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Answer :
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given PW = AW×( ((1+i)^n - 1) ÷ (i×(1+i)^n) )
where we're finding i
and we know
PW = $10000
AW = $2940
n = 6
because we get i = 0.190981
and using that we find FW in
FW = -PW×(1+i)^n + AW×( ((1+i)^n - 1) ÷ i)
where
PW = $10000
AW = $2940
n = 6
we get FW = $0.03 approx $0
where i = 0, n is infinity... just look at the equations. if there is a discount rate of 0%, the discounted payback period is infinite
given PW = AW×( ((1+i)^n - 1) ÷ (i×(1+i)^n) )
where we're finding i
and we know
PW = $10000
AW = $2940
n = 6
because we get i = 0.190981
and using that we find FW in
FW = -PW×(1+i)^n + AW×( ((1+i)^n - 1) ÷ i)
where
PW = $10000
AW = $2940
n = 6
we get FW = $0.03 approx $0
where i = 0, n is infinity... just look at the equations. if there is a discount rate of 0%, the discounted payback period is infinite
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