CHAPTER 4CHAPTER 4

ECONOMIC EVALUATION OF ECONOMIC EVALUATION OF ALTERNATIVESALTERNATIVES

TOPICS IN CHAPTER 4TOPICS IN CHAPTER 4• BASES FOR COMPARISON OF BASES FOR COMPARISON OF

ALTERNATIVESALTERNATIVES

• PRESENT WORTH AMOUNTPRESENT WORTH AMOUNT

• CAPITALISED EQUIVALENT AMOUNTCAPITALISED EQUIVALENT AMOUNT

• ANNUAL EQUIVALENT AMOUNTANNUAL EQUIVALENT AMOUNT

• FUTURE WORTH AMOUNTFUTURE WORTH AMOUNT

• CAPITAL RECOVERY WITH RETURNCAPITAL RECOVERY WITH RETURN

• RATE OF RETURN APPROACHRATE OF RETURN APPROACH

• INCREMENTAL APPROACHINCREMENTAL APPROACH

OBJECTIVESOBJECTIVES• TO SELECT THE BEST ALTERNATIVE TO SELECT THE BEST ALTERNATIVE

ECONOMICALLYECONOMICALLY

• UNDERSTAND THE VARIOUS BASESUNDERSTAND THE VARIOUS BASES

• KNOW HOW REGARDING USAGE OF KNOW HOW REGARDING USAGE OF VARIOUS METHODSVARIOUS METHODS

PRESENT WORTH AMOUNT METHODPRESENT WORTH AMOUNT METHOD• THE PRESENT WORTH IS A NET EQUIVALENT AMOUNT THE PRESENT WORTH IS A NET EQUIVALENT AMOUNT

AT THE PRESENT THAT REPRESENTS THE DIFFERENCE AT THE PRESENT THAT REPRESENTS THE DIFFERENCE BETWEEN EQUIVALENT DISBURSEMENTS AND BETWEEN EQUIVALENT DISBURSEMENTS AND EQUIVALENT RECEIPTS OF AN INVESTMENT CASH EQUIVALENT RECEIPTS OF AN INVESTMENT CASH FLOW FOR A SELECTED INTEREST RATEFLOW FOR A SELECTED INTEREST RATE

• IT CONSIDERS THE TIME VALUE OF MONEYIT CONSIDERS THE TIME VALUE OF MONEY• IN A COST DOMINATED CASH FLOW DIAGRAM THE IN A COST DOMINATED CASH FLOW DIAGRAM THE

COST WILL BE ASSIGNED WITH POSITIVE SIGN AND THE COST WILL BE ASSIGNED WITH POSITIVE SIGN AND THE PROFIT ,REVENUE WILL BE ASSIGNED WITH NEGATIVE PROFIT ,REVENUE WILL BE ASSIGNED WITH NEGATIVE SIGNSIGN

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

PRESENT WORTH AMOUNT METHODPRESENT WORTH AMOUNT METHOD• IN REVENUE DOMINATED CASH FLOW DIAGRAM THE IN REVENUE DOMINATED CASH FLOW DIAGRAM THE

ALTERNATIVE WITH THE MAXIMUM PRESENT WORTH ALTERNATIVE WITH THE MAXIMUM PRESENT WORTH AMOUNT SHOULD BE SELECTED AS THE BEST AMOUNT SHOULD BE SELECTED AS THE BEST ALTERNATIVEALTERNATIVE

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

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

PROBLEM 1PROBLEM 1

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

PROBLEM 2PROBLEM 2

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

PROBLEM 3PROBLEM 3

• A PLANT MANAGER IS TRYING TO DECIDE BETWEEN THE A PLANT MANAGER IS TRYING TO DECIDE BETWEEN THE MACHINES DETAILED BELOW : TAKE I = 15%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 4PROBLEM 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 %

PROPOSAL FIRST COST NET INCOME YEARS 1 TO 10

A1 -10,000 2000

A2 - 12,000 2100

B1 -20,000 3100

B2 - 30,000 5000

C1 - 35,000 4500

ANNUAL EQUIVALENT ANNUAL EQUIVALENT METHODMETHOD

• 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 5PROBLEM 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 ?

MACHINE A MACHINE B

INITIAL COST 3400 6500

SERVICE LIFE 3 YEARS 6 YEARS

SALVAGE VALUE

100 500

NET OPERATING CASH FLOW AFTER TAXES

2000/YEAR 1800/YEAR

PROBLEM 6PROBLEM 6COMPARE THE FOLLOWING MACHINES ON THE

BASIS OF THEIR EUAC ,USE I = 18 % PER YEAR

MACHINE X MACHINE Y

FIRST COST 44,000 23,000ANNUAL OPERATING COST

7000 9000

ANNUAL REPAIR COST

210 350

OVERHAUL EVERY 2 YEARS

---------- 1900

OVERHAUL EVERY 5 YEARS

2500 --------

SALVAGE VALUE

4000 3000

LIFE (YEARS) 15 8

PROBLEM 7PROBLEM 7THE 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 FUTURE WORTH AMOUNT METHODMETHOD

• 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 8PROBLEM 8• CONSIDER THE FOLLOWING 2 CONSIDER THE FOLLOWING 2

ALTERNATIVES ALTERNATIVES

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

END OF YEAR 0 1 2 3 4

ALTERNATIVE

A(RS) -50 lacs 20 lacs 20 lacs 20 lacs 20 lacs

B(RS) -45 lacs 18 lacs 18 lacs 18 lacs 18 lacs

PROBLEM 9PROBLEM 9• M/S KRISNA CASTINGS LTD IS PLANNING TO REPLACE ITS M/S KRISNA CASTINGS LTD IS PLANNING TO REPLACE ITS

ANNEALING FURNACE. IT HAS RECEIVED TENDERS FROM 3 ANNEALING FURNACE. IT HAS RECEIVED TENDERS FROM 3 DIFFERENT ORIGINAL MANUFACTURERS OF ANNEALING FURNACE. DIFFERENT ORIGINAL MANUFACTURERS OF ANNEALING FURNACE. THE DETAILS ARE AS FOLLOWS : THE DETAILS ARE AS FOLLOWS :

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

MANUFACTURER

1 2 3

INITIAL COST

80 LACS 70 LACS 90 LACS

LIFE 12 12 12

ANNUAL OPERATION COST

8 LACS 9 LACS 8.5LACS

SALVAGE VALUE

5 LACS 4 LACS 7LACS

PROBLEM 10PROBLEM 10A COMPANY MUST DECIDE WHETHER TO BUY M/C A OR A COMPANY MUST DECIDE WHETHER TO BUY M/C A OR

M/C B . THE DETAILS ARE AS FOLLOWSM/C B . THE DETAILS ARE AS FOLLOWS

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

MACHINE A MACHINE B

INITIAL COST RS 4 LACS RS 8 LACS

LIFE 4 YEARS 4 YEARS

SALVAGE VALUE

2 LACS 5.5 LACS

ANNUAL MAINTENANCE COST

40,000 0

RATE OF RETURN METHODRATE OF RETURN METHOD• IN THIS METHOD OF COMPARISON THE RATE OF IN THIS METHOD OF COMPARISON THE RATE OF

RETURN FOR EACH ALTERNATIVE IS COMPUTED.RETURN FOR EACH ALTERNATIVE IS COMPUTED.• THE ALTERNATIVE WITH THE HIGHEST RATE OF RETURN THE ALTERNATIVE WITH THE HIGHEST RATE OF RETURN

IS SELECTED AS THE BEST ALTERNATIVEIS SELECTED AS THE BEST ALTERNATIVE• IN THIS TYPE EXPENDITURES ARE ALWAYS ASSIGNED IN THIS TYPE EXPENDITURES ARE ALWAYS ASSIGNED

WITH A NEGATIVE SIGN AND REVENUES WITH A WITH A NEGATIVE SIGN AND REVENUES WITH A POSITIVE SIGNPOSITIVE SIGN

• START WITH AN INTUITIVE VALUE OF I SATISFYING THE START WITH AN INTUITIVE VALUE OF I SATISFYING THE RELATIONRELATION

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

PROBLEM 11PROBLEM 11• A PROJECT REQUIRES AN INITIAL INVESTMENT OF RS A PROJECT REQUIRES AN INITIAL INVESTMENT OF RS

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

PROBLEM 12PROBLEM 12• A COMPANY IS TRYING TO DIVERSIFY ITS BUSINESS IN A A COMPANY IS TRYING TO DIVERSIFY ITS BUSINESS IN A

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

PROBLEM 13PROBLEM 13• A COMPANY IS PLANNING TO EXPAND ITS PRESENT A COMPANY IS PLANNING TO EXPAND ITS PRESENT

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

ALTERNATIVES INITIAL INVESTMENT

YEARLY REVENUE

ALTERNATIVE 1 5,00,000 1,70,000

ALTERNATIVE 2 8,00,000 2,70,000

CAPITAL RECOVERY WITH CAPITAL RECOVERY WITH RETURN METHODRETURN METHOD

• CR(i) = (P-F) (A/P,I,N) + FI CR(i) = (P-F) (A/P,I,N) + FI

WHERE P = FIRST COST OF THE ASSET WHERE P = FIRST COST OF THE ASSET

F= ESTIMATED SALVAGE VALUE F= ESTIMATED SALVAGE VALUE

N= LIFE IN YEARSN= LIFE IN YEARS

PROBLEM 14PROBLEM 14• A FIRM REQUIRES POWER SHOVELS FOR ITS OPEN PIT A FIRM REQUIRES POWER SHOVELS FOR ITS OPEN PIT

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

PROBLEM 15PROBLEM 15• A COMPANY CAN INVEST IN ONE OF THE TWO A COMPANY CAN INVEST IN ONE OF THE TWO

ALTERNATIVES . THE LIFE OF BOTH THE ALTERNATIVES ALTERNATIVES . THE LIFE OF BOTH THE ALTERNATIVES IS ESTIMATED TO BE 5 YEARS WITH THE FOLLOWING IS ESTIMATED TO BE 5 YEARS WITH THE FOLLOWING INITIAL INVESTMENTS AND SALVAGE VALUES.INITIAL INVESTMENTS AND SALVAGE VALUES.

A)A) WHAT IS CR (i) FOR EACH ALTERNATIVE ? WHAT IS CR (i) FOR EACH ALTERNATIVE ?

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

ALTERNATIVE A B

INVESTMENT 10,000 12,000

SALVAGE VALUE

1500 3500

CAPITALISED COST CAPITALISED COST METHOD METHOD

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

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

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

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

CAPITALISED COST CAPITALISED COST METHOD METHOD

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

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

PROBLEM 16 PROBLEM 16 • RS 8000 IS TO BE WITHDRAWN FROM A SAVINGS RS 8000 IS TO BE WITHDRAWN FROM A SAVINGS

ACCOUNT AT THE END OF EVERY 10 YEARS. CALCULATE ACCOUNT AT THE END OF EVERY 10 YEARS. CALCULATE THE CAPITALISED EQUIVALENT AMOUNT OR THE SINGLE THE CAPITALISED EQUIVALENT AMOUNT OR THE SINGLE AMOUNT TO BE DEPOSITED NOW AT AN INTEREST RATE AMOUNT TO BE DEPOSITED NOW AT AN INTEREST RATE OF 10 % PER YEAR ? OF 10 % PER YEAR ?

PROBLEM 17 PROBLEM 17 • CALCULATE THE CAPITALISED COST OF A PROJECT CALCULATE THE CAPITALISED COST OF A PROJECT

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

INCREMENTAL APPROACH INCREMENTAL APPROACH • TWO TYPES OF INCREMENTAL APPROACH – THEY ARE TWO TYPES OF INCREMENTAL APPROACH – THEY ARE

PRESENT WORTH ON INCREMENTAL APPROACH & RATE PRESENT WORTH ON INCREMENTAL APPROACH & RATE OF RETURN ON INCREMENTAL INVESTMENT.OF RETURN ON INCREMENTAL INVESTMENT.

• PRESENT WORTH ON INCREMENTAL INVESTMENTPRESENT WORTH ON INCREMENTAL INVESTMENT

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

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

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

INCREMENTAL APPROACH INCREMENTAL APPROACH • SELECT AS THE INITIAL CURRENT BEST SELECT AS THE INITIAL CURRENT BEST

ALTERNATIVE ,THE ONE REQUIRING THE SMALLEST ALTERNATIVE ,THE ONE REQUIRING THE SMALLEST FIRST COST. IN MOST CASES IT IS THE “DO NOTHING “ FIRST COST. IN MOST CASES IT IS THE “DO NOTHING “ ALTERNATIVE.ALTERNATIVE.

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

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

INCREMENTAL APPROACH INCREMENTAL APPROACH • IF THE PRESENT WORTH OF THE INCREMENTAL CASH IF THE PRESENT WORTH OF THE INCREMENTAL CASH

FLOW IS NEGATIVE THEN THE CURRENT BEST FLOW IS NEGATIVE THEN THE CURRENT BEST ALTERNATIVE REMAINS UNCHANGED AND THE ALTERNATIVE REMAINS UNCHANGED AND THE CHALLENGER IS REJECTED.CHALLENGER IS REJECTED.

• THE PROCESS OF COMPARISON AND SELECTION IS THE PROCESS OF COMPARISON AND SELECTION IS REPEATED.REPEATED.

PROBLEM 18 PROBLEM 18 • SIX MUTUALLY EXCLUSIVE ALTERNATIVES ARE SHOWN SIX MUTUALLY EXCLUSIVE ALTERNATIVES ARE SHOWN

BELOW . SELECT THE BEST ALTERNATIVE USING THE BELOW . SELECT THE BEST ALTERNATIVE USING THE INCREMENTAL APPEOACH IF THE MARR IS 10% AND THE INCREMENTAL APPEOACH IF THE MARR IS 10% AND THE LIFE OF THE ALTERNATIVE IS 10 YEARS. LIFE OF THE ALTERNATIVE IS 10 YEARS.

A B C D E F

INVESTMENT

900 1500 2500 4000 5000 7000

ANNUAL REVENUE

150 276 400 925 1125 1425

PROBLEM 19 PROBLEM 19 • SELECT THE BEST ALTERNATIVE FROM A SET OF SELECT THE BEST ALTERNATIVE FROM A SET OF

ALTERNATIVES SHOWN BELOW IF THE MARR IS 10 %. ALTERNATIVES SHOWN BELOW IF THE MARR IS 10 %. USE PRESENT WORTH ON INCREMENTAL INVESTMENT.USE PRESENT WORTH ON INCREMENTAL INVESTMENT.

END OF YEAR

A B C D

0 -25,000 -30,000 -30,000 -37,500

1 -6250 -3750 -3000 -1000

2 -6250 -3750 -3000 -1000

3 2500 3750 3750 7500

PROBLEM 20 PROBLEM 20 THREE MUTUALLY EXCLUSIVE ALTERNATIVES ARE THREE MUTUALLY EXCLUSIVE ALTERNATIVES ARE SHOWN BELOW . IF THE MARR IS 15 % PER YERAR AND SHOWN BELOW . IF THE MARR IS 15 % PER YERAR AND THE ALTERNATIVE HAVE DIFFERENT LIVES SELECT THE THE ALTERNATIVE HAVE DIFFERENT LIVES SELECT THE BEST ALTERNATIVE USING PRESENT WORTH ON BEST ALTERNATIVE USING PRESENT WORTH ON INCREMENTAL INVESTMENT METHOD.INCREMENTAL INVESTMENT METHOD.

A B C

INITIAL COST

- 21,000 - 24,500 -31,500

SALVAGE VALUE

0 700 1050

CASH FLOW

7000 10,500 10,500

LIFE 3 4 6

RATE OF RETURN ON RATE OF RETURN ON INCREMENTAL APPROACH INCREMENTAL APPROACH

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

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

PROBLEM 21 PROBLEM 21 FOUR DESIGNS OF A PRODUCT WITH THEIR ASSOCIATED REVENUE FOUR DESIGNS OF A PRODUCT WITH THEIR ASSOCIATED REVENUE

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

A B C D

INITIAL INVESTMENT

1,70,000 2,60,000 3,00,000 3,30,000

ANNUAL RECEIPTS

1,14,000 1,20,000 1,30,000 1,47,000

ANNUAL DISBURSEMENTS

70,000 71,000 64,000 79,000

PROBLEM 22 PROBLEM 22 A COMPANY IS GOING TO INSTALL A NEW PLASTIC A COMPANY IS GOING TO INSTALL A NEW PLASTIC

EXTRUDING MACHINE . FOUR DIFFERENT TYPES ARE EXTRUDING MACHINE . FOUR DIFFERENT TYPES ARE AVAILABLE. THE COSTS ASSOCIATED WITH EACH AVAILABLE. THE COSTS ASSOCIATED WITH EACH MACHINES ARE SHOWN BELOW :MACHINES ARE SHOWN BELOW :

A B C D

INVESTMENT

5,25,000 6,65,000 10,85,000 11,37,500

ANNUAL DIS BURSEMENTS

POWER 59,500 59,500 1,05,000 1,10,250

LABOUR 5,77,500 5,25,000 3,67,500 3,23,750

MAINTENANCE

35,000 39,375 56,875 43,750

INSURANCE

10,500 13,300 21,700 22,750

LIFE 5 5 5 5

SUMMARY SUMMARY • PRESENT WORTH IS AN EQUIVALENCE METHOD OF ANALYSIS IN PRESENT WORTH IS AN EQUIVALENCE METHOD OF ANALYSIS IN

WHICH A PROJECT’S CASH FLOW ARE DISCOUNTED TO SINGLE WHICH A PROJECT’S CASH FLOW ARE DISCOUNTED TO SINGLE PRESENT VALUE.PRESENT VALUE.

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

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

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

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

PROBABLE QUESTIONS PROBABLE QUESTIONS • WRITE DOWN THE FORMULA FOR CAPITAL RECOVERY WRITE DOWN THE FORMULA FOR CAPITAL RECOVERY

WITH RETURN.WITH RETURN.• ALL THE WORKED OUT PROBLEMS BY ALL THE METHODS ALL THE WORKED OUT PROBLEMS BY ALL THE METHODS

IN THE CLASS.IN THE CLASS.

FURTHER READING FOR FURTHER READING FOR CHAPTER 4 CHAPTER 4

• ENGINEERING ECONOMY BY TED G ESCHENBACH, OXFORD ENGINEERING ECONOMY BY TED G ESCHENBACH, OXFORD PUBLICATIONS.PUBLICATIONS.

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

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

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

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