A company wants to have $20,000 at the beginning of each 6-month period for the next $4\frac{1}{2}$ years. The annuity earns $6.61\%$ compounded semiannually. We need to find how much must be invested now if the annuity is set up for this purpose, recognizing that this is an annuity due.

Applied MathematicsFinanceAnnuityPresent ValueCompound Interest
2025/3/11

1. Problem Description

A company wants to have 20,000atthebeginningofeach6monthperiodforthenext20,000 at the beginning of each 6-month period for the next 4\frac{1}{2}years.Theannuityearns years. The annuity earns 6.61\%$ compounded semiannually. We need to find how much must be invested now if the annuity is set up for this purpose, recognizing that this is an annuity due.

2. Solution Steps

Since the company wants to have $20,000 at the *beginning* of each period, this is an annuity due.
The formula for the present value of an annuity due is:
PV=PMT×1(1+i)ni×(1+i)PV = PMT \times \frac{1 - (1+i)^{-n}}{i} \times (1+i)
Where:
PVPV = Present Value (the amount to be invested now)
PMTPMT = Payment per period ($20,000)
ii = Interest rate per period
nn = Number of periods
First, we need to find the interest rate per period. The annual interest rate is 6.61%6.61\%, and it's compounded semiannually, so the interest rate per period is:
i=0.06612=0.03305i = \frac{0.0661}{2} = 0.03305
Next, we need to find the number of periods. The annuity lasts for 4124\frac{1}{2} years, and payments are made semiannually, so the number of periods is:
n=4.5×2=9n = 4.5 \times 2 = 9
Now we can plug these values into the formula for the present value of an annuity due:
PV=20000×1(1+0.03305)90.03305×(1+0.03305)PV = 20000 \times \frac{1 - (1+0.03305)^{-9}}{0.03305} \times (1+0.03305)
PV=20000×1(1.03305)90.03305×(1.03305)PV = 20000 \times \frac{1 - (1.03305)^{-9}}{0.03305} \times (1.03305)
PV=20000×10.7456330.03305×1.03305PV = 20000 \times \frac{1 - 0.745633}{0.03305} \times 1.03305
PV=20000×0.2543670.03305×1.03305PV = 20000 \times \frac{0.254367}{0.03305} \times 1.03305
PV=20000×7.708517×1.03305PV = 20000 \times 7.708517 \times 1.03305
PV=20000×7.963252PV = 20000 \times 7.963252
PV=159265.04PV = 159265.04
Therefore, the amount that must be invested now is $159265.
0
4.

3. Final Answer

$159265.04

Related problems in "Applied Mathematics"

The problem asks us to find the pressure exerted by a cylinder on the floor. We are given the volume...

PhysicsPressureCylinderVolumeAreaUnits Conversion
2025/4/1

The problem asks us to calculate the pressure exerted by a storage tank on the ground, given that th...

PhysicsPressureAreaForceUnits
2025/4/1

The problem asks for the magnitude and direction of the net force acting on the car. There are four...

PhysicsForcesNet ForceVector Addition
2025/3/31

The problem involves evaluating spreadsheet formulas. Question 30: Given that cell A1 contains the n...

Spreadsheet FormulasFunctionsData AnalysisCOUNTAVERAGEConcatenation
2025/3/31

The problem provides the volume of an acetic acid solution in a bottle ($500 \, \text{cm}^3$), the d...

DensityPercentageUnits ConversionSolution Chemistry
2025/3/30

An object falls from a height towards the ground under gravity. It takes 5 seconds to reach the grou...

PhysicsKinematicsFree FallAccelerationVelocityDistance
2025/3/30

The problem describes the relationship between the amount of time a car is parked and the cost of pa...

FunctionsLinear FunctionsGraphingModeling
2025/3/28

The problem requires us to find the node voltages $V_1$, $V_2$, and $V_3$ in the given circuit using...

Circuit AnalysisNodal AnalysisKirchhoff's Current Law (KCL)Linear EquationsElectrical Engineering
2025/3/27

The problem asks us to find the current $i_o$ in the given circuit using source transformation. The ...

Circuit AnalysisSource TransformationOhm's LawKirchhoff's Laws
2025/3/27

A DC motor takes an armature current of $I_a = 110$ A at a voltage of $V = 480$ V. The armature circ...

Electrical EngineeringDC MotorTorqueBack EMFPhysics
2025/3/27