Financial decisions require consideration of projected revenues and expenditure over period of time being considered. Performing a Cost/Benefit Analysis is essential to sound financial decision making. A critical part of a Cost Benefit Analysis is determining the value of money over time.
Future value measures what today's money would be worth at a specified time in the future assuming a certain discount rate
Present value measure what money at a specified period of time in the future would be worth if valued in terms of today's money.
A capital project must provide a return that exceeds a minimum level established by the organization. The minimum level is reflected in a firm's Minimum Attractive Rate of Return (MARR) or hurdle rate. MARR is treated as the interest rate chargedd by the source of the capital. In theory, the MARR is the interest rate that could be received if the funds were invested elsewhere. In practice, it is determined by the top-management and depends on:
the amount of money available for investment
the source and cost of these funds
the number of good project available for investment opportunities
the cost of administering investments
the type of organization
The most used method is the present worth (PW) method. It is a function of i% (found by discounting all cash inflows and outflows to the present time at an interest rate that is generally the MARR. A positive PW for an investment project means that the project is acceptable.
Net Present Worth (NPW) is the future stream of benefits and cost converted into equivalent values today. Program with a positive NPW are generally cost effective and the another way round. NPW is a criterion for deciding whether a project can be justified on economic principles.
PW(i%) = F(1+i)^-k
Example: Which is more financially sound project? Project A produce $5000 in 2006, Project B produces $5200 in 2007... (i assumed 4.5%)
2006 is PV and 2007 is FV.
PV of 2007 = $5200/(1+0.45) = $4976
Hence Project A is chosen.
Example2: Consider a project that has an initial investment of $50000 and that returns $18000 per year for the next four years. If the MARR is 12%, is this a good investment?
PW = -50000 + 18000(P/A, 12%, 4) = $4671.40
This is a good investment.
The commercial value of a bond is the PW of all future net cash flows expected to be received the period dividend [face or par value (Z) times the bond or nominal rate per interest period,r], and the redemption of disposal price (C), all discounted to the present at the bond;s ield rate per period or effective interest rate, i%.
Vn = C(P/F, i%, N) + rZ(P/A, i%, N)
Example: What is the current value (PW) of a 6% bond rate, 10-year bond with a par (and redemption) value of $20000 that pays dividends semi-annually, if the purchaser wishes to earn an 8% return?
N = 10 x 2 = 20 period before redemption (semi-annually)
r = 6%/2 = 3% per period
i = 8%/2 = 4% per period
C = Z = $20000
Vn = C(P/F, i%, N) + rZ(P/A, i%, N)
= $20000(P/F,4%,20)+(0.03)$20000(P/A,4%,20) = $17282.18
Capitalized worth (CW) is a special variation of present worth of all revenues or expenses over an infinite length of time. The capitalized worth method is especially useful in problems involving endowments and public projects with indefinite lives.
Capitalized worth calculation
1. Draw a cash-flow diagram showing all nonrecurring costs and at least twoo cucles of all recurring (periodic) costs and receipts.
2. Find the present worth of all nonrecurring amounts.
3. Find the equivalent annual worth through one cycle of recurring amounts and add this to all other uniform amounts occurring in years 1 through infinity. This results in a total equivalent uniform annual worth (AW).
4. Divide AW obtained in step 3 by the interest rate to get its capitalized worth.
5. Ass the CW values obtained in steps 2 and 4.
Example: Construction cost: $2000000, Annual maintenance cost: $50000, renovation cost: $500000 every 15 years, Planning horizon: infinite period, Interest rate: 5%...
P = 2000000 + 50000/0.05 + 500000(A/F, 5%, 15)/0.05 = $3464423
Future worth (FW) method is alternative to PW method. FW is based on the equivalent worth of all cash inflows and outflows at the end of the study period at an interest rate that is generally the MARR. Decision made using FW and PW will be the same.
Example: A $4500 investment in a new conveyor system is projected to improve throughout and increasing revenue by $14000 per year for five years. The conveyor will have an estimated market value of $4000 at the end of five years. Using FW and a MARR of 12%, is this good investment?
FW = -$45000(F/P, 12%, 5 + $14000(F/A, 12%, 5) + $4000 = $13635.70
Hence a good investment.
Example2: A company purchased a store chain for $75 million three years ago. There was a et loss of $10 million at the end of year 1 of ownership. Net cash flow is increasing with an arithmetic gradient of $5 million per year starting the second year, and thi pattern is expected to continue for the foreseeable future. Expected MARR of 25% per year...
The comapany has just been offered $159.5 million to sell the store. Use FW analysis to determine if the MARR will be realized at this selling price...
FW = -75(F/P, 25%, 3) - 10((F/P, 25%, 2) - 5(F/P, 25%, 1) + 159.5
= -8.86million
The MARR of 25% will not be realized if the $159.5million offer is accepted.
If the company continues to own the chain, what selling price must be obtained at he end of 5 years of ownership to make the MARR?
FW = -75(F/P, 25%, 5) - 10(F/A, 25%, 5) + 5(A/G, 25%, 5)(F/A, 25%, 5)
= -$246.81 million
The offer must be for at least $246.81 million to make the MARR.
Annual worth is an equal periodic series of dollar amounts that is equivalent to the cash inflows and outflows, at an interest rate that is generally the MARR.
AW(i%) = Revenue - Expenses - CR(i%)
Annual worth analysis measure an investment worth on annual basis. It helps to seek consistency of report format, determine the unit cost and facillitate the unequal project life comparison. AW is easily understood as the results are reported in $/time period.
Example: Consider a project with $30 annual operating cost and a $5000 inverstment required each 5 years. i = 10%
For 1 cycle, EAC = 3000 + 5000(A/P, 10%, 5) = $4.319/yr
For 2 cycle, EAC = 3000 + 5000[1 + (P/F, 10%, 5)](A/P, 10%,10)
= $4319/yr
Capital Recovery (CR) is the annual equivalent cost of the capital invested. The CR covers loss in value of the asset and interest on invested capital (at the MARR). The CR distributes the initial cost(I) and the salvage value (S) across the life of the asset.
CR(i%) = I(A/P, i%, N) - S(A/F, i%, N)
Example: A project requires an initial investment of $45000, has a salvage value of $12000 after six years, incurs annual expenses of $6000 and provides an annual revenue of $18000. Using a MARR of 10%, determine the AW of this project.
AW(10%) = R - E - CR(10%)
CR(10%) = 45000(A/P, 10%, 6) - 12000(A/F, 10%, 6) = 8777
AW(10%) = 18000 - 6000 - 8777 = $3223
Since the AW is positive, it's a good investment.
Example: Land and building cost: $3500000, Annual upkeep cost: $150000, Property taxes and insurance: 5% of total investment, Study period: 25 years, Salvage value: only land cost can be recovered in full...
Ownership cost: CR(15%) = (3500000-1000000)(A/P, 15%, 25)
+ (1000000)(0.15) = $536749
Annual O&M Cost = 0.05(3500000) + 150000 = 325000
Total Equivalent Annual Cost, AEC(15%) = 536749 + 325000 = 861749
Required Monthly Charge = 861749/(12 x 50 x 0.85) = $1690
Internal Rate of Return (IRR) is the discounted rate that equates the present value of a projected cash inflows to the present value of the project's costs. The discount rate which sets the NPV of all cash flows equal to 0. Helps to determine the YIELD on an investment.
NPV = 0 = initial investment + Cash flow year 1/(1+IRR) + ... so on
Example: An investment of a new machine requires $345000 and the estimated market value of the machine after 6 years is $115000. Annual revenue attributable to the new machine will be $120000. Whereas additional annual expenses will be $22000. Determine the IRR if the corporation ;s MARR is 20%...
PW = 0 = -345000+(120000-22000)(P/A,i%,6)+(120000-22000)(P/A,i%,6)
when i=20%, PW=19413
when i=25%, PW=-25621
hence i=22.16%
External Rate of Return (ERR) takes into account the interest rate, external to a project at which net cash flows generated by a project over its life can be reinvested. This is usually the MARR.
Future value measures what today's money would be worth at a specified time in the future assuming a certain discount rate
Present value measure what money at a specified period of time in the future would be worth if valued in terms of today's money.
A capital project must provide a return that exceeds a minimum level established by the organization. The minimum level is reflected in a firm's Minimum Attractive Rate of Return (MARR) or hurdle rate. MARR is treated as the interest rate chargedd by the source of the capital. In theory, the MARR is the interest rate that could be received if the funds were invested elsewhere. In practice, it is determined by the top-management and depends on:
the amount of money available for investment
the source and cost of these funds
the number of good project available for investment opportunities
the cost of administering investments
the type of organization
The most used method is the present worth (PW) method. It is a function of i% (found by discounting all cash inflows and outflows to the present time at an interest rate that is generally the MARR. A positive PW for an investment project means that the project is acceptable.
Net Present Worth (NPW) is the future stream of benefits and cost converted into equivalent values today. Program with a positive NPW are generally cost effective and the another way round. NPW is a criterion for deciding whether a project can be justified on economic principles.
PW(i%) = F(1+i)^-k
Example: Which is more financially sound project? Project A produce $5000 in 2006, Project B produces $5200 in 2007... (i assumed 4.5%)
2006 is PV and 2007 is FV.
PV of 2007 = $5200/(1+0.45) = $4976
Hence Project A is chosen.
Example2: Consider a project that has an initial investment of $50000 and that returns $18000 per year for the next four years. If the MARR is 12%, is this a good investment?
PW = -50000 + 18000(P/A, 12%, 4) = $4671.40
This is a good investment.
The commercial value of a bond is the PW of all future net cash flows expected to be received the period dividend [face or par value (Z) times the bond or nominal rate per interest period,r], and the redemption of disposal price (C), all discounted to the present at the bond;s ield rate per period or effective interest rate, i%.
Vn = C(P/F, i%, N) + rZ(P/A, i%, N)
Example: What is the current value (PW) of a 6% bond rate, 10-year bond with a par (and redemption) value of $20000 that pays dividends semi-annually, if the purchaser wishes to earn an 8% return?
N = 10 x 2 = 20 period before redemption (semi-annually)
r = 6%/2 = 3% per period
i = 8%/2 = 4% per period
C = Z = $20000
Vn = C(P/F, i%, N) + rZ(P/A, i%, N)
= $20000(P/F,4%,20)+(0.03)$20000(P/A,4%,20) = $17282.18
Capitalized worth (CW) is a special variation of present worth of all revenues or expenses over an infinite length of time. The capitalized worth method is especially useful in problems involving endowments and public projects with indefinite lives.
Capitalized worth calculation
1. Draw a cash-flow diagram showing all nonrecurring costs and at least twoo cucles of all recurring (periodic) costs and receipts.
2. Find the present worth of all nonrecurring amounts.
3. Find the equivalent annual worth through one cycle of recurring amounts and add this to all other uniform amounts occurring in years 1 through infinity. This results in a total equivalent uniform annual worth (AW).
4. Divide AW obtained in step 3 by the interest rate to get its capitalized worth.
5. Ass the CW values obtained in steps 2 and 4.
Example: Construction cost: $2000000, Annual maintenance cost: $50000, renovation cost: $500000 every 15 years, Planning horizon: infinite period, Interest rate: 5%...
P = 2000000 + 50000/0.05 + 500000(A/F, 5%, 15)/0.05 = $3464423
Future worth (FW) method is alternative to PW method. FW is based on the equivalent worth of all cash inflows and outflows at the end of the study period at an interest rate that is generally the MARR. Decision made using FW and PW will be the same.
Example: A $4500 investment in a new conveyor system is projected to improve throughout and increasing revenue by $14000 per year for five years. The conveyor will have an estimated market value of $4000 at the end of five years. Using FW and a MARR of 12%, is this good investment?
FW = -$45000(F/P, 12%, 5 + $14000(F/A, 12%, 5) + $4000 = $13635.70
Hence a good investment.
Example2: A company purchased a store chain for $75 million three years ago. There was a et loss of $10 million at the end of year 1 of ownership. Net cash flow is increasing with an arithmetic gradient of $5 million per year starting the second year, and thi pattern is expected to continue for the foreseeable future. Expected MARR of 25% per year...
The comapany has just been offered $159.5 million to sell the store. Use FW analysis to determine if the MARR will be realized at this selling price...
FW = -75(F/P, 25%, 3) - 10((F/P, 25%, 2) - 5(F/P, 25%, 1) + 159.5
= -8.86million
The MARR of 25% will not be realized if the $159.5million offer is accepted.
If the company continues to own the chain, what selling price must be obtained at he end of 5 years of ownership to make the MARR?
FW = -75(F/P, 25%, 5) - 10(F/A, 25%, 5) + 5(A/G, 25%, 5)(F/A, 25%, 5)
= -$246.81 million
The offer must be for at least $246.81 million to make the MARR.
Annual worth is an equal periodic series of dollar amounts that is equivalent to the cash inflows and outflows, at an interest rate that is generally the MARR.
AW(i%) = Revenue - Expenses - CR(i%)
Annual worth analysis measure an investment worth on annual basis. It helps to seek consistency of report format, determine the unit cost and facillitate the unequal project life comparison. AW is easily understood as the results are reported in $/time period.
Example: Consider a project with $30 annual operating cost and a $5000 inverstment required each 5 years. i = 10%
For 1 cycle, EAC = 3000 + 5000(A/P, 10%, 5) = $4.319/yr
For 2 cycle, EAC = 3000 + 5000[1 + (P/F, 10%, 5)](A/P, 10%,10)
= $4319/yr
Capital Recovery (CR) is the annual equivalent cost of the capital invested. The CR covers loss in value of the asset and interest on invested capital (at the MARR). The CR distributes the initial cost(I) and the salvage value (S) across the life of the asset.
CR(i%) = I(A/P, i%, N) - S(A/F, i%, N)
Example: A project requires an initial investment of $45000, has a salvage value of $12000 after six years, incurs annual expenses of $6000 and provides an annual revenue of $18000. Using a MARR of 10%, determine the AW of this project.
AW(10%) = R - E - CR(10%)
CR(10%) = 45000(A/P, 10%, 6) - 12000(A/F, 10%, 6) = 8777
AW(10%) = 18000 - 6000 - 8777 = $3223
Since the AW is positive, it's a good investment.
Example: Land and building cost: $3500000, Annual upkeep cost: $150000, Property taxes and insurance: 5% of total investment, Study period: 25 years, Salvage value: only land cost can be recovered in full...
Ownership cost: CR(15%) = (3500000-1000000)(A/P, 15%, 25)
+ (1000000)(0.15) = $536749
Annual O&M Cost = 0.05(3500000) + 150000 = 325000
Total Equivalent Annual Cost, AEC(15%) = 536749 + 325000 = 861749
Required Monthly Charge = 861749/(12 x 50 x 0.85) = $1690
Internal Rate of Return (IRR) is the discounted rate that equates the present value of a projected cash inflows to the present value of the project's costs. The discount rate which sets the NPV of all cash flows equal to 0. Helps to determine the YIELD on an investment.
NPV = 0 = initial investment + Cash flow year 1/(1+IRR) + ... so on
Example: An investment of a new machine requires $345000 and the estimated market value of the machine after 6 years is $115000. Annual revenue attributable to the new machine will be $120000. Whereas additional annual expenses will be $22000. Determine the IRR if the corporation ;s MARR is 20%...
PW = 0 = -345000+(120000-22000)(P/A,i%,6)+(120000-22000)(P/A,i%,6)
when i=20%, PW=19413
when i=25%, PW=-25621
hence i=22.16%
External Rate of Return (ERR) takes into account the interest rate, external to a project at which net cash flows generated by a project over its life can be reinvested. This is usually the MARR.
