1. CHAPTER 4 ECONOMIC EVALUATION OF ALTERNATIVES

2. TOPICS IN CHAPTER 4 BASES FOR COMPARISON OF ALTERNATIVESBASES FOR COMPARISON OF ALTERNATIVES 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

3. OBJECTIVES TO SELECT THE BEST ALTERNATIVE ECONOMICALLYTO SELECT THE BEST ALTERNATIVE ECONOMICALLY UNDERSTAND THE VARIOUS BASESUNDERSTAND THE VARIOUS BASES KNOW HOW REGARDING USAGE OF VARIOUS METHODSKNOW HOW REGARDING USAGE OF VARIOUS METHODS

4. 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 RATETHE 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 MONEYIT 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 SIGNIN 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 SIGNIN A REVENUE DOMINATED CASH FLOW DIAGRAM THE PROFIT,REVENUE WILL BE ASSIGNED WITH POSITIVE SIGN AND CASH OUTFLOWS WITH NEGATIVE SIGN

5. PRESENT WORTH AMOUNT METHOD IN REVENUE DOMINATED CASH FLOW DIAGRAM THE ALTERNATIVE WITH THE MAXIMUM PRESENT WORTH AMOUNT SHOULD BE SELECTED AS THE BEST ALTERNATIVEIN 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.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 ALTERNATIVESWE HAVE TWO SITUATIONS EQUAL LIVED ALTERNATIVES & UNEQUAL LIVED ALTERNATIVES

6. 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.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.

7. 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.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.

8. PROBLEM 3 A PLANT MANAGER IS TRYING TO DECIDE BETWEEN THE MACHINES DETAILED BELOW : TAKE I = 15%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

9. 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 % PROPOSALFIRST COSTNET INCOME YEARS 1 TO 10 A1A1 -10,0002000 A2A2 - 12,0002100 B1B1 -20,0003100 B2B2 - 30,0005000 C1C1 - 35,0004500

10. 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.

11. 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 ? MACHINE AMACHINE B INITIAL COST34006500 SERVICE LIFE3 YEARS6 YEARS SALVAGE VALUE100500 NET OPERATING CASH FLOW AFTER TAXES 2000/YEAR1800/YEAR

12. PROBLEM 6 COMPARE THE FOLLOWING MACHINES ON THE BASIS OF THEIR EUAC,USE I = 18 % PER YEAR MACHINE XMACHINE Y FIRST COST 44,00023,000 ANNUAL OPERATING COST 70009000 ANNUAL REPAIR COST 210350 OVERHAUL EVERY 2 YEARS ----------1900 OVERHAUL EVERY 5 YEARS 2500-------- SALVAGE VALUE 40003000 LIFE (YEARS) 158

13. 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 %.

14. 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

15. PROBLEM 8 CONSIDER THE FOLLOWING 2 ALTERNATIVESCONSIDER THE FOLLOWING 2 ALTERNATIVES AT I = 18%,SELECT THE BEST ALTERNATIVE BASED ON FUTURE WORTH METHOD OF COMPARISON. END OF YEAR01234 ALTERNATIVE A(RS)-50 lacs20 lacs B(RS)-45 lacs18 lacs

16. 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 :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 % MANUFACT URER 123 INITIAL COST 80 LACS70 LACS90 LACS LIFE12 ANNUAL OPERATION COST 8 LACS9 LACS8.5LACS SALVAGE VALUE 5 LACS4 LACS7LACS

17. 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. MACHINE AMACHINE B INITIAL COSTRS 4 LACSRS 8 LACS LIFE4 YEARS SALVAGE VALUE2 LACS5.5 LACS ANNUAL MAINTENANCE COST 40,0000

18. RATE OF RETURN METHOD IN THIS METHOD OF COMPARISON THE RATE OF RETURN FOR EACH ALTERNATIVE IS COMPUTED.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 ALTERNATIVETHE 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 SIGNIN 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 RELATIONSTART WITH AN INTUITIVE VALUE OF I SATISFYING THE RELATION THE RATE OF RETURN IS DETERMINED BY INTERPOLATION IN THE RANGE VALUES OF ITHE RATE OF RETURN IS DETERMINED BY INTERPOLATION IN THE RANGE VALUES OF I

19. 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.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.

20. 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.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.

21. 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.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. ALTERNATIVESINITIAL INVESTMENT YEARLY REVENUE ALTERNATIVE 15,00,0001,70,000 ALTERNATIVE 28,00,0002,70,000

22. CAPITAL RECOVERY WITH RETURN METHOD CR(i) = (P-F) (A/P,I,N) + FICR(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 YEARS N= LIFE IN YEARS

23. 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.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.

24. 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.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. A)WHAT IS CR (i) FOR EACH ALTERNATIVE ? B) 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 % ALTERNATIVEAB INVESTMENT10,00012,000 SALVAGE VALUE15003500

25. CAPITALISED COST METHOD THIS METHOD IS OTHERWISE CALLED AS CAPITALISED EQUIVALENT AMOUNT 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.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 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.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.

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

27. 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 ?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 ?

28. 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.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.

29. INCREMENTAL APPROACH TWO TYPES OF INCREMENTAL APPROACH – THEY ARE PRESENT WORTH ON INCREMENTAL APPROACH & RATE OF RETURN ON INCREMENTAL INVESTMENT.TWO TYPES OF INCREMENTAL APPROACH – THEY ARE PRESENT WORTH ON INCREMENTAL APPROACH & RATE OF RETURN ON INCREMENTAL INVESTMENT. PRESENT WORTH ON INCREMENTAL INVESTMENTPRESENT 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 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.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 INVESTMENTLIST THE ALTERNATIVES IN THE ASCENDING ORDER OF THEIR INITIAL INVESTMENT

30. 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.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.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.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.

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

32. 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.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. ABCDEF INVESTM ENT 90015002500400050007000 ANNUAL REVENUE 15027640092511251425

33. PROBLEM 19 SELECT THE BEST ALTERNATIVE FROM A SET OF ALTERNATIVES SHOWN BELOW IF THE MARR IS 10 %. USE PRESENT WORTH ON INCREMENTAL INVESTMENT.SELECT THE BEST ALTERNATIVE FROM A SET OF ALTERNATIVES SHOWN BELOW IF THE MARR IS 10 %. USE PRESENT WORTH ON INCREMENTAL INVESTMENT. END OF YEAR ABCD 0-25,000-30,000 -37,500 1-6250-3750-3000-1000 2-6250-3750-3000-1000 325003750 7500

34. 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. ABC INITIAL COST - 21,000- 24,500-31,500 SALVAGE VALUE 07001050 CASH FLOW 700010,500 LIFE346

35. 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 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.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.

36. 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. 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. ABCD INITIAL INVESTMENT 1,70,0002,60,0003,00,0003,30,000 ANNUAL RECEIPTS 1,14,0001,20,0001,30,0001,47,000 ANNUAL DISBURSEMENTS 70,00071,00064,00079,000

37. 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 : 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 : ABCD INVESTM ENT 5,25,0006,65,00010,85,00011,37,500 ANNUALDISBURSEM ENTS POWER59,500 1,05,0001,10,250 LABOUR5,77,5005,25,0003,67,5003,23,750 MAINTEN ANCE 35,00039,37556,87543,750 INSURAN CE 10,50013,30021,70022,750 LIFE5555

38. SUMMARY PRESENT WORTH IS AN EQUIVALENCE METHOD OF ANALYSIS IN WHICH A PROJECT’S CASH FLOW ARE DISCOUNTED TO SINGLE PRESENT VALUE.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 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.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.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.THE CAPITAL RECOVERY FACTOR ALLOWS MANAGERS TO CALCULATE AN ANNUAL EQUIVALENT COST OF CAPITAL FOR EASE OF ITEMIZATION WITH ANNUAL OPERATING COSTS.

39. PROBABLE QUESTIONS WRITE DOWN THE FORMULA FOR CAPITAL RECOVERY WITH RETURN.WRITE DOWN THE FORMULA FOR CAPITAL RECOVERY WITH RETURN. ALL THE WORKED OUT PROBLEMS BY ALL THE METHODS IN THE CLASS.ALL THE WORKED OUT PROBLEMS BY ALL THE METHODS IN THE CLASS.

40. FURTHER READING FOR CHAPTER 4 ENGINEERING ECONOMY BY TED G ESCHENBACH, OXFORD PUBLICATIONS.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 PUBLICATIONPRINCIPLES 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 EDUCATIONENGINEERING ECONOMY BY WILLIAM.G.SULLIVAN, JAMES.A. BONTADELLI, ELIN.M.WICKS, PEARSON EDUCATION ENGINEERING ECONOMY BY G.J.THUESEN,W.J.FABRYCKY,PHI PUBLICATIONENGINEERING ECONOMY BY G.J.THUESEN,W.J.FABRYCKY,PHI PUBLICATION CONTEMPORARY ENGINEERING ECONOMICS BY CHAN.S PARK,PHI PUBLICATIONCONTEMPORARY ENGINEERING ECONOMICS BY CHAN.S PARK,PHI PUBLICATION