Post on 27-Jul-2018
PV & FV of Annuities
An annuity is a series of equal
regular payment amounts
made for a fixed number of
periods
2
Problem
• An engineer deposits P1,000 in a savings
account at the end of each year for 5 years.
How much money can he withdraw at the end
of 5 years if the bank pays interest at the rate
of 6% p.a., compounded annually?
An engineer deposits P1,000 in a savings account at the
end of each year for 5 years. How much money can he
withdraw at the end of 5 years if the bank pays interest
at the rate of 6% p.a. compounded annually?
FV = PV ( 1 + i ) ⁿ
FV = 1000(1.06)⁴ + 1000(1.06)³
+ 1000(1.06)² + 1000(1.06) + 1000
= P5,637.09
1 2 3 4 5
FV
An engineer deposits P1,000 in a savings account at the
end of each year for 5 years. How much money can he
withdraw at the end of 5 years if the bank pays interest
at the rate of 6% p.a. compounded annually?
FV = FV( rate, nper, pmt, pv, type)
FV = FV (.06,4,0,-1000,0) + FV (.06,3,0,-1000,0)
+ FV (.06,2,0,-1000,0) + FV(.06,1,0,-1000,0)
+ FV(.06,0,0,-1000,0)
= P5,637.09
1 2 3 4 5
FV
An engineer deposits P1,000 in a savings account at the
beginning of each year for 5 years. How much money
can he withdraw at the end of 5 years if the bank pays
interest at the rate of 6% p.a., compounded annually?
FV = PV ( 1 + i ) ⁿ
FV = 1000(1.06)⁵ + 1000(1.06)⁴ + 1000(1.06)³ + 1000(1.06)² + 1000(1.06)
= P5,637.09 (1.06) = P5,975.32
FV
1 2 3 4 5
An engineer deposits P1,000 in a savings account at the
beginning of each year for 5 years. How much money
can he withdraw at the end of 5 years if the bank pays
interest at the rate of 6% p.a., compounded annually?
FV = FV (rate, nper, pmt, pv, type)
FV=FV(.06,5,0,-1000,0) + FV(.06,4,0,-1000,0)
+ FV(.06,3,0,-1000,0) + FV(.06,2,0,-1000,0)
+ FV(.06,1,0,-1000,0) = P5,975.32
FV
1 2 3 4 5
FV = PV ( 1 + i ) ⁿ
FV = 1000 (1.06)⁵ + 1000(1.06)⁴ + 1000(1.06)³ + 1000(1.06)²+1000(1.06)
FV = Pmt (1 + i)ⁿ + Pmt ( 1+ i)ⁿ⁻ᶦ + Pmt (1 + i)ⁿ⁻² + Pmt(1 + i)ⁿ⁻³
+ Pmt (1 + i)ⁿ⁻⁴ + ….. + Pmt (1 + i)ⁿ⁻ⁿ
1000 = Pmt 0.06 = i
Geometric series: a + ax + ax² + + ax³ + ... + ax ⁿ⁻¹
Summation = Sn = a (1-xⁿ)
(1-x)
x = ratio of successive terms = [Pmt (1+i)ⁿ⁻ᶦ] / [Pmt (1+i)ⁿ⁻ᶦ] = 1 / (1+i)
a = first term = Pmt(1+i)ⁿ
FV = {Pmt(1+i)ⁿ} {1- [1/(1+i)]ⁿ} / {1 – [1/(1+i)]}
= Pmt (1+i)ⁿ [(1+i)ⁿ- 1)/(1+i)ⁿ] /[(1+ i – 1) / (1+i)]
FV = {Pmt (1+i) [(1+i)ⁿ- 1)]} / i annuity due formula
1 2 3 4 5
Basic Formula to UseBasic Formula to Use
B Loan Balance after n payments = FV Compounding – FV annuity
(5) B = PV (1 + i ) ⁿ - Pmt [ (1 + i)ⁿ - 1]
i
(1) FV compounding = PV ( 1 + i ) ⁿ
single transaction
(2) PV discounting = FV ( 1+ i )⁻ⁿsingle transaction
(3) FV annuity = Pmt [ (1 + i)ⁿ - 1]
ordinary annuity i
(4) PV annuity = Pmt [1 – (1 + i )⁻ⁿ]
ordinary annuity i 9
Basic Formula to UseBasic Formula to Use
B Loan Balance after n payments = FV Compounding – FV annuity
(5) B = PV (1 + i ) ⁿ - Pmt [ (1 + i)ⁿ - 1]
i
(1) FV compounding = PV ( 1 + i ) ⁿ
= FV (rate, nper, 0, pv, 0)
(2) PV discounting = FV ( 1+ i )⁻ⁿ= PV (rate, nper, 0, fv, 0)
(3) FV annuity = Pmt [ (1 + i)ⁿ - 1]
ordinary annuity i
= FV (rate, nper, pmt, pv, 0)
(4) PV annuity = Pmt [1 – (1 + i )⁻ⁿ]
ordinary annuity i
= PV (rate, nper, pmt, fv, 0) 10
An engineer deposits P1,000 in a savings account at the end of each
year for 5 years. How much money can he withdraw at the end of 5
years if the bank pays interest at the rate of 6% p.a., compounded
annually?
Method 1: Single transaction FV = PV ( 1 + i ) ⁿ
= 1000(1.06)⁴ + 1000(1.06)³ + 1000(1.06)²
+ 1000(1.06) + 1000 = P5,637.09
Method 2: FV annuity = Pmt [ (1 + i)ⁿ - 1]
i
= [1000] [(1.06)⁵ - 1] / [0.06]
Method 3: = FV (rate, nper, pmt, pv, type)
= FV ( 0.06, 5, -1000, 0, 0 )
1 2 3 4 5
B. Annuities:
A series of equal payments
A. Ordinary Annuity: Regular
deposits are made at the end of
the period.
B. Annuities Due: Regular deposits
are made at the beginning of the
period
12
Annuity Due = Ordinary Annuity x (1 + i) [Present value same formula]
� Future Value of an investment three years after, for a $3000 annuity at 8%.
Ordinary Annuity $3000 $3000 $3000
Year 1 Year 2 Year 3 FV
Annuity Due
$3000 $3000 $3000
Year 1 Year 2 Year 3 FV13
I. Ordinary Annuity:
Future Value
Toby Martin invests $2,000 at the end of each year for 10 years at 11% p.a., compounded annually. What is the final value of Toby’s investment at the end of year 10?
1) By Table:Periods = 10 x 1 = 10 ; Rate = 11 % / 1 = 11%
Table 13-1 Table Factor = 16.7220
Future Value = $2,000 X 16.7220 = $ 33,444
14
Toby Martin invests $2,000 at the end of each year for 10 years at 11% p.a., compounded annually. What is the final value of Toby’s investment at the end of year 10?
2) FV = Pmt [(1 + i )ⁿ - 1]i
= 2000 x {[(1.11¹⁰) -1] / [0.11]}= 2000 x (2.839 – 1)/ (0.11) = $33,444
3) = FV (rate, nper, pmt, pv, type) Excel = FV( 0.11, 10, -2000, 0, 0) = $33,444
15
II. Annuity Due
= Ordinary Annuity x (1 + i)
Tony invests $3000 at the start of each year at 8% p.a., compounded annually. Find its value at the end of three years.
� Future Value: Ordinary Annuity
FV Ordinary = Pmt [ (1 + i )ⁿ - 1]i
= 3000 [(1.08)³ - 1] / [.08] = $9,739.20
� Future Value: Annuity Due
FVDUE = $9,739.20 x (1.08) = $10,518.34
16
Annuity Due
By Table 13-1 of textbook:
� Future Value: Ordinary Annuity
n = 3, i = 8% , table factor = 3.2464
FV = 3000 x 3.2464 = $9,739.20
� Future Value: Annuity Due
Add one period & subtract one payment
n = 4, i = 8% , table factor = 4.5061
FV = [3000 x 4.5061] – 3000 = $10,518.30
17
Annuity Due
� Future Value: Ordinary Annuity
n = 3, i = 8%
Excel: = FV (0.08,3,-3000,0,0) = $9,739.20
� Future Value: Annuity Due
n = 3, i = 8%
Excel: =FV (0.08,3,-3000,0,1) = $10,518.34
18
# 9
• # 13.21. At the beginning of each 6-month
period for 10 years, Merl Agnes invests $500
at 6% p.a., compounded semi-annually. What
would be its cash value at the end of year 10?
1) Formula: FV = 500 [(1.03)²⁰ - 1] (1.03)
0.03
= [500 [0.806111/ 0.03] (1.03)
= $13,838.24
2) = FV ( 0.03, 20, -500, 0, 1 ) = $13,838.24
19
Annuities Due: Future Value
# 13.21. At the beginning of each 6-month period for 10 years, Merl Agnes invests $500 at 6% p.a., compounded semi-annually. What would be its cash value at the end of year 10?
Step 1: Calculate the number of periods and the rate per period. Add one extra period.
Periods = 10 X 2 = 20 , add 1 extra , 20 +1 = 21 ; Rate = 6% / 2 = 3 %
Step 2: Look up in an Ordinary Annuity Table 13.1 the table factor based on the above computed periods and rate. This is the future value of $1.
Table Factor = 28.6765
Step 3: Multiply payment(deposit) each period by the table factor:
$500 X 28.6765 = $14,338.25
Step 4: Subtract one payment from Step3 to get Future Value of Annuity Due
Future Value = $14,338.25 - $500 = $13,838.25
20
II. Ordinary Annuity: Present value
On Joe’s graduation from college, his uncle promised him a gift of $12,000 in cash, or $900 every quarter for the next 4 years after graduation. If money could be invested at 8% p.a., compounded quarterly, which offer is better for Joe?
1) By Table:
Periods = 4 X 4 = 16 Rate = 8% / 4 = 2 %
Table 13 – 2 Factor = 13.5777
PV = $ 900 X 13.5777 = $ 12,219.93
Annuity is better than $ 12,000 cash
21
PV Ordinary Annuity
On Joe’s graduation from college, Joe’s uncle promised him a gift of $12,000 in cash, or $900 every quarter for the next 4 years after graduation. If money could be invested at 8% p.a., compounded quarterly, which offer is better for Joe?
2) Formula: PV = Pmt [1 – (1 + i )⁻ⁿ]i
= 900 x [1 – (1.02)⁻⁴*⁴]/ [0.02]= $12,219.94
3) Excel: = PV ( 0.02, 16, 900, 0,0)= - $12,219.94
22
Sinking Fund – annuity where the equal
periodic payments are determined
Jeff Associates plans to setup a sinking fund to repay $30,000 at the end of 8 years. Assume an interest rate of 12% p.a., compounded semi-annually.
1) By Table 13-3:
n = 8 x 2 = 16; rate = 12% / 2 = 6%
Table 13 – 3 factor = 0.0390
Sinking Fund = $ 30,000 X .0390
= $1,170
23
Sinking Fund
Jeff Associates plans to setup a sinking fund to repay $30,000 at the end of 8 years. Assume an interest rate of 12% p.a., compounded semi-annually.
� Formula: Pmt = FV [ i ]
[ (1 + i)ⁿ - 1 ]
= 30000 * {.06/[(1.06)⁸*² - 1]}
= $1,168.56
� Excel: = PMT(rate, nper, pv, fv, type)
Excel: = PMT (0.06, 16, 0, 30000,0) = - $1,168.56
24
Monthly payment & Payoff amountMonthly payment & Payoff amount
• Mr. Joson buys a car for P1M. A down payment of
20% of the car price is paid in cash, with the
balance to be paid in 36 months. The interest rate
is 14.25% p.a., compounded monthly. After 24
months of paying, Mr. Joson would like to know
his payoff amount.
i = 14.25% / 12 = 0.011875 per month
PV = P1M – P200K = P800,000
n = 36 months
• First step: Find monthly payment amount
Monthly Payment
• PV = [Pmt/i ] [ 1 – (1 + i )⁻ⁿ]
Pmt = [i PV ] / [1 – (1 + i)⁻ⁿ]
= [(0.011875)(800,000) = P27,439.34
[ 1 – (1.011875)⁻³⁶ ]
Using Excel: pmt(rate, nper, pv, fv, type)
= pmt (0.011875, 36, -800000, 0, 0)
= P27,439.34
• Second Step: Find the payoff amount
Payoff AmountB Loan Balance after n payments = FV Compounding – FV annuity
B = PV (1 + i ) ⁿ - Pmt [ (1 + i)ⁿ - 1]
i
= 800,000 ( 1.011875)²⁴
- 27,439.34 [ (1.011875)²⁴ - 1]
0.011875
= 1,062,025.08 – 756,820.62
= P305,204.46
Excel: = FV ( rate, nper, pmt, pv, type)
Method 1: = FV(0.011875, 24, 0, -800000, 0)
- FV(0.011875, 24, -27439.34, 0, 0)
Method 2: = FV(0.011875, 24, 27439.34, -800000, 0)
= P305,204.46 amount still to be paid
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