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BMFG 4623 ENGINEERING ECONOMY AND MANAGEMENT

TUTORIAL : Present Worth Analysis / Future Worth Analysis 1. A company that manufactures magnetic membrane switches is investigating two production options that have the estimated cash flows shown (RM 1 million units). Suggest which alternative should be selected with 10% interest rate using present worth analysis. In house 30 5 14 2 5

First cost, RM Annual Cost, RM per year Annual income, RM per year Salvage value, RM Life, years

Contract 0 2 3.1 5

Answer: PWin house = RM 5.359 (5,359,000) PWcontract = RM 4.170 (4,170,000) The in house contract option is selected because it has the numerically largest PW value, or larger profit.

2. The manager of a canned food processing plant must decide between two different labelling machines. Machine A will have a first cost of RM 42,000, an annual operating cost of RM 28,000, and a service life of 4 years. Machine B will cost RM 51,000 to buy and will have an annual operating cost of RM 17,000 during its 4-year life. At an interest rate of 10% per year, select the best machine on the basis of a present worth analysis. Answer: PWA = RM – 130,757 PWB = RM – 104,888 Select machine B because the PW of its costs is the lowest; it has the numerically largest PW value.

3. A mechanical engineer is considering two materials for use in an electrical vehicle. All estimates are made. Material X Material Y First cost, RM 15,000 35,000 Maintenance Cost, RM per year 9,000 7,000 Salvage value, RM 2,000 20,000 Life, years 5 5 a) Suggest which alternative should be selected with 10% interest rate. b) Find out at what first cost the method not selected above will become the more economics preference. Answer: a. PWX = RM – 47,875 PWY = RM – 49,117 Select material X

b. First cost of Y, PY= RM –33,758 Select Y, if first cost of Material Y ≤ 33,758

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by: Nor Akramin Mohamad

BMFG 4623 ENGINEERING ECONOMY AND MANAGEMENT

4. The supervisor of a community swimming pool has developed two methods for chlorinating the pool. If gaseous chlorine is added, a chlorinator will be required that has an initial cost of RM 8,000 and a useful life of 5 years. The chlorine will cost RM 650 per year, and the labor cost will be RM 800 per year. Alternatively, dry chlorine can be added manually at a cost of RM 1,000 per year for chlorine and RM 1, 900 per year for labor. Recommend the best option the basis of a present worth analysis if the interest rate is 10% per year. Answer: PWGas = RM – 13,497 PWDry = RM – 10,993 Select Dry chlorine because

5. An electric switch manufacturing company has to choose one of three different assembly methods. Method A will have a first cost of RM 40,000, an annual operating cost of RM 9,000, and a service life of 2 years. Method B will cost RM 80,000 to buy and will have an annual operating cost of RM 6,000 over its 4-year service life. Method C will cost RM130,000 initially with an annual operating cost of RM 4,000 over its 8-year life. Methods A and B will have no salvage value, but method C will have some equipment worth an estimated RM 12,000. At an interest rate of 10% per year, select the best method on the basis of a present worth analysis. Answer: PWA = RM – 170,970 PWB = RM – 166,649 PWC = RM – 145,742 The method C is selected since the PW of its costs is the lowest; it has the numerically largest PW value.

6. Machines that have the following costs are under consideration for a automatic welding process. Using an interest rate of 10% per year, determine which machine should be selected on the basis of a present worth analysis. First cost, RM Annual Operating Cost, RM per year Salvage value, RM Life, years

Machine X 250,000 60,000 70,000 3

Machine Y 430,000 40,000 95,000 6

Answer: PWX = RM – 607,037 PWY = RM – 550,585

7. Chemical processing corporation is considering three methods to dispose of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are shown. 2

by: Nor Akramin Mohamad

BMFG 4623 ENGINEERING ECONOMY AND MANAGEMENT

First cost, RM Annual Operating Cost, RM per year Salvage value, RM Life, years

Land Application 130,000 95,000 25,000 3

Incineration

Contract

900,000 60,000 300,000 6

0 120,000 0 2

a. Determine which has the least cost on the basis of a present worth comparison at 10% per year. b. If the contract award cost increases by 20% every 2-year renewal, analyse whether the decision will change. Answer: a.

b.

PWLand = RM – 608,528 PWIncinerator = RM – 991,968 PWContract = RM – 522,636 Select disposal contract PWContract = RM – 619,615 Select land application, the selection changed

8. An industrial engineer is considering two robots for purchase by a fiber-optic manufacturing company. Robot X will have a first cost of $80,000, an annual maintenance and operation (M&O) cost of $30,000, and a $40,000 salvage value. Robot Y will have a first cost of $97,000, an annual M&O cost of $27,000, and a $50,000 salvage value. Analyse which should be selected on the basis of a future worth comparison at an interest rate of 15% per year using a 3-year study period. Answer: FWX = RM – 185,847 FWY = RM – 191,285

9. Compare the alternatives shown below on the basis of a future worth analysis, using an interest rate of 8% per year. First cost, RM Annual Operating Cost, RM per year Salvage value, RM Life, years

Alternative P -23,000 -4,000 3,000 3

Alternative Q -30,000 -2,500 1,000 6

Answer: FWP = RM – 88,036 FWQ= RM – 64,947 Select Alternative Q because….

10. A wealthy businessman wants to start a permanent fund for supporting research directed toward sustainability. The donor plans to give equal amounts of money for each of the next 5 years, plus one now (i.e., six donations) so that RM100,000 per year can be 3

by: Nor Akramin Mohamad

BMFG 4623 ENGINEERING ECONOMY AND MANAGEMENT

withdrawn each year forever, beginning in year 6. If the fund earns interest at a rate of 8% per year, determine how much money must be donated each time. Answer: A = RM 170,400

11. Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per year. First cost, RM Annual Operating Cost, RM per year Salvage value, RM Life, years

Alternative M 150,000 50,000 8,000 5

Alternative N 800,000 12,000 1,000,000 ∞

Answer: CCM = RM – 882,600 CCN = RM – 920,000 Select Alternative M

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by: Nor Akramin Mohamad