + i) (P/A.i.n)- (1 + i) - 1 i(1 + i) (F/A,1,n) = (1 + i) – 1 (P/G,1,n)= (1 DICE _(1 + i) - in - 1 if (1 +

Solve without Using excel/Matlab.
solve using GIVEN formulas whenever APPLICABLE. ((Do not try to solve by assuming each cash flow as a single cash flow at its corresponding year))

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Answer #1

OPTION1:

Uniform Series Present Worth Factor (USPWF)=(P/A,i,n)=(((1+i)^n)-1)/(i*((1+i)^n))

i=Monthly interest rate =30/12=2.5%=0.025

n=Number of months of payment=2*12=24

USPWF=(P/A,2.5%,24)=((1.025^24)-1)/(0.025*(1.025^24))=17.88499

Present Value of cash flow at end of Month 5=3000*USPWF=3000*17.88499=53654.96

Present Value NOW=53654/((1+i)^n)

n=5, i=0.025

Present Value NOW=53654/((1.025^5)=47423.16

OPTION 2:

i=2.5%=0.025

n=24

USPWF=(P/A,i,n)=(P/A,2.5%,24)=17.88499

Present Value of Cash flow now=1000*17.88499=17884.99

BETTER OPTION:

OPTION 2

CASH FLOW DIAGRAM: OPTION 1 Month Cash Flow 0 -47423.16 2 6 3,000 7 3,000 8 3,000 93,000 10 3,000 11 3,000 12 3,000 13 3,000CASH FLOW-OPTION-1 10000 023 LLLLLLLLLLLLLLLLLLLLLLLLL 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29CASH FLOW DIAGRAM: OPTION 2 Month Cash Flow 0 -17884.99 1 1000 2 1,000 3 1,000 1,000 1,000 1,000 1,000 1,000 1,000 10 1,000 1CASH FLOW-OPTION-2 5000 LLLLLLLLLLLLLLLLLLLLL 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 1 23 24 25 -5000 -10000

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