Quiz 2: Factors

Uniform series involving F/A and A/F

Cash flow occurs in consecutive interest periods. Last cash flow occurs in same period as F. The Sinking fund factor (A/F) determines uniform annual series that is equivalent to a given future amount F. Its equation: A = F[i/((1+i)^n-1}] Factor Notation: A = F(A/F,i,n) The Uniform series compound factor (F/A) yields future worth of uniform series. Its equation: F = A[((1+i)^n-1)/i)] Factor Notation: F = A(F/A,i,n)

Uniform Series Involving P/A and A/P

Here the cash flow occurs in consecutive interest periods. Cash Flow amount is same in each interest period. The P/A factor calculates equivalent P value in year 0 for a uniform end-of-period series of A values beginning at end of period 1 & extending for n periods. Its formula: P = A[( (1+i)^n-1 )/(i (1+i)^n)] , i ≠ 0 Factor Notation: P = A(P/A, i, n) The A/P factor calculates equivalent uniform annual worth. Its formula: A = P[(i(1+i)^n/((1+i)^n-1)] Factor Notation: A = P(A/P,i,n)

Single Payment Factors

Single payment compound amount factor (F/P). This gives you the amount after n years: F = P(1+i)^n Single payment present worth factor (P/F). This determine present value for a given amount F: P = F[1/(1+i)^n] Terms in ( ) or [ ] are called factors. Values for i and n are in tables. Factors are represented in standard factor notation such as: (F/P, i, n) The Symbol to the left of / is what is being sought, symbols to the right of / is what is given. On spreadsheets: Future Value = FV(i%, n ,, P) Present Value = PV(i%, n,,F)

Factor values for untabulated i or n

There is 3 ways to find factor values for untabulated i or n values: 1. Use formulas 2. Use spreadsheet function with corresponding P, F, or A value set to 1 3. Linearly interpolate in interest tables Find the boundary factor values for the percentage values do not write them as decimals. Leave them as %'s

Unknown Interest Rate and Recovery period n

These problems involve solving for i given n & 2 other values (P,F, or A) usually requiring a trial and error solution or interpolation in interest tables. To solve set up equations with all symbols and solve for i. These problems involve solving for n given i and 2 other values (P,F,or A). Solve them as you would solve unknown interest rate problems.

Geometric Gradients

is a cash flow series that increases or decreases by a constant % each period. The uniform change is called the rate of change. - g starts between periods 1 & 2 - there are no tables for geometric factors Equation given that g ≠ i P_g = A₁[(1-((1+g)/(1+i))^n/(i-g)] A₁ = cash flow before the gradient starts g = rate of change Equation given that g = i P_g = A₁ = (n/(1+i)) Here A₁ is the first deposit before the gradient starts Note: If g is - change signs in front of both g values. Looking for the 1st deposit given the nth year deposit? Use this: nth year deposit = A₁(1+g)^(n-1)

Arithmetic Gradients

Is a cash flow series that either increases or decreases by a constant amount each period. The amount of change is called the gradient. Cash flow diagrams for the P_G of an arithmetic gradient: - G starts between 1 & 2, this is because cash flow in yr. 1 is usually not equal to G & is handled separately as a base amount. Standard Factor Notation: P_G = G(P/G,i,n) P_T = P_A ± P_G P_T = Present Worth for an arithmetic gradient P_A = A(P/A, i, n) Where A is the cash flow in the first consecutive payment before the gradient starts P_G = G(P/G, i, n) Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n). The general equation when base amount is involved: A_T = A_A + A_G A_T = (A) or Consecutive amount A_A = Base amount (Amount in year 1) A_G = G(A/G, i, n) For decreasing gradients use - instead of + Note: the value for A is the value of the consecutive payment one year before the gradient starts.