riaology
Formula & reasoning reference

A compact map.
Not a pile of formulas.

Build a useful reference around decisions, dates, and assumptions. Connect each formula to the cash-flow pattern it describes, then return to a live model whenever the setup is unclear.

Put these five decisions at the top.

  1. Whose cash flow? Draw + receipts and − expenses; place time 0.
  2. What are the payment periods? Match effective i and count N.
  3. What pattern? Single sum, uniform series, deferred series, arithmetic or geometric gradient.
  4. What comparison? Independent acceptance, exclusive choice, incremental upgrade, or constrained portfolio.
  5. What horizon? Equal life, repeatability/common multiple, or specified study period with terminal values.

A factor table you can interrogate.

Study how factors change with rate and period count. Compare the factor values with their formulas and notice how the rate and timing change the result.

F/P1.6105
P/F0.6209
F/A6.1051
P/A3.7908
A/P0.2638
A/F0.1638

All six factors are dimensionless. At zero interest, series factors become N or 1/N. Multiply by the given amount only after choosing the requested/given direction.

Formula and trap index

Make a defensible engineering decision

ΔNB=(BB−BA)−(CB−CA)\Delta NB=(B_B-B_A)-(C_B-C_A)

A benefit-cost difference is not a benefit-cost ratio. A cost common to both designs cancels in the comparison, but a cost that occurs only under one design does not.

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Sort costs before calculating

C(q)=F+vq,Cˉ=Fq+v,MC=vC(q)=F+vq,\qquad \bar C=\frac{F}{q}+v,\qquad MC=v

Total variable cost is vq, not v. At the starting example, $10,000 + $0.60 × 70,000 = $52,000, not $42,000.

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Find where costs and revenue meet

qBE=Fp−v,qπmax⁡=a−v2bq_{BE}=\frac{F}{p-v},\qquad q_{\pi\max}=\frac{a-v}{2b}

A crossover outside the feasible demand range does not justify changing machines. Dominance means one alternative is at least as good in every relevant criterion and strictly better in at least one.

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Estimate a cost model from observations

v^=n∑qkCk−(∑qk)(∑Ck)n∑qk2−(∑qk)2,F^=Cˉ−v^qˉ\hat v=\frac{n\sum q_kC_k-(\sum q_k)(\sum C_k)}{n\sum q_k^2-(\sum q_k)^2},\qquad \hat F=\bar C-\hat v\bar q

A high R² does not establish causality or accuracy outside the observed range. An estimated negative fixed cost may indicate the linear approximation is only local.

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Scale a plant without scaling every dollar

C2=C1(K2/K1)b,b=ln⁡(C2/C1)ln⁡(K2/K1)C_2=C_1(K_2/K_1)^b,\quad b=\frac{\ln(C_2/C_1)}{\ln(K_2/K_1)}

An exponent below one describes cost scaling, not diminishing production output. Doubling capacity reduces unit-capacity cost, not necessarily total cost.

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Watch interest earn interest

Is=PNi,Fs=P(1+Ni),Fc=P(1+i)NI_s=PNi,\quad F_s=P(1+Ni),\quad F_c=P(1+i)^N

F is principal plus interest. To report interest alone, subtract P. A statement that annual interest is paid out rather than reinvested changes the compounding model.

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Move money to a common date

Vt=Vs(1+i)t−s,FN=P∏k=1N(1+ik),i=(F/P)1/N−1,N=ln⁡(F/P)ln⁡(1+i)V_t=V_s(1+i)^{t-s},\quad F_N=P\prod_{k=1}^N(1+i_k),\quad i=(F/P)^{1/N}-1,\quad N=\frac{\ln(F/P)}{\ln(1+i)}

Never add amounts at different dates before converting. Equivalent P and F represent the same cash flow and should not both be counted as separate receipts.

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Follow the loan balance

A=Pi(1+i)N(1+i)N−1,Bk=Bk−1(1+i)−AA=P\frac{i(1+i)^N}{(1+i)^N-1},\quad B_k=B_{k-1}(1+i)-A

Do not apply interest to the balance after the payment when calculating that period’s interest. Keep full precision and round displayed payments only as instructed.

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Choose the right factor

FP=(1+i)N,  PA=1−(1+i)−Ni,  FA=(1+i)N−1i,  AP=i(1+i)N(1+i)N−1,  AF=i(1+i)N−1\frac FP=(1+i)^N,\;\frac PA=\frac{1-(1+i)^{-N}}i,\;\frac FA=\frac{(1+i)^N-1}i,\;\frac AP=\frac{i(1+i)^N}{(1+i)^N-1},\;\frac AF=\frac{i}{(1+i)^N-1}

A/P is not P/A. A sinking-fund deposit is A=F(A/F), not F(A/P). Count the actual payments, not only a final calendar year.

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Put a deferred annuity on the right date

P0=A(P/A,i,N)(P/F,i,s−1)P_0=A(P/A,i,N)(P/F,i,s-1)

The most common error is discounting by s instead of s−1, or using the final year as N. Draw the first and last payment before choosing factors.

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Separate a base series from a gradient

(P/G)=1i[1−(1+i)−Ni−N(1+i)−N],(A/G)=1i−N(1+i)N−1,(F/G)=(F/A)−Ni(P/G)=\frac1i\left[\frac{1-(1+i)^{-N}}i-N(1+i)^{-N}\right],\quad (A/G)=\frac1i-\frac{N}{(1+i)^N-1},\quad (F/G)=\frac{(F/A)-N}{i}

The standard gradient has no G payment at year 1. If a series is shifted, find its local equivalent and then shift the whole value.

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Distinguish growth from interest

P={A11−[(1+g)/(1+i)]Ni−g,i≠g,NA11+i,i=g.P=\begin{cases}A_1\frac{1-[(1+g)/(1+i)]^N}{i-g},&i\ne g,\\ \frac{NA_1}{1+i},&i=g.\end{cases}

Do not use A(1+g)ᴺ for the year-N payment when A is the year-1 amount. Infinite growth perpetuities require i>g; a finite series does not.

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Read between factor-table entries

f(i)≈fL+i−iLiH−iL(fH−fL)f(i)\approx f_L+\frac{i-i_L}{i_H-i_L}(f_H-f_L)

Do not interpolate rates in a nominal-rate table while using an effective rate in the problem. A wide bracket usually increases approximation error.

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Convert nominal into effective interest

ieff=(1+r/m)m−1,iquarter=(1+imonth)3−1i_{eff}=(1+r/m)^m-1,\quad i_{quarter}=(1+i_{month})^3-1

APR terminology can include regulatory conventions and fees; these study problems use a stated nominal rate with the specified compounding schedule.

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Match interest to payment periods

ip=(1+r/m)m/p−1,Np=pTi_p=(1+r/m)^{m/p}-1,\quad N_p=pT

A monthly rate with an annual payment count is inconsistent. Convert BOTH rate and period count. Cash flows within a posting interval need an explicit interest-crediting assumption.

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Separate continuous interest from deposits

F=PerT,Fseries=AerT−1er/p−1F=P e^{rT},\quad F_{series}=A\frac{e^{rT}-1}{e^{r/p}-1}

Do not divide continuous nominal r by p and treat it as an effective periodic rate. Use exp(r/p)−1; at r=0 use the limit.

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Set an acceptance threshold

PW(MARR)≥0for an acceptable independent investmentPW(MARR)\ge0\quad\text{for an acceptable independent investment}

There is no universal instruction to add borrowing rate and lending rate. MARR is given or established from a clearly specified model; IRR is solved from project cash flows.

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Make PW, FW, and AW tell the same story

PW=∑t=0NCFt(1+i)−t,FW=PW(1+i)N,AW=PW(A/P,i,N)PW=\sum_{t=0}^N CF_t(1+i)^{-t},\quad FW=PW(1+i)^N,\quad AW=PW(A/P,i,N)

Changing the horizon or omitting salvage can change the decision. Equivalent methods agree only when they represent the same underlying cash flows.

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Recover capital and cover operation

CR=I(A/P,i,N)−S(A/F,i,N),AW=R−E−CR,CE=I−S(P/F,i,N),CR=CE(A/P,i,N)CR=I(A/P,i,N)-S(A/F,i,N),\quad AW=R-E-CR,\quad CE=I-S(P/F,i,N),\quad CR=CE(A/P,i,N)

Salvage is a positive terminal receipt, so it reduces ownership cost. Depreciation schedules are not automatically cash-flow schedules.

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Value service that continues forever

CC=I+Ei+I(1+i)N−1,EAC∞=iCCCC=I+\frac{E}{i}+\frac{I}{(1+i)^N-1},\quad EAC_\infty=iCC

Do not place the first replacement at year 0 twice. A salvage value, changed replacement cost, delayed recurring cost, or growth requires modifying the model.

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Find the zero-worth interest rate

0=∑t=0NCFt(1+i∗)t0=\sum_{t=0}^N\frac{CF_t}{(1+i^*)^t}

The highest individual IRR is not necessarily the project with the greatest PW. A small highly efficient investment can create less total value than a larger acceptable one.

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Refine IRR in two interpolation steps

i1=iL+PWLPWL−PWH(iH−iL)i_1=i_L+\frac{PW_L}{PW_L-PW_H}(i_H-i_L)

The bracket must straddle zero. Interpolating two negative PW values is not a bracketed root calculation. Use rates consistently as percentages or decimals.

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Spot an ambiguous rate of return

0=−1000(1+i)2+2300(1+i)−13200=-1000(1+i)^2+2300(1+i)-1320

More than one sign change is a warning, not proof of exactly two positive IRRs. A mathematically real rate below −100% is outside the usual financial domain.

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Make the external rate explicit

PE=∑CFt<0−CFt(1+ϵ)t,FR=∑CFt>0CFt(1+ϵ)N−t,ERR=(FR/PE)1/N−1P_E=\sum_{CF_t<0}\frac{-CF_t}{(1+\epsilon)^t},\quad F_R=\sum_{CF_t>0}CF_t(1+\epsilon)^{N-t},\quad ERR=(F_R/P_E)^{1/N}-1

Do not accumulate expenses and receipts to inconsistent dates. Negative net flows use positive magnitudes in the expense bucket; the original cash-flow diagram retains their negative signs.

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Track recovery without confusing it with profit

∑t=0kCFt≥0  (simple),∑t=0kCFt(1+i)t≥0  (discounted)\sum_{t=0}^{k}CF_t\ge0\;\text{(simple)},\qquad \sum_{t=0}^{k}\frac{CF_t}{(1+i)^t}\ge0\;\text{(discounted)}

Both methods ignore value after the selected recovery date. A short payback can conflict with PW; discounted payback may never occur within the life.

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Compare dollars, returns, and liquidity carefully

Sk=∑jwjsjk,wj≥0,∑jwj=1S_k=\sum_j w_js_{jk},\quad w_j\ge0,\quad\sum_jw_j=1

Do not add raw dollar PW, percentage IRR, and years of payback. A scoring model should make the preference tradeoffs visible.

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Choose among alternatives, not only projects

Optional selection: max⁡{0,PWA,PWB,…}\text{Optional selection: }\max\{0,PW_A,PW_B,\ldots\}

Comparing investment size or individual IRR alone can mis-rank alternatives. If neither optional project has nonnegative PW, doing nothing is economically preferred.

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Ask whether the extra investment earns its keep

ΔCFt=CFB,t−CFA,t,0=∑tΔCFt(1+Δi∗)−t\Delta CF_t=CF_{B,t}-CF_{A,t},\quad 0=\sum_t\Delta CF_t(1+\Delta i^*)^{-t}

Never subtract initial cost B−A but later receipts A−B. Individual IRRs cannot replace the incremental test. Compare cash flows over a justified common service horizon.

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Give unequal lives a fair comparison

EAC=I(A/P,i,N)+E−S(A/F,i,N),PWcommon=EAC(P/A,i,H)EAC=I(A/P,i,N)+E-S(A/F,i,N),\qquad PW_{common}=EAC(P/A,i,H)

Annual-worth comparison of different lives requires repeatability or an explicitly equivalent service model. A longer life is not automatically preferable.

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Choose a feasible portfolio

max⁡∑jPWjxjsubject to budget and relationship constraints\max\sum_j PW_jx_j\quad\text{subject to budget and relationship constraints}

A large project’s standalone return does not tell you which feasible package maximizes worth. Treat dependencies as constraints, not an afterthought.

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Final reasonableness checks

Discounting a positive future receipt at a positive rate reduces its present equivalent. For a conventional investment, increasing MARR lowers PW. Ordinary annuity P sits one period before the first A. PW, FW, and AW signs agree on a common basis. Salvage reduces ownership cost. Higher cost requires justified incremental benefit. A short payback does not prove a good investment.