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CHAPTER 4. ECONOMIC EVALUATION OF ALTERNATIVES. TOPICS IN CHAPTER 4. BASES FOR COMPARISON OF ALTERNATIVES PRESENT WORTH AMOUNT CAPITALISED EQUIVALENT AMOUNT ANNUAL EQUIVALENT AMOUNT FUTURE WORTH AMOUNT CAPITAL RECOVERY WITH RETURN RATE OF RETURN APPROACH INCREMENTAL APPROACH.

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CHAPTER 4

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Chapter 4

CHAPTER 4

ECONOMIC EVALUATION OF ALTERNATIVES


Topics in chapter 4

TOPICS IN CHAPTER 4

  • BASES FOR COMPARISON OF ALTERNATIVES

  • PRESENT WORTH AMOUNT

  • CAPITALISED EQUIVALENT AMOUNT

  • ANNUAL EQUIVALENT AMOUNT

  • FUTURE WORTH AMOUNT

  • CAPITAL RECOVERY WITH RETURN

  • RATE OF RETURN APPROACH

  • INCREMENTAL APPROACH


Objectives

OBJECTIVES

  • TO SELECT THE BEST ALTERNATIVE ECONOMICALLY

  • UNDERSTAND THE VARIOUS BASES

  • KNOW HOW REGARDING USAGE OF VARIOUS METHODS


Present worth amount method

PRESENT WORTH AMOUNT METHOD

  • THE PRESENT WORTH IS A NET EQUIVALENT AMOUNT AT THE PRESENT THAT REPRESENTS THE DIFFERENCE BETWEEN EQUIVALENT DISBURSEMENTS AND EQUIVALENT RECEIPTS OF AN INVESTMENT CASH FLOW FOR A SELECTED INTEREST RATE

  • IT CONSIDERS THE TIME VALUE OF MONEY

  • IN A COST DOMINATED CASH FLOW DIAGRAM THE COST WILL BE ASSIGNED WITH POSITIVE SIGN AND THE PROFIT ,REVENUE WILL BE ASSIGNED WITH NEGATIVE SIGN

  • IN A REVENUE DOMINATED CASH FLOW DIAGRAM THE PROFIT ,REVENUE WILL BE ASSIGNED WITH POSITIVE SIGN AND CASH OUTFLOWS WITH NEGATIVE SIGN


Present worth amount method1

PRESENT WORTH AMOUNT METHOD

  • IN REVENUE DOMINATED CASH FLOW DIAGRAM THE ALTERNATIVE WITH THE MAXIMUM PRESENT WORTH AMOUNT SHOULD BE SELECTED AS THE BEST ALTERNATIVE

  • IN COST DOMINATED CASH FLOW DIAGRAM THE ALTERNATIVE WITH MINIMUM PRESENT WORTH SHOULD BE SELECTED AS THE BEST ALTERNATIVE.

  • WE HAVE TWO SITUATIONS EQUAL LIVED ALTERNATIVES & UNEQUAL LIVED ALTERNATIVES


Problem 1

PROBLEM 1

  • A COMPANY IS CONSIDERING A PLANT TO MANUFACTURE A PARTICULAR PRODUCT. THE LAND COSTS RS 3,00,000 AND THE BUILDING COSTS RS 6,00,000 THE EQUIPMENT COSTS RS 2,50,000 AND RS 1,00,000 WORKING CAPITAL IS REQUIRED. IT IS EXPECTED THAT THE PRODUCT WILL RESULT IN SALES OF RS 7,50,000 PER YEAR FOR TEN YEARS AT WHICH TIME THE LAND CAN BE SOLD FOR RS 4,00,000 THE BUILDING FOR RS 3,50,000 THE EQUIPMENT FOR RS 50,000 AND ALL OF THE WORKING CAPITAL IS RECOVERED. THE ANNUAL OUT OF POCKET EXPENSES FOR LABOUR,MATERIALS AND ALL OTHER ITEMS ARE ESTIMATED TO TOTAL OF RS 4,75,000. IF THE INTEREST RATE IS 25 % PER YEAR DETERMINE WHETHER THE COMPANY SHOULD INVEST IN THE NEW PRODUCT LINE . USE PRESENT WORTH METHOD.


Problem 2

PROBLEM 2

  • TWO MACHINES ARE UNDER CONSIDERATION BY A METAL FABRICATING COMPANY. MACHINE “A” WILL HAVE A FIRST COST OF RS 15,000 AN ANUUAL MAINTENANCE AND OPERATION COST OF RS 3000 AND RS 3000 SALVAGE VALUE. MACHINE “B” WILL HAVE A FIRST COST OF RS 22,000 AN ANNUAL COST OF RS 1500 AND A RS 5000 SALVAGE VALUE. IF BOTH MACHINES ARE EXPECTED TO LAST FOR 10 YEARS DETERMINE WHICH MACHINE SHOULD BE SELECTED ON THE BASIS OF PRESENT WORTH METHOD USING AN INTEREST RATE OF 12 % PER YEAR.


Problem 3

PROBLEM 3

  • A PLANT MANAGER IS TRYING TO DECIDE BETWEEN THE MACHINES DETAILED BELOW : TAKE I = 15%

    MACHINE “A”

    FIRST COST - RS 49,500

    ANNUAL OPERATING COST – RS 15,750

    SALVAGE VALUE - RS 4500

    LIFE - 6 YEARS

    MACHINE “B”

    FIRST COST - RS 63,000

    ANNUAL OPERATING COST – RS 13,950

    SALVAGE VALUE - RS 9000

    LIFE - 9 YEARS


Problem 4

PROBLEM 4

  • THE FOLLOWING DATA REFERS TO THE CASH FLOWS OF FIVE INVESTMENT PROPOSALS. THE TOTAL MONEY AVAILABLE FOR THE COMPANY TO INVEST IS RS 35,000. THE COMPANY CAN ACCEPT ANY ONE OF THE PROPOSALS WITH DIFFERENT LETTERS SUBJECT TO BUDGET AVAILABLE. SELECT THE BEST ALTERNATIVE BASED ON THE PRESENT WORTH ON TOTAL INVESTMENT CRITERION . THE RATE OF RETURN IS 8 %


Annual equivalent method

ANNUAL EQUIVALENT METHOD

  • THIS IS OTHERWISE CALLED AS EQUIVALENT UNIFORM ANNUAL WORTH OR EQUIVALENT UNIFORM ANNUAL COST

  • AS ITS NAME SUGGESTS ANNUAL EQUIVALENT WORTH ANALYSIS IS ALSO A METHOD BY WHICH WE CAN DETERMINE THE EQUIVALENT ANNUAL RATHER THAN OVERALL PRESENT OR FUTURE WORTH OF A PROJECT

  • THE ANNUAL WORTH CRITERION PROVIDES A BASIS FOR MEASURING WORTH BY DETERMINING EQUAL PAYMENTS ON AN ANNUAL BASIS.


Problem 5

PROBLEM 5

  • A FIRM IS CONSIDERING THE PURCHASE OF ONE OF THE TWO NEW MACHINES . THE DATA ON EACH ARE DESCRIBED BELOW : - IF THE FIRM MARR IS 12 % WHICH M/C SHOULD BE SELECTED USING EUAW METHOD ?


Problem 6

PROBLEM 6

COMPARE THE FOLLOWING MACHINES ON THE BASIS OF THEIR EUAC ,USE I = 18 % PER YEAR


Problem 7

PROBLEM 7

THE HEAT LOSS THROUGH THE EXTERIOR WALLS OF A BUILDING COSTS RS 215 PER YEAR . INSULATION THAT WILL REDUCE THE HEAT LOSS COST BY 93% CAN BE INSTALLED FOR RS 127 AND INSULATION THAT WILL REDUCE THE HEAT LOSS COST BY 89% CAN BE INSTALLED FOR RS 90. DETERMINE WHICH INSULATION IS MOST DESIRABLE IF THE BUILDING IS TO BE USED FOR 6 YEARS AND IF THE INTEREST RATE IS 12 % .


Future worth amount method

FUTURE WORTH AMOUNT METHOD

  • IN THE FUTURE WORTH METHOD OF COMPARISON OF ALTERNATIVES THE FUTURE WORTH OF VARIOUS ALTERNATIVES WILL BE COMPUTED.

  • THE ALTERNATIVE WITH THE MAXIMUM FUTURE WORTH OF NET REVENUE OR WITH THE MINIMUM FUTURE WORTH OF NET COST WILL BE SELECTED AS THE BEST ALTERNATIVE FOR IMPLEMENTATION


Problem 8

PROBLEM 8

  • CONSIDER THE FOLLOWING 2 ALTERNATIVES

    AT I = 18% ,SELECT THE BEST ALTERNATIVE BASED ON FUTURE WORTH METHOD OF COMPARISON.


Problem 9

PROBLEM 9

  • M/S KRISNA CASTINGS LTD IS PLANNING TO REPLACE ITS ANNEALING FURNACE. IT HAS RECEIVED TENDERS FROM 3 DIFFERENT ORIGINAL MANUFACTURERS OF ANNEALING FURNACE. THE DETAILS ARE AS FOLLOWS :

    WHICH IS THE BEST ALTERNATIVE BASED ON FUTURE WORTH METHOD AT I = 20 %


Problem 10

PROBLEM 10

A COMPANY MUST DECIDE WHETHER TO BUY M/C A OR M/C B . THE DETAILS ARE AS FOLLOWS

AT 12 % INTEREST RATE WHICH M/C SHOULD BE SELECTED ? USE FUTURE WORTH METHOD OF COMPARISON.


Rate of return method

RATE OF RETURN METHOD

  • IN THIS METHOD OF COMPARISON THE RATE OF RETURN FOR EACH ALTERNATIVE IS COMPUTED.

  • THE ALTERNATIVE WITH THE HIGHEST RATE OF RETURN IS SELECTED AS THE BEST ALTERNATIVE

  • IN THIS TYPE EXPENDITURES ARE ALWAYS ASSIGNED WITH A NEGATIVE SIGN AND REVENUES WITH A POSITIVE SIGN

  • START WITH AN INTUITIVE VALUE OF I SATISFYING THE RELATION

  • THE RATE OF RETURN IS DETERMINED BY INTERPOLATION IN THE RANGE VALUES OF I


Problem 11

PROBLEM 11

  • A PROJECT REQUIRES AN INITIAL INVESTMENT OF RS 2,50,000 AND GENERATES NET CASH FLOWS OF RS 95,000 , 95,000 ,1,00,000 AND 1,12,500 IN THE FIRST ,SECOND,THIRD AND FOURTH YEAR RESPECTIVELY. CALCULATE THE INTERNAL RATE OF RETURN OF THE PROJECT.


Problem 12

PROBLEM 12

  • A COMPANY IS TRYING TO DIVERSIFY ITS BUSINESS IN A NEW PRODUCT LINE . THE LIFE OF THE PRODUCT IS 10 YEARS WITH NO SALVAGE VALUE AT THE END OF ITS LIFE, THE INITIAL OUTLAY OF THE PROJECT IS RS 20,00,000. THE ANNUAL NET PROFIT IS RS 3,50,000. FIND THE RATE OF RETURN FOR THE NEW BUSINESS.


Problem 13

PROBLEM 13

  • A COMPANY IS PLANNING TO EXPAND ITS PRESENT BUSINESS ACTIVITY. IT HAS TWO ALTERNATIVES FOR THE EXPANSION PROGRAMME AND THE CORRESPONDING CASH FLOWS ARE TABULATED BELOW . EACH ALTERNATIVE HAS A LIFE OF 5 YEARS AND A NEGLIBILE SALVAGE VALUE. SUGGEST THE BEST ALTERNATIVE TO THE COMPANY.


Capital recovery with return method

CAPITAL RECOVERY WITH RETURN METHOD

  • CR(i) = (P-F) (A/P,I,N) + FI

    WHERE P = FIRST COST OF THE ASSET

    F= ESTIMATED SALVAGE VALUE

    N= LIFE IN YEARS


Problem 14

PROBLEM 14

  • A FIRM REQUIRES POWER SHOVELS FOR ITS OPEN PIT MINING OPERATION. THE FIRST COST IS RS 2,50,000 AND THE SALVAGE VALUE IS RS 35,000 AT THE END OF 10 YEARS OF SERVICE. IF THE FIRM USES A RATE OF INTEREST OF 12 % FOR THE PROJECT EVALUATION HOW MUCH MUST BE EARNED ON AN EQUIVALENT ANNUAL BASIS SO THAT THE FIRM RECOVERS ITS INVESTED CAPITAL PLUS EARNS A RETURN ON THE CAPITAL COMMITTED TO THE EQUIPMENT DURING ITS LIFE TIME.


Problem 15

PROBLEM 15

  • A COMPANY CAN INVEST IN ONE OF THE TWO ALTERNATIVES . THE LIFE OF BOTH THE ALTERNATIVES IS ESTIMATED TO BE 5 YEARS WITH THE FOLLOWING INITIAL INVESTMENTS AND SALVAGE VALUES.

  • WHAT IS CR (i) FOR EACH ALTERNATIVE ?

  • DETERMINE THE SALVAGE VALUE AT THE END OF PROJECT LIFE FOR ALTERNATIVE B WHICH WILL RESULT IN SAME CR (i) IN BOTH ALTERNATIVES . ASSUME I = 15 %


Capitalised cost method

CAPITALISED COST METHOD

  • THIS METHOD IS OTHERWISE CALLED AS CAPITALISED EQUIVALENT AMOUNT METHOD.

  • DRAW THE CFD SHOWING ALL NON-RECURRING CASH FLOW AND ATLEAST TWO CYCLES OF ALL RECURRING CASH FLOWS.

  • FIND THE PRESENT WORTH OF ALL NON- RECURRING CASH FLOWS USING SINGLE PAYMENT PRESENT WORTH RELATIONSHIPS.

  • FIND THE EQUIVALENT UNIFORM ANNUAL AMOUNT A OVER ONE LIFE CYCLE FOR ALL RECURRING CASH FLOWS AND DIVIDE THAT AMOUNT BY THE INTEREST RATE TO GET THE CAPITALISED COST OF RECURRING CASH FLOWS.


Capitalised cost method1

CAPITALISED COST METHOD

  • DIVIDE ALL THE UNIFORM CASH FLOWS OCCURING FROM YEAR 1 TO YEAR ∞ BY THE INTEREST RATE TO GET THE CAPITALISED COST OF THOSE UNIFORM CASH FLOWS.

  • ADD THE VALUES OBTAINED IN THE ABOVE STEPS TO GET THE TOTAL CAPITALISED COST OF THE GIVEN INVESTMENT.


Problem 16

PROBLEM 16

  • RS 8000 IS TO BE WITHDRAWN FROM A SAVINGS ACCOUNT AT THE END OF EVERY 10 YEARS. CALCULATE THE CAPITALISED EQUIVALENT AMOUNT OR THE SINGLE AMOUNT TO BE DEPOSITED NOW AT AN INTEREST RATE OF 10 % PER YEAR ?


Problem 17

PROBLEM 17

  • CALCULATE THE CAPITALISED COST OF A PROJECT THAT HAS AN INITIAL COST OF RS 1,50,000 AND AN ADDITIONAL INVESTMENT OF RS 50,000 AFTER 10 YEARS. THE ANNUAL OPERATING COST WILL BE RS 5000 FOR THE FIRST FOUR YEARS AND RS 8000 THEREAFTER. IN ADDITION THERE IS EXPECTED TO BE A RECURRING MAJOR REWORK COST OF RS 15,000 EVERY 13 YEARS . ASSUME I = 15 % PER YEAR.


Incremental approach

INCREMENTAL APPROACH

  • TWO TYPES OF INCREMENTAL APPROACH – THEY ARE PRESENT WORTH ON INCREMENTAL APPROACH & RATE OF RETURN ON INCREMENTAL INVESTMENT.

  • PRESENT WORTH ON INCREMENTAL INVESTMENT

  • IF THE NEGATIVE CASH FLOWS ARE LESS THAN THE POSITIVE CASH FLOWS THEN “DO NOTHING” ALTERNATIVE MUST BE CONSIDERED.

  • IF THE NUMBER OF NEGATIVE CASH FLOWS ARE MORE THAN THE POSITIVE CASH FLOWS THEN THERE IS NO NEED TO CONSIDER THE “ DO NOTHING “ ALTERNATIVE.

  • LIST THE ALTERNATIVES IN THE ASCENDING ORDER OF THEIR INITIAL INVESTMENT


Incremental approach1

INCREMENTAL APPROACH

  • SELECT AS THE INITIAL CURRENT BEST ALTERNATIVE ,THE ONE REQUIRING THE SMALLEST FIRST COST. IN MOST CASES IT IS THE “DO NOTHING “ ALTERNATIVE.

  • COMPARE THE INITIAL BEST ALTERNATIVE AND FIRST CHALLENGING ALTERNATIVE. THE CHALLENGER IS ALWAYS THE NEXT HIGHEST ALTERNATIVE IN THE ORDER OF FIRST COST. THE COMPARISON IS ACCOMPLISHED BY EXAMINING THE DIFFERENCES BETWEEN 2 CASH FLOWS.

  • IF THE PRESENT WORTH OF THE INCREMENTAL CASH FLOW IS GREATER THAN ZERO THE CURRENT BEST ALTERNATIVE IS REJECTED AND THE CHALLENGER BECOMES THE NEXT CURRENT BEST ALTERNATIVE.


Incremental approach2

INCREMENTAL APPROACH

  • IF THE PRESENT WORTH OF THE INCREMENTAL CASH FLOW IS NEGATIVE THEN THE CURRENT BEST ALTERNATIVE REMAINS UNCHANGED AND THE CHALLENGER IS REJECTED.

  • THE PROCESS OF COMPARISON AND SELECTION IS REPEATED.


Problem 18

PROBLEM 18

  • SIX MUTUALLY EXCLUSIVE ALTERNATIVES ARE SHOWN BELOW . SELECT THE BEST ALTERNATIVE USING THE INCREMENTAL APPEOACH IF THE MARR IS 10% AND THE LIFE OF THE ALTERNATIVE IS 10 YEARS.


Problem 19

PROBLEM 19

  • SELECT THE BEST ALTERNATIVE FROM A SET OF ALTERNATIVES SHOWN BELOW IF THE MARR IS 10 %. USE PRESENT WORTH ON INCREMENTAL INVESTMENT.


Problem 20

PROBLEM 20

THREE MUTUALLY EXCLUSIVE ALTERNATIVES ARE SHOWN BELOW . IF THE MARR IS 15 % PER YERAR AND THE ALTERNATIVE HAVE DIFFERENT LIVES SELECT THE BEST ALTERNATIVE USING PRESENT WORTH ON INCREMENTAL INVESTMENT METHOD.


Rate of return on incremental approach

RATE OF RETURN ON INCREMENTAL APPROACH

  • IF THE RATE OF RETURN RESULTING FROM INCREMENTAL CASH FLOW IS GREATER THAN MARR THE INCREMENT OF INVESTMENT IS CONSIDERED DESIRABLE AND THE CURRENT BEST ALTERNATIVE IS DROPPED AND THE CHALLENGER BECOMES THE NEW CURRENT BEST ALTERNATIVE.

  • IF THE RATE OF RETURN ON INCREMENT INVESTMENT IS LESS THAN MARR THE CURRENT BEST ALTERNATIVE REMAINS UNCHANGED. IT HAS TO BE COMPARED WITH THE NEXT CHALLENGER IN THE ORDER OF INITIAL INVESTMENT REQUIREMENT.


Problem 21

PROBLEM 21

FOUR DESIGNS OF A PRODUCT WITH THEIR ASSOCIATED REVENUE AND COST ESTIMATES HAVE BEEN PRODUCED TO THE TOP MANAGEMENT FOR A DECISION. A 10 YEAR STUDY PERIOD WAS USED. A MARR OF 10 % IS REQUIRED. . BASED ON THE FOLLOWING PROJECTED CASH FLOWS WHICH OF THE FOUR ALTERNATIVE DESIGN APPEAR MOST ATTRACTIVE . USE INCREMENTAL RATE OF RETURN APPROACH.


Problem 22

PROBLEM 22

A COMPANY IS GOING TO INSTALL A NEW PLASTIC EXTRUDING MACHINE . FOUR DIFFERENT TYPES ARE AVAILABLE. THE COSTS ASSOCIATED WITH EACH MACHINES ARE SHOWN BELOW :


Summary

SUMMARY

  • PRESENT WORTH IS AN EQUIVALENCE METHOD OF ANALYSIS IN WHICH A PROJECT’S CASH FLOW ARE DISCOUNTED TO SINGLE PRESENT VALUE.

  • THE MARR OR MINIMUM ATTRACTIVE RATE OF RETURN IS THE INTEREST RATE AT WHICH A FIRM CAN ALWAYS EARN OR BORROW MONEY.

  • THE TERM MUTUALLY EXCLUSIVE MEANS THAT WHEN ONE OF SEVERAL ALTERNATIVES THAT MEET THE SAME NEED IS SELECTED ,OTHERS WILL BE REJECTED.

  • AE ANALYSIS YIELDS THE SAME DECISION AS PRESENT WORTH ANALYSIS.

  • THE CAPITAL RECOVERY FACTOR ALLOWS MANAGERS TO CALCULATE AN ANNUAL EQUIVALENT COST OF CAPITAL FOR EASE OF ITEMIZATION WITH ANNUAL OPERATING COSTS.


Probable questions

PROBABLE QUESTIONS

  • WRITE DOWN THE FORMULA FOR CAPITAL RECOVERY WITH RETURN.

  • ALL THE WORKED OUT PROBLEMS BY ALL THE METHODS IN THE CLASS.


Further reading for chapter 4

FURTHER READING FOR CHAPTER 4

  • ENGINEERING ECONOMY BY TED G ESCHENBACH, OXFORD PUBLICATIONS.

  • PRINCIPLES OF ENGINEERING ECONOMY ANALYSIS BY JOHN .A WHITE , KENNETH .E.CASE ,DAVID .B.PRATT,MARVIN.H.AGEE ,WILEY PUBLICATION

  • ENGINEERING ECONOMY BY WILLIAM .G.SULLIVAN , JAMES .A. BONTADELLI, ELIN .M .WICKS, PEARSON EDUCATION

  • ENGINEERING ECONOMY BY G.J.THUESEN ,W.J.FABRYCKY ,PHI PUBLICATION

  • CONTEMPORARY ENGINEERING ECONOMICS BY CHAN .S PARK ,PHI PUBLICATION


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