ECE 192 - Cash Flow Analysis

single cash flow definition/examples =

non-recurring costs:

• purchasing equipment, salvage value, investments

compound amount factor =

(F/P, i, N) = (1 + i)^N

gives the future amount F given:

• P = present amount

• i = interest rate

• N = # periods to go to F

present worth factor =

P(F, i, N) = 1/(1 + i)^N

gives the present amount P given:

• F = future amount

• i = interest rate

• N = # periods to go BACK to P

compound amount factor is the inverse of =

present worth factor

annuity definition/examples =

uniform series of cash flows over N periods (starts at end of 1st period)

• receipts (money in)

  • rent income

  • stock dividends

  • pension

• disbursements (money out)

  • insurance premiums

  • mortgages

  • operations and maintenance costs

NOTE: receipt/disbursements represent a SINGLE payment out of the ANNUITY

uniform series compound amount factor =

(F/A, i, N) = [(1 + i)^N - 1]/i

gives the future value F for an annuity given:

• A = amount of a single payment

• i = interest rate

• N = # of periods

sinking fund factor

(A/F, i, N) = i/[(1 + i)^N - 1]

gives the size of the receipt/disbursement A that is equivalent to F given:

• F = future amount

• i = interest rate

• N = # of periods

uniform series compound factor is the inverse of =

sinking fund factor

capital recovery factor =

(A/P, i, N) = [i(1 + i)^N] / [(1 + i)^N - 1]

gives the size of receipt/disbursement A equivalent to P given:

• P = present amount

• i = interest rate

• N = # of periods

“What N payments of A is worth the same as a lump sum of P today, given i% interest yearly?”

series present worth factor =

(P/A, i, N) = [(1 + i)^N - 1] / [i(1 + i)^N]

gives the present amount P equivalent to A given:

• A = size of receipt/disbursement

• i = interest rate

• N = # periods

“What lump sum P today is worth the same as N payments of A, given i% interest yearly?”

capital recovery factor is the inverse of =

series present worth factor

bisection and linear approximation method to find i_s =

get function for f(i_s) = P(i_s) - P

bisection method

• find two intervals that have f(i_s1) > 0 and f(i_s2) < 0 respectively

• evaluate f([i_s2 + i_s1]/ 2)

  • if > 0, then lower bound is [i_s2 + i_s1]/ 2

  • if < 0, then upper bound is [i_s2 + i_s1]/ 2

• repeat until f(i_s) ≈ 0

linear approximation method:

• x* = x₁ + (x₂ − x₁) × [y* − y₁] / [y₂ − y₁]

  • x = Compound interest rate i_s

  • y = f(x) = f(i_s)

  • y* = 0

can use bisection → linear approximation to save time

arithmetic gradient series =

series of constantly increasing receipts/disbursements

• end of 1st period = 0

arithmetic gradient present value factor =

(P/G, i, N) = [(1+i)^N − iN − 1] / [i²(1+i)^N]

gives the present amount P equivalent to constantly increasing G per period given:

• G = constant increase in receipt/disbursement

• i = interest rate

• N = # of periods

“What lump sum P is worth the same as a series of payments that starts at 0 in period 1 and increases by a constant amount G each period after, given interest rate i per period over N periods?"

arithmetic gradient uniform series factor =

(A/G, i, N) = [1/i] − [N/((1+i)^N − 1)]

gives the annuity A equivalent to arithmetic gradient series G given:

• G = constant increase in receipt/disbursements

• i = interest rate

• N = # of periods

“What annuity A is worth the same as a series of payments that start at 0 in period 1 and increases by a constant G each period after, given interest rate i per period over N periods?”

combined annuity =

A_T = A' + G[1/i − N/((1+i)^N − 1)]

gives the annuity A_T equivalent to an annuity A’ and arithmetic gradient series G given:

• A’ = annuity

• G = constant increase in receipts/disbursements

• i = interest rate

• N = # of periods

“What total annuity is worth the same as the sum of an annuity and a series of payments that start at 0 and increase by a constant G each period after, given interest rate i and per period over N periods?”

geometric gradient series =

series of cash flows that increase/decrease by a % each period

• ex.) inflation/deflation, productivity improvement

base value = A

rate of growth = g

growth adjusted interest rate =

i° = (1+i)/(1+g) − 1

interest rate with growth rate accounted for:

• can be used to convert to annuity problem

geometric gradient → present worth factor =

i° = (1+i)/(1+g) − 1

(P/A, g, i, N) = [(P/A, i°, N)] / (1+g) = [((1+i°)^N − 1) / (i°(1+i°)^N)] × [1/(1+g)]

gives the present amount P equivalent to a geometric gradient series g given:

• A = base value of geometric gradient series

• g = rate of growth

• i = interest rate

• N = # of periods

“What lump sum P is worth the same as a geometric gradient series with interest rate i and per period over N periods?”

what happens if compounding and payment periods dont match =

non-standard annuities/gradients

• ex.)

  • interest compounds monthly, but you only make payments quarterly

  • interest compounds daily, but you make payments annually

3 methods to solve non-standard annuities/gradients =

• treat each cash flow in the annuity/gradient individually

• convert to standard annuity/gradient by changing compounding period

• convert to standard annuity/gradient by finding equivalent standard annuity/gradient for the compounding period