Interests on Transactions Chapter 10 13 PV & FV of Annuities
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 1 2 3 4 5 FV = 1000(1.06)⁴ + 1000(1.06)³ + 1000(1.06)² + 1000(1.06) + 1000 = P5,637.09 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) 1 2 3 4 5 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 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 1 2 3 4 5 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
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) 1 2 3 4 5 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
FV = PV ( 1 + i ) ⁿ 1 2 3 4 5 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
Basic 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) PVdiscounting = FV ( 1+ i ) ⁿ single 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 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) PVdiscounting = 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 1 2 3 4 5 = 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 )
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 FV 13
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 = 8x 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 semiannually. 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 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 Amount B 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 Add Timeline Drawing
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