Learning path 4 · Topic 19 of 30

Recover capital and cover operation

Capital recovery is the equivalent annual cost of owning capital, accounting for terminal salvage and the required return on unrecovered investment.

Change a scenario. Watch the consequence.
01234Receipts (+)Expenses (−)

Period 0: −$10,000.00. This is the time-zero cash flow. Periods use the payment interval stated in the model.

equivalent annual cost including operation · $/yearCE=I−S(P/F)=8633.973089,CR=2723.76643,EAC=CR+E=3223.76643CE=I-S(P/F)=8633.973089,\quad CR=2723.76643,\quad EAC=CR+E=3223.76643

Salvage is $2,000.00 received at the end of year 4. EAC is a positive cost magnitude.

Follow the reasoning

Capital recovery accounts for both the return OF capital and the required return ON capital.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

01

What the model means

Let initial investment I, salvage S at N, operating cost E per year. CR=I(A/P)−S(A/F). The identity A/P−A/F=i leads to CR=(I−S)(A/P)+Si, or CR=(I−S)(A/F)+Ii.

02

Build the calculation

Annual cost is EAC=CR+E. With annual revenue R, net annual worth is AW=R−E−CR. A lower EAC is economically preferred only when alternatives deliver equivalent required service over a justified study horizon. The capital equivalent CE is the present equivalent of ownership capital costs, I−S(P/F,i,N). Multiplying CE by A/P yields capital recovery.

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)
Practice with feedback

Set up. Calculate. Explain.

Money: enter dollars

Rebuild the starting scenario

First cost: 10000 $; Salvage fraction: 0.2 fraction of first cost; Annual operating cost: 500 $/year; Interest rate: 10 % per year; Number of years: 4 years. Calculate the equivalent annual cost including operation.

Setup hint

Capital recovery accounts for both the return OF capital and the required return ON capital.

Open the visual solutionCE=I−S(P/F)=8633.973089,CR=2723.76643,EAC=CR+E=3223.76643CE=I-S(P/F)=8633.973089,\quad CR=2723.76643,\quad EAC=CR+E=3223.76643
  1. Capital recovery accounts for both the return OF capital and the required return ON capital.
  2. Annualize the first cost and subtract annualized terminal salvage.
  3. Equivalent identity: CR=(I−S)(A/P)+Si. Add operating cost; a net-revenue AW would subtract this cost.
01234Receipts (+)Expenses (−)

Period 0: −$10,000.00. This is the time-zero cash flow. Periods use the payment interval stated in the model.

equivalent annual cost including operation · $/yearCE=I−S(P/F)=8633.973089,CR=2723.76643,EAC=CR+E=3223.76643CE=I-S(P/F)=8633.973089,\quad CR=2723.76643,\quad EAC=CR+E=3223.76643

Salvage is $2,000.00 received at the end of year 4. EAC is a positive cost magnitude.

Visual solution

Capital recovery accounts for both the return OF capital and the required return ON capital.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.

Money: enter dollars

Transfer the idea to a new scenario

First cost is $18,000, salvage is 20% of first cost at year 6, operation costs $1,200/year, i=8%. Find EAC.

Setup hint

Capital recovery accounts for both the return OF capital and the required return ON capital.

Open the visual solutionCE=I−S(P/F)=15731.389343,CR=3402.941562,EAC=CR+E=4602.941562CE=I-S(P/F)=15731.389343,\quad CR=3402.941562,\quad EAC=CR+E=4602.941562
  1. Capital recovery accounts for both the return OF capital and the required return ON capital.
  2. Annualize the first cost and subtract annualized terminal salvage.
  3. Equivalent identity: CR=(I−S)(A/P)+Si. Add operating cost; a net-revenue AW would subtract this cost.
0123456Receipts (+)Expenses (−)

Period 0: −$18,000.00. This is the time-zero cash flow. Periods use the payment interval stated in the model.

equivalent annual cost including operation · $/yearCE=I−S(P/F)=15731.389343,CR=3402.941562,EAC=CR+E=4602.941562CE=I-S(P/F)=15731.389343,\quad CR=3402.941562,\quad EAC=CR+E=4602.941562

Salvage is $3,600.00 received at the end of year 6. EAC is a positive cost magnitude.

Visual solution

Capital recovery accounts for both the return OF capital and the required return ON capital.

Cash-flow dates and signs are part of the answer. Model values retain precision; displays round money to cents. Illustrative assumptions exclude taxes and inflation unless stated.

Checks allow 0.1% numerical rounding (minimum $0.05 or 0.001 in the stated nonmonetary unit). Interpolated factors and rates are approximations; keep table precision consistent.

Explain the distinction

Why does salvage appear with a minus sign in capital recovery cost?

Can you defend your setup?

Can you place each amount on the timeline, match the interest period, calculate the equivalent quantity, and state the decision or limitation?

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