Comparing alternatives презентация

Alternatives

to select the best alternative out of a set of mutually exclusive alternatives (MEA)
The cash-flow analysis methods (previously described) used in this process:

Present Worth ( PW )
Annual Worth ( AW )
Future Worth ( FW )
Internal Rate of Return ( IRR )
External Rate of Return ( ERR )

most one project out of the group can be chosen, i.e. when one project is chosen all the other ones are excluded
Example: suppose you are shopping for a car. You consider several cars, but will only buy one from a mutually exclusive set of choices
2. Independent
The choice of a project is independent of the choice of any other project in the group, so that all or none of the projects may be selected or some number in between
3. Contingent
The choice of the project is conditional on the choice of one or more other projects

the MARR for each dollar invested

Basic Rule:

Spend the least amount of capital possible unless the extra capital can be justified by the extra savings or benefits


In other words, any increment of capital spent (above the minimum) must be able to pay its own way

producing positive cash flows resulting from increased revenues, reduced costs, or both
"Do nothing" (DN) is usually an implicit investment alternative
If ∑positive cash flows > ∑ negative cash flows, then IRR exists
If PW(MARR) ≥0, investment is profitable
if PW(MARR)<0, do nothing (DN) is better

Cost Alternatives have all negative cash flows except for the salvage value (if applicable)
These alternatives represent “must do” situations, and DN is not an option
IRR not defined for cost alternatives. Can you explain why?

Period (or planning horizon): The time interval over which service is needed to fulfill a specified function or the interval over which the alternatives are compared
Useful Life: The period during which an asset is kept in productive operation

We will discuss two cases:
Case 1: Study period = Useful life
Case 2: Study period ≠ Useful life
Fundamental Principle: Compare MEAs over the same period of time

MEA using the EW method:
Compute the equivalent worth of each alternative, using the MARR as the interest rate
For Investment Alternatives: Select the alternative having the greatest equivalent worth
Note: If all equivalent worths are < 0 for investment alternatives, then "do nothing" is the best alternative
For Cost Alternatives: Select the alternative having the smallest equivalent cost (the one that is least negative)

All three equivalent worth methods (PW, AW, FW) will identify the same "best" alternative

PW method to select the best alternative

Nothing) (12%) = 0
PWI (12%) = -$10,897
PWII (12%) = +$28.241
PWIII (12%) = +$23,672
PWIV (12%) = +$20,923
Select Alternative ___ to maximize PW.

5.6502

0.3220

the best alternative

7) - $7,400 = -$26,155
AW B = = -$25,947
AW C = = -$25,781

Assuming one must be chosen (i.e., DN is not an option), select alternative C to minimize annual equivalent costs

0.2191

MARR = 20%, would you rather have A or B if comparable risk is involved?
If MARR = 20%, PW(A) = $733 and PW(B )= $2,500
Never simply maximize the IRR!
Never simply select the MEA that maximizes the IRR! Do not maximize the rate of return. Look at the increment
Never compare the IRR to anything except the MARR!
IRR(A→B): PW(B-A) = 0 = -9,900 + 14,000(P/F, i'%, 1)
9,900/14,000 = (P/F, i'%, 1)
i' = 41.4% > MARR . Select B

year
Use the Incremental IRR procedure to choose the best alternative.

– 0 =-$16,000
∆ Annual receipts = $3,300 – 0 = $3,300
∆ Salvage value = 0 – 0 = 0
Compute ∆ IRR(2-DN)
PW(∆i') = 0 = -$16,000 + $3,300(P/A, ∆i'%, 10)
i‘(2-DN) ≈ 15.9%

Step 3. Since ∆i' > MARR, keep alt. 2 (higher FC) as current best alternative. Drop DN from further consideration

Rank in order alternatives from low capital investment to high capital investment

∆ cash flow

-$23,500 – (-16,000) =-$7,500
∆ Annual receipts = $ 4,800 - 3,300 = $1,500
∆ Salvage value = $500 – 0 = $500
Compute ∆ IRR(3-2)

PW(∆ i') = 0 = -$7,500 + $1,500(P/A, ∆i'%, 10) + $500(P/F, ∆i'%, 10)
i‘(3-2) ≈ 15.5%
Since ∆i' > MARR, keep alt. 3 (higher FC) as current best alternative
Drop alt. 2 from further consideration

∆ cash flow

= -$28,000 – (-23,500) =-$4,500
∆ Annual receipts = $ 5,500 - 4,800 = $700
∆ Salvage value = $1,500 -500 = $1,000
Compute ∆ IRR(1-3)
PW(∆ i') = 0 = -$4,500 + $700(P|A, ∆i'%, 10) + $1,000(P/F, ∆i'%, 10)
∆ i‘(1-3) ≈ 10.9%
Since ∆i' < MARR, keep alt. 3 (lower FC) as current best alternative. Drop alt. 1 from further consideration
Step 5. All alternatives have been considered
Recommend alternative 3 for investment

∆ cash flow

lives have been the same length. The study period is frequently taken to be a common multiple of the alternatives’ lives when study period ≠ useful life
Repeatibility Assumption
Conditions:
Study period is either indefinitely long or equal to a common multiple of the lives of the alternative
The cash flows associated with an alternative's initial life span are representative of what will happen in succeeding life spans

Actual situations in engineering practice seldom meet these two conditions, which limits the use of the repeatability assumption

= 15% per year, which alternative is preferred based on the repeatability assumption?

Hint: first draw a cash flow diagram

each Alternative

AW(A )= -14,000(A/P, 15, 5)- 14,000 + 8,000(A/F, 15, 5) = -16,990
Life 1: AW (1-5) = -16,990
Life 2: AW (6-10) = -14,000(A/P, 15, 5)- 14,000 + 8,000(A/F, 15, 5) = -16,990
Life 3: AW (11-15 )= -16,990
Life 4: AW (16-20) = -16,990
AW(B)= -65,000(A/P, 15, 20)- 9,000 + 13,000(A/F, 15, 20) = -19.262
AW(A )> AW(B)
Shortcut: If the study period equals a common multiple of the alternatives' lives, simply compare AW computed over the respective useful lives (assuming repeatability is valid)

= -14,000 - 14,000(P/A,15,20) + 8,000(P/F, 15, 20) - 6,000(P/F, 15, 5) -6,000(P/F, 15, 10) -6,000(P/F, 15, 15) = -$106,345

PW (B) = -65,000 - 9,000(P/A, 15, 20) + 13,000(P/F, 15, 20) = -$120,539

Select A to minimize costs
Also,
AW(A) = -$106,345(A/P, 15, 20) = -$16,990
AW(B)= -$120,539(A/P, 15, 20) = -$19,262

the alternatives‘ lives or repeatability is not applicable?

An appropriate study period need to be selected (coterminated assumption)

The cash flows of the alternatives need to be adjusted so that all the alternatives are compared over the same study period
Case 1: Suppose study period > useful life

Cost alternatives: Since each cost alternative has to provide the same level of service over the study period, contracting for the service or leasing the needed equipment for the remaining years may be appropriate

Investment alternatives: Assume all cash flows will be reinvested at the MARR to the end of the study period (i.e., calculate FW at end of useful life and move this to the end of the study period using the MARR). PW can also be used.
See Examples 5-7 and 5-8, page 210

explicitly stated to be shorter than the useful life, use the Cotermination assumption

Procedure: The cash flows of the alternatives need to be adjusted to terminate at the end of the study period
Truncate the alternative at the end of the study period using an estimated Market Value

life of B = 40 years > study period
Assume Market Value (B) @ EOY20 = $50,000
The MARR is 10% per year.
Which boiler do you recommend?

= - 6,000 –[120,000(A/P, 10, 20) -50,000(A/F, 10, 20)] = -19,225

Still select boiler A to minimize costs

What would the market value of Boiler B @ EOY 20 have to be in order to select Boiler B instead of A?

?

Set AW(A) = AW(B) , let X be the unknown market value

-14,700 = -6,000 - {120,000(A/P, 10, 20) - X(A/F,10, 20)}
-8,700 = -{14,100 - 0.0175X}
5,400 = 0.0175X
X = $308,571 therefore, MV(B) > $308,571 to favor B

Such a value is very unlikely because X is more than the initial cost of Boiler B