Other Analysis Techniques, Slides of Economic Analysis

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Other Analysis Techniques

 Future Worth Analysis (FWA)

 Benefit-Cost Ratio Analysis (BCRA)

 Payback Period

Techniques for Cash Flow Analysis

 Present Worth Analysis

 Annual Cash Flow Analysis

 Rate of Return Analysis

 Incremental Analysis

 Other Techniques:

 Future Worth Analysis

 Benefit-Cost Ration Analysis

 Payback Period Analysis

 Chapter 5

 Chapter 6

 Chapter 7

 Chapter 8

 Chapter 9

Example 9-

Ron Jamison, a 20-year-old college student, consumes about a carton of

cigarettes a week. He wonders how much money h~ could accumulate by

age 65 if he quit smoking now and put his cigarette money into a savings

account. Cigarettes cost $35 per carton. Ron expects that a savings

account would earn 5% interest, compounded semiannually. Compute

Ron's future worth at age 65.

Semiannual saving $35/carton x 26 weeks = $

Future worth (FW)= A(FjA,2.5%, 90) = 910(329.2) = $ 299,

Benefit-cost ratio comparison criteria

Situation Criterion

Fixed input Amount of^ money or other input

resources are fixed

Maximize B/C

Fixed output Fixed benefit^ Maximize B/C

Neither input nor

output fixed

Neither amount of money or other

inputs not amount of benefits or

other outputs are fixed

Compute incremental benefit-

cost ratio (∆B/∆C) on the

increment of investment

between the alternatives. If

∆B/∆C ≥ 1, choose the higher-

cost alternative; otherwise,

choose the lower-cost

alternative.

Example 9-

Year Device A Device B

We have solved this problem using the present worth analysis, the

annual cash flow analysis, and the rate of return analysis. Now, we

will use the Benefit-Cost Ratio analysis.

Since this is a fixed cost case, the criterion is to maximize the B/C

ratio.

Using an interest rate of 7%, choose the best alternative using the

benefit-cost ratio analysis

Example 9-

Machine X Machine Y

Initial cost $200 $

Uniform annual benefit 95 120

End of Useful live salvage value 50 150

Useful life (years) 6 12

Using an interest rate of 10%, choose the best machine using the

benefit-cost ratio analysis

D B A C E F

Cost (^) $1000 $2000 $4000 $6000 $9000 $ PWB (^) $1340 $4700 $7330 $8730 $9000 $ B/C 1.34 2.35 1.83 1.46 1.00 0. NPV 340 2700 3330 1730 0 - …Example 9-

4) Comparing B/C with 1 for consecutive alternatives select the best

alternative.

Increment B-D is attractive; therefore B is preferred to D.

Increment A-B is attractive; therefore B is preferred to A.

Increment C-A is not attractive, as B/C = 0.76 < 1 ; therefore A is preferred to C.

Increment E-A is not attractive, as B/C = 0.33 < 1 ; therefore A is preferred to E.

►►Finally A is the best project.

B-D A-B C - A E-A

Incremental Cost $1000 $2000 $2000 $ Incremental Benefit $3360 $2630 $1400 $ B/C 3.36 1.32 0.76 0. Comparison result B preferred

A

preferred

A

preferred

A

preferred

 A, B, C, and D are above the 45-degree line; their B/C ratio is > 1.  F is below the line: B/C ratio is < 1.  We can discard F if we wish. Benefit/Cost Ratio Analysis – Graphical representation PWB/PWC = 1 PWB PWC D B A C E F B-D A-B C - A E-A Incremental Cost $1000 $2000 $2000 $ Incremental Benefit $3360 $2630 $1400 $ Incr.B/Incr. C 3.36 1.32 0.76 0.

Examine each separable increment of

investment.

B/C < 1  increment is not attractive

B/C  1  increment is desirable.

Compare D to B: Slope or  B/  C > 1. ►►B preferred Compare B to A: Slope or  B/  C > 1. ►►A preferred Compare A to C: Slope or  B/  C < 1; discard C. Compare A to E: Slope or  B/  C < 1; discard E. F was discarded earlier since its B/C is less than 1 Therefore A is the best alternative. Note: Despite the fact that Alt. B has the highest B/C ratio, we should not make the mistake of choosing B since the economic criterion here should be based on  B/  C (since neither input nor output is fixed)