## How to Calculate and Solve for Annual Worth | Gradient Series | Economic Equivalence

The image above represents annual worth.

To compute for annual worth, three essential parameters are needed and these parameters are Gradient Amount (G), Interest Rate (i) and Number of Years (N).

The formula for calculating annual worth:

A = G[((1 + i)N) – iN – 1] / [((1 + i)N) – 1]

Where:

A = Annual Amount or Worth (Conversion Factor)
i = Interest Rate
N = Number of Years

Let’s solve an example;
Find the annual worth when the gradient amount is 11, the interest rate is 0.2 and the number of years is 12.

This implies that;

G = Gradient amount = 11
i = Interest Rate = 0.2
N = Number of Years = 12

A = G[((1 + i)N) – iN – 1] / [((1 + i)N) – 1]
A = 11[((1 + 0.2)12) – 0.2(12) – 1] / [((1 + 0.2)12) – 1]
A = 11[((1.2)12) – 2.40 – 1] / [((1.2)12) – 1]
A = 11[8.916 – 2.40 – 1] / [8.916 – 1]
A = 11[5.516] / [7.916]
A = 11 x 0.696
A = 7.66

Therefore, the annual worth is ₦7.66.

## How to Calculate and Solve for Present Worth | Geometric Gradient | Economic Equivalence

The image above represents present worth.

To compute for present worth, four essential parameters are needed and these parameters are First Payment (A1), Interest Rate (i), Number of Year(s) (N) and Percentage change in payment (g).

The formula for calculating present worth:

P = A₁[1 – ((1 + g)N)((1 + i)-N)] / [i – g]

Where:

P = Present Worth or Amount
A₁ = First Payment
i = Interest Rate
g = Percentage change in payment
N = Number of Years

Let’s solve an example;
Find the present worth when the first payment is 10, the interest rate is 0.2, the percentage change in payment is 20 and the number of years is 5.

This implies that;

A₁ = First Payment = 10
i = Interest Rate = 0.2
g = Percentage change in payment = 0.1
N = Number of Years = 5

P = A₁[1 – ((1 + g)N)((1 + i)-N)] / [i – g]
P = 10[1 – ((1 + 0.1)5)((1 + i)-(5))] / [0.2 – 0.1]
P = 10[1 – ((1.1)5)((1.2)-5)] / [0.1]
P = 10[1 – (1.61051)(0.401)] / [0.1]
P = 10[1 – 0.6472278485082308] / [0.1]
P = 10[0.3527721514917692] / [0.1]
P = 10 x 3.527721514917692
P = 35.277215149176925

Therefore, the present worth is ₦35.28.

## How to Calculate and Solve for Present Worth | Geometric Gradient | Economic Equivalence

The image above represents present worth.

To compute for present worth, three essential parameters are needed and these parameters are First Payment (A1), Interest Rate (i) and Number of Year(s) (N).

The formula for calculating present worth:

P = A₁[N / (1 + i)]

Where:

P = Present Worth or Amount
A₁ = First Payment
i = Interest Rate
N = Number of Years

Let’s solve an example;
Find the present worth when the first payment is 8, the interest rate is 0.3 and the number of years is 22.

This implies that;

A₁ = First Payment = 8
i = Interest Rate = 0.3
N = Number of Years = 22

P = A₁[N / (1 + i)]
P = 8[22 / (1 + 0.3)]
P = 8[22 / 1.3]
P = 8 x 16.92
P = 135.38

Therefore, the present worth is ₦135.38.

## How to Calculate and Solve for Present Worth | Gradient Series | Economic Equivalence

The image above represents present worth.

To compute for present worth, three essential parameters are needed and these parameters are Gradient Amount (G), Interest Rate (i) and Number of Years (N).

The formula for calculating present worth:

P = G[((1 + i)N) – iN – 1] / [i²((1 + i)N)]

Where:

P = Present Amount or Worth
i = Interest Rate
N – Number of Years

Let’s solve an example;
Find the present worth when the gradient amount is 22, the interest rate is 0.2 and the number of years is 2.

This implies that;

G = Gradient Amount = 22
i = Interest Rate = 0.2
N – Number of Years = 2

P = G[((1 + i)N) – iN – 1] / [i²((1 + i)N)]
P = 22[((1 + 0.2)2) – 0.2(2) – 1] / [0.2²((1 + 0.2)2)]
P = 22[((1.2)2) – 0.4 – 1] / [0.0400000000000001((1.2)2)]
P = 22[1.44 – 0.4 – 1] / [0.04000000000000001 x 1.44]
P = 22[0.040000000000000036] / [0.05760000000000001]
P = 22 x 0.6944444444444449
P = 15.27

Therefore, the present worth is ₦15.27.