Engineering Economy Computations and Analysis

Decisions regarding the choice of technology or process design require investment of capital. Therefore, these decisions utilize financial analysis of discounted cash flows or present values to determine the economic worth. In the segment, several methods of financial analysis will be described, particularly as they relate to decisions in operations.

Typical operations decisions that require detailed financial analysis are

1. The purchase of new equipment or facilities2. The replacement of existing equipment or facilities

An example of each of these decisions will be described later.

TIME VALUE OF MONEYIn evaluating investments, we should consider the time value of money. We would rather have a dollar now than a dollar a year from now because we could invest the current dollar and earn a return on it for a year. Therefore, any future cash flows have less value to us than current cash flows. As a result, future cash flows must be discounted or reduced in value to their present values in order for future dollars to be comparable to present dollars.

Discounting of future cash flows is based on the idea of compound interest. If we have P dollars at the present time and invest it at an interest rate of i, the future value in 1 year will be

F1 = P + iP = P(1 + i)

In n years, at compound interest, the value of our P dollars will be

Fn = P(1 + i)n

This assumes that the interest is reinvested each year as it is earned.

If we divide the above equation by (1 + i)n, we will have

By turning the compound interest equation around, we see that the present value of an amount Fn paid in n years is simply P. We can, therefore, discount Fn to its present value by multiplying by the quantity 1/(1 + i)n. This quantity is known as the discount factor of the present value of $1 in year n. Values of the discount factor are tabulated in Appendix A. These factors can be used to convert any future cash flow to a present value amount.

For example, suppose an investment has an annual cash flow of $1000 after taxes for 5 years. The present value of this cash stream at 10 percent interest is $3790.

N Present

Year Return Value1 $1000 .909 $ 9092 1000 .826 8263 1000 .751 7514 1000 .683 6835 1000 .621 621

$3790

In this case, each future cash amount was converted to a present value and the present values were then added. As a result, if we want to earn 10 percent on our money, we would be willing to invest $3790, the present value, now so as to get the future earnings of $1000 a year for 5 years.

In discounting future cash flows, it is also convenient to know the present value of a $1 annuity each year for n years. The annuity's present value is

Here we have discounted $1 each year back to the present time and added. The resulting values of P are tabulated in Appendix B for various interest rates and numbers of years.

We can solve the above problem directly by using the annuity's present values. For example, the present value of $1 a year for 5 years at 10 percent interest, from Appendix B, is 3.791. If $1000 is earned each year for 5 years, the present value is

P = $1000(3.791) = $3791

This is the same figure we arrived at above by adding the present values for each year (with the exception of round-off error in the last digit). The annuity table, therefore, saves time when uniform annual payments are present. The modern calculator and computer, however, may be even more convenient than the table.

In some investment problems, it is necessary to calculate an internal rate of return (IRR). The IRR is the interest rate which will just make the present investment equal to the future stream of earnings. In mathematical notation, suppose that an investment I has after-tax cash flows C1 in year 1, C2 in year 2, . . . , Cn in year n. The IRR is obtained by solving the following equation for i:

In general, the value of i is obtained by trial and error or iteration using the above equation. Suppose, for example, that we invest $5000 and earn $3000 in the first year, $2000 in the second year, and $2500 in the third year. Arbitrarily, assume i = 20 percent and solve the equation

Since $5335 is larger than the $5000 investment, we need a larger value of i to reduce the right-hand side. Assuming i = .30, we have

Since i = $4629 is smaller than the investment of $5000, the true value of i lies between 20 and 30 percent. Using linear interpolation, we can estimate that

Using i = .2475 in the equation to check the result, we have

Since this result, I = $4977, is slightly below $5000, we should reduce the interest rate a little more, perhaps to .24. By successive approximation, we will finally arrive at the interest rate to any desired degree of approximation.

In cases where the annual payments are equal, we can use the annuity tables (Appendix B) to find the IRR directly. For example, if the $5000 investment earns $2500 each year for 3 years, the following equation must be solved:

The annuity factor is therefore . From Appendix B, the interest rate for 3 years

will be between 22 and 24 percent. The estimated figure, by interpolation, is 23.4 percent.

CHOOSING INVESTMENT PROJECTS

Now that we know how to calculate present values and internal rates of return, we can apply these ideas to choosing investment projects. Suppose there is a portfolio of

investment alternatives. How should we choose among these alternatives or rank them in order of preference?

In general, there are three ways to make the choice; payback, present value, and IRR.

Payback. According to the payback method, a payback period for each investment is calculated as follows:

where N = payback period in yearsI = investmentS = salvage valueA = annual cash flow after tax

The investments in the portfolio are then ranked in order of their payback periods.For example, suppose a $10,000 investment will earn $2000 a year after taxes, and

there is no salvage value. The payback period for this investment is then 5 years.The payback method has several shortcomings. First, the length of the earning

period of the investment is not taken into account. Two investments could have the same payback period but drastically different lifetimes. The second problem with the payback method is that it does not consider the time value of money. Thus different earnings streams are not evaluated differently. Finally, the above formula requires a constant annual cash flow. This assumption could easily be relaxed, however, by determining the time required for earnings to equal the investment.

Although the payback method has serious weaknesses, it is still quite popular because it gives a sense of time to recover the investment. Nevertheless, it is being replaced by the next two methods as ways to rank investment alternatives.

Net Present Value. Whenever a hurdle rate or cost of capital is specified for investment comparisons, the investments can be compared through the use of present values. The given hurdle rate is used as the “interest” rate, and all future cash flows are discounted to the present time. The net present value (NPV) is then computed as follows:

Where I = investment requiredPj = present value of cash flow for year j

Whenever the net present value exceeds zero, the investment is worthwhile at the specified hurdle rate. If capital is limited, the investments can be ranked in terms of NPV from largest to smallest and funded in order of priority until capital is exhausted.

Internal Rate of Return. The IRR can also be used to rank investments and select those for funding from the portfolio. Figure T2.1 shows several investments ranked by IRR

and the cost of capital as a function of the amount invested. Notice how the cost of capital increases when large amounts of investment are required. As a result, the IRR falls below the cost of capital for

alternatives E and F. In this case, alternatives A, B, C, and D should be funded because their IRR exceeds the cost of capital, and alternatives E and F should not be funded.

The NPV and IRR methods are the opposite of each other. With NPV, the``interest'' rate or cost of capital is used to compute NPVs; a positive NPV indicates a worthwhile investment. For IRR, the “interest” rate is not given but is computed and compared with the cost of capital. An IRR greater than the cost of capital is considered a worthwhile investment.

If two investments have equal lifetimes, both IRR and NPV methods will yield the same result. If investment lifetimes vary, however, these methods require additional assumptions to yield a correct answer. The additional assumptions must specify what is done after the lifetime of the shortest investment. Is the capital invested in a riskless investment, a like investment, a technologically superior alternative, or what? The answer to these questions will affect the ranking of the alternatives.

A series of applications of the above investment methods will be described next. In each case, not only the numerical analysis but also other factors in the decision will be discussed.

PURCHASE OF A NEW MACHINE

The operations department is considering the installation of a machine to reduce the labor required in one of its processes. The machine will cost $50,000 and have a 5-year life, with a salvage value of $10,000 at the end of 5 years. The pretax cash-flow savings in labor which will accrue over the cost of operating the machine is $11,000 per year. Assume a 50

FIGURE T2.1

percent tax bracket, straight-line depreciation, and a 10 percent investment tax credit. What is the NPV of the investment at a 15 percent hurdle rate after tax? What IRR does the investment provide?

In all investment problems first the cash flow must be determined on an annual basis. In this case, the annual cash flow is

Cash flow -- pretax $11,000Depreciation 8,000Net income $ 3,000Additional taxes 1,500

Since the additional taxes paid are $1500 per year, the cash flow after tax is $9500 per year ($11,000--$1500). In the first year, there is an additional tax credit of $5000 (10 percent of $50,000). The net cash flow in the first year is therefore $14,500 = ($9500 + $5000). The after-tax cash flows - assuming that all of these occur at the end of the year - are shown in Figure T2.2. Here, investments and cash outflows are shown as negative numbers while cash inflows and salvage values are shown as positive numbers. It is always helpful to draw one of these cash-flow diagrams prior to making NPV or IRR calculations.

The NPV at 15 percent cost of capital is

-$50,000

0 1 2 3 4 5

$14,500$9,500 $9,500 $9,500

$19,500

Figure T2.2 Cash flow, example 1.

Since the NPV is negative, the investment is not worthwhile at 15 percent.The IRR is obtained by inserting i in place of .15 in the above equation and solving

for an NPV = 0. Thus the IRR must satisfy the following equation:

Since the NPV at 15 percent was negative, we know that i < .15. As a trial value select i = .10 and plug into the right-hand side of the above equation. At i = .10, we have NPV = - $3232. Since NPV is still negative, try a smaller interest rate, say 5 percent, which yields an NPV of $3727. Since this NPV is positive, the interest rate must lie between 5 and 10 percent. By interpolation:

The IRR is thus estimated to be 7.7 percent.In this case, there are other factors to be considered in the decision, such as a

possible loss of flexibility after converting to the machine and more consistent quality due to the machine. Since the ROI is so low, these factors will probably not be overriding.

MACHINE REPLACEMENT

The second example is the well-known machine-replacement problem, where the decision is whether or not to replace a current machine with a new model. Suppose for the sake of this example that we have a 5-year-old motorcycle and are considering whether to replace it with a ``new'' one (only 2 years old). If the motorcycle is not replaced now, assume it will be driven for another 3 years. For each alternative, the following costs are given:

YearKeep Old Motorcycle 1 2 3Maintenance $ 200 $250 $ 300Tires 200 --- ---

Gas and oil 600 600 600Insurance and license 150 125 100

Total $1150 $975 $1000

Buy New MotorcycleMaintenance $ 50 $150 $ 200Tires --- --- ---Gas and oil 400 400 400Insurance and license 250 200 150

Total $ 700 $750 $ 750

The new motorcycle is expected to get better gas mileage, as reflected in the above costs of gas and oil. The new motorcycle will have higher insurance and license costs but lower maintenance and tire costs, also as shown in the above numbers. The net result is that the new motorcycle will be less expensive to operate than the old one.

Assume that the new motorcycle will cost $5000 ($2000 with the trade-in) and it will be worth $3000 at the end of 3 years. Also assume that the old motorcycle is worth $3000 now and will be worth $1400 in 3 more years.

The cash-flow pattern for the difference between these two alternatives is shown in Figure T2.3. The new-motorcycle alternative requires a new investment of $2000 at time zero, including the trade-in. The new-motorcycle alternative will save $450 in the first year, $225 in the second year, and $250 in the third year. In addition, the salvage value of the new motorcycle will be $1600 ($3000 - $1400) more than the old motorcycle's salvage value at the end of 3 years. Since this is a personal decision, there will be no depreciation and no tax consequences.

As in the previous example, two questions can be asked: (1) is the new motorcycle worth the investment at some given rate of interest, perhaps 15 percent, and (2) what rate of interest does the investment earn? The first question can be answered by computing the net present value of the cash flow shown in Figure T2.3.

Since the net present value of the investment is negative, it is not worthwhile to buy the new motorcycle at 15 percent return on capital.

0 1 2 3

Figure T2.3 Cash flow, example 2. The cash flows reflect the net difference between purchasing the new motorcycle and keeping the old one.

$450$225 $250

$1600

-$2000

The IRR is the interest rate which makes the above net present value equal to zero. By iteration, we find that IRR = 9.6 percent.

In this example, there are several additional questions of interest. First, what is the value of capital to the motorcycle buyer? If the buyer takes the money out of a savings account, the after-tax value of capital might be about 3 or 4 percent. At this rate the investment would clearly be worthwhile, since the IRR is 9.6 percent. If the capital were diverted from other personal investments, which could earn, say, 15 percent after tax, then the investment would not be worthwhile.

The new-motorcycle buyer would also want to consider the intangible value of having a new motorcycle. The buyer might have fewer repair problems, there might be less squeaks and rattles, and there would be an aesthetic value to owning the new motorcycle. It would be difficult to put a dollar figure on such intangible benefits, and they would probably have to be incorporated subjectively in the decision. One way to evaluate these intangibles, however, is to compute their dollar value. If capital is worth 15 percent, the intangibles would have to be worth more than $222 (the negative amount of present value) to go ahead with the decision.

Similar problems are encountered in process technology choice when a current machine is being replaced by a new one. The new machine will require additional investment, but it will probably reduce annual operating costs. The new machine will also provide intangible benefits, as does the new motorcycle, such as fewer production disruptions through greater reliability.

The examples illustrate two different types of decisions on choice of process technology. The first example was a choice between an existing manual technology and a proposed new automated technology. The second example was one of technology replacement. In practice, there are other variations of these problems, but the basic principles still apply.

QUESTIONS1. What are the advantages and disadvantages in using the payback method?2. Under what conditions would you use the IRR or NPV method?3. What problems are created by unequal investment lives? How are these problems

handled?4. Precisely how does depreciation affect cash flow?5. Under what conditions would you choose an investment which does not meet the hurdle

rate?

PROBLEMS1. A factory is considering the installation of a new machine which will replace two

workers who have been doing the job by manual methods. The combined wages and benefits of the two workers are $45,000 per year. The new machine will be run by a single operator paid $25,000 a year in wages and benefits; it will also require $5000 a year for maintenance and utilities expense. The machine will provide the same output as the manual method, but the investment and installation will be $60,000. For tax

purposes, the machine can be depreciated over a 5-year period using a double declining balance. Assume a 50 percent tax rate, a 10 percent investment credit, and no salvage value.a. Is the machine a good investment at a 15 percent interest rate? Use a 10-year

machine life.b. Calculate the IRR for this decision.c. What other issues are important in this process technology decision?

2. A laundromat is considering replacing its washers and dryers. Two options are available. Option A: The new machines will cost $11,000 initially and $12,000 per year to operate. Expected life for taxes and operations purposes is 5 years. Option B: The new machines will cost $14,000 initially and $11,000 per year to operate. Expected life for taxes and operations purposes is 10 years.a. Using straight-line depreciation, determine which option is the best at a 10 percent

interest rate. Assume a 50 percent tax rate.b. What is the break-even interest rate, where the two options have equal present

values?3. A machine has been used in production for 5 years and is being considered for

replacement. At the present time the machine is fully depreciated but could be sold for $8000. The firm pays tax on half of all capital gains. The new machine being considered will cost $30,000 and will have a 5-year life for both tax and cash-flow considerations. The new machine will require one less operator at a savings of $22,000 per year in wages and benefits. Assume no salvage value, a 50 percent corporate income tax, and straight-line depreciation.a. What is the NPV of the investment at 10 percent, 15 percent, and 20 percent?b. What is the IRR at these same percentages?c. What additional factors would you consider in this decision?

4. The ABC Company is considering whether to lease or own its automobiles. If an automobile is purchased, it will cost an average of $22,000. It will be driven for 20,000 miles a year and can be sold for $10,000 at the end of 3 years. Assume that the out-of-pocket expenses for gasoline, oil, and repairs are 22 cents per mile. Also assume that straight-line depreciation is used with a 3- year life and $10,000 salvage value. If the leasing option is used, the lease will cost 30 cents per mile driven. This cost includes the car plus all gas, oil, and repairs.a. If the cost of capital is 12 percent, should the company buy or lease the cars?

Assume a 50 percent tax rate.b. What is the IRR of this investment?c. What are the intangible factors in this decision?

5. A company is considering installation of a new flexible manufacturing system. The system will be able to make some of the company's existing parts at lower costs, thereby saving $700,000 per year. The product line will also be redesigned to use fewer parts at a saving of $200,000 per year. Faster throughput and higher quality will also be possible. The expected profits from the additional business gained from these advantages will be $800,000 per year. The FMS will cost $4 million to install and $800,000 per year to operate. An expected lifetime of 10 years can be assumed. Straight-line depreciation is used, and corporate income taxes are 35 percent. Assume a 5 percent future inflation rate per year in future savings and costs.

a. At a hurdle rate of 15 percent, is the investment worthwhile?b. What IRR does this project earn?c. If the profits from the additional business are ignored, is the investment still

worthwhile at 15 percent?6. A factory that makes piano parts has been operating at the same site for 50 years.

While the equipment has been upgraded (the last time was 10 years ago), management thinks it is now time to build a new factory. Competition has recently entered the market and is now producing piano parts more cheaply and selling them at lower prices. As a result, the company has lowered its prices and is taking lower profit margins than in the past. If the new factory is not built, management projects that market share and profitability will continue to erode, ultimately putting the company out of business. The new factory will have a lower manufacturing cost (expressed as a percentage of sales) due to modernization. It will also have higher sales (in millions of dollars) due to retention of the market, as shown below:

Old Factory New FactoryYear Sales Mfg. Cost Sales Mfg. Cost

1 100 70% 100 60%2 80 75% 90 65%3 60 80% 80 70%4 40 90% 70 70%5 0 0% 70 70%

a. The new factory will cost $20 million to build. After 5 years it will have a salvage value of $10 million. The old factory has no salvage value at the present time. Should the new factory be built at a hurdle rate of 20 percent? Assume straight-line depreciation and income tax rates of 35 percent.

b. Instead of using a salvage value for the new factory, assume that its lifetime is 20 years at the same sales level and manufacturing cost as at the end of year 5. At the end of 20 years there is no salvage value. Should the new factory be built on the basis of these assumptions?

7. The ACE machine shop is considering reorganizing its production into a cell using the principles of group technology. As a result of this change the following savings are estimated. A savings of $1 million in inventory initially and continuing savings of $300,000 per year in inventory carrying costs. The parts made in the cell will have the same direct labor costs, but overhead will be much less, resulting in a savings of $100,000 per year. The overhead savings is due to simplified production control, less supervision, and better quality control. The use of the cell will also permit much faster throughput of production orders. As a result, new business will be gained, resulting in additional profits of $100,000 per year. To organize the cell will require some additional equipment at a cost of a $2 million initial investment and $150,000 per year operating cost. The equipment can be depreciated over a 5-year period, and corporate tax rates are 35 percent. Assume a 5 percent inflation rate per year in future savings and costs. Use straight-line depreciation with no salvage value.a. Is it worthwhile to form this cell using a 15 percent hurdle rate? Also, try a 20

percent hurdle rate.

b. What is the internal rate of return from this project?c. The additional business gained from the cell is considered a ``soft'' estimate. What

effect does leaving this estimate out of the calculations have on the answers to part a?

8. Two different machines are being considered for a production job, with the following characteristics:

Machine A Machine BInitial cost $10,000 $7000Op. cost/year $ 2,000 $1500Lifetime 10 years 5 years

Which machine is the better buy at a 15 percent discount rate? Assume that machine B is replaced by another one of the same type at the end of 5 years. Also assume that depreciation is straight line, income taxes are 35 percent, and inflation is 5 percent per year.

SELECTED BIBLIOGRAPHYAnthony, Robert N., and Glenn A. Welsch: Fundamentals of Management Accounting,

Homewood, Ill.: Irwin, 1974.Grant, Eugene L., W. Grant Ireson, and Richard Leavenworth: Principles of Engineering

Economy, 7th ed., New York: Wiley, 1982.Van Horne, James C.: Financial Management and Policy, 5th ed., Englewood Cliffs, N.J.:

Prentice-Hall, 1980.Welsch, Glenn A., and Robert N. Anthony: Fundamentals of Financial Accounting, rev.

ed., Homewood, Ill.: Irwin, 1977.

Appendix A Present – Value Factors For Future Single Payments

Periods Until Pay-ments

1% 2% 4% 6% 8% 10% 12% 14% 15% 16% 18% 20% 22% 24% 25% 26% 28% 30%

12345

0.9900.9800.9710.9610.951

0.9800.9610.9420.9240.906

0.9620.9250.8890.8550.822

0.9430.8900.8400.7920.747

0.9260.8570.7940.7350.681

0.9090.8260.7510.6830.621

0.8930.7970.7120.6360.567

0.8770.7690.6750.5920.519

0.8700.7560.6580.5720.497

0.8620.7430.6410.5520.476

0.8470.7180.6090.5160.437

0.8330.6940.5790.4820.402

0.8200.6720.5510.4510.370

0.8060.6500.5240.4230.341

0.8000.6400.5120.4100.328

0.7940.6300.5000.3970.315

0.7810.6100.4770.3730.291

0.7690.5920.4550.3500.269

678910

0.9420.9330.9230.9140.905

0.8880.8710.8530.8370.820

0.7900.7600.7310.7030.676

0.7050.6650.6270.5950.558

0.6300.5830.5400.5000.463

0.5640.5130.4670.4240.386

0.5070.4530.4040.3610.322

0.4560.4000.3510.3080.270

0.4320.3760.3270.2840.247

0.4100.3540.3050.2630.227

0.3700.3140.2660.2250.191

0.3350.2790.2330.1940.162

0.3030.2490.2040.1670.137

0.2750.2220.1790.1440.116

0.2620.2100.1680.1340.107

0.2500.1980.1570.1250.099

0.2270.1780.1390.1080.085

0.2070.1590.1230.0940.073

1112131415

0.8960.8870.8790.8700.861

0.8040.7880.7730.7580.743

0.6500.6250.6010.5770.555

0.5270.4970.4690.4420.417

0.4290.3970.3680.3400.315

0.3500.3190.2900.2630.239

0.2870.2570.2290.2050.183

0.2370.2080.1820.1600.140

0.2150.1870.1630.1410.123

0.1950.1680.1450.1250.108

0.1620.1370.1160.0990.084

0.1350.1120.0930.0780.065

0.1120.0920.0750.0620.051

0.0940.0760.0610.0490.040

0.0860.0690.0550.0440.035

0.0790.0620.0500.0390.031

0.0660.0520.0400.0320.025

0.0560.0430.0330.0250.020

1617181920

0.8530.8440.8360.8280.820

0.7280.7140.7000.6860.673

0.5340.5130.4940.4750.456

0.3940.3710.3500.3310.312

0.2920.2700.2500.2320.215

0.2180.1980.1800.1640.149

0.1630.1460.1300.1160.104

0.1230.1080.0950.0830.073

0.1070.0930.0810.0700.061

0.0930.0800.0690.0600.051

0.0710.0600.0510.0430.037

0.0540.0450.0380.0310.026

0.0420.0340.0280.0230.019

0.0320.0260.0210.0170.014

0.0280.0230.0180.0140.012

0.0250.0200.0160.0120.010

0.0190.0150.0120.0090.007

0.0150.0120.0090.0070.005

2122232425

0.8110.8030.7950.7880.780

0.6600.6470.6340.6220.610

0.4390.4220.4060.3900.375

0.2940.2780.2620.2470.233

0.1990.1840.1700.1580.146

0.1350.1230.1120.1020.092

0.0930.0830.0740.0660.059

0.0640.0560.0490.0430.038

0.0530.0460.0400.0350.030

0.0440.0380.0330.0280.024

0.0310.0260.0220.0190.016

0.0220.0180.0150.0130.010

0.0150.0130.0100.0080.007

0.0110.0090.0070.0060.005

0.0090.0070.0060.0050.004

0.0080.0060.0050.0040.003

0.0060.0040.0030.0030.002

0.0040.0030.0020.0020.001

2627282930

0.7720.7640.7570.7490.742

0.5980.5860.5740.5630.552

0.3610.3470.3330.3210.308

0.2200.2070.1960.1850.174

0.1350.1250.1160.1070.099

0.0840.0760.0690.0630.057

0.0530.0470.0420.0370.033

0.0330.0290.0260.0220.020

0.0260.0230.0200.0170.015

0.0210.0180.0160.0140.012

0.0140.0110.0100.0080.007

0.0090.0070.0060.0050.004

0.0060.0050.0040.0030.003

0.0040.0030.0020.0020.002

0.0030.0020.0020.0020.001

0.0020.0020.0020.0010.001

0.0020.0010.0010.0010.001

0.0010.0010.0010.001

Appendix B Present – Value Factors For Annuities

Periods Until Pay-ments

1% 2% 4% 6% 8% 10% 12% 14% 15% 16% 18% 20% 22% 24% 25% 26% 28% 30%

12345

0.9901.9702.9413.9024.853

0.9801.9422.8843.8084.713

0.9621.8862.7753.6304.452

0.9431.8332.6733.4654.212

0.9261.7832.5773.3123.993

0.9091.7362.4873.1703.791

0.8931.6902.4023.0373.605

0.8771.6472.3222.9143.433

0.8701.6262.2832.8553.352

0.8621.6052.2462.7983.274

0.8471.5662.1742.6903.127

0.8331.5282.1062.5892.991

0.8201.4922.0422.4942.864

0.8061.4571.9812.4042.745

0.8001.4401.9522.3622.689

0.7941.4241.9232.3202.635

0.7811.3921.8682.2412.532

0.7691.3611.8162.1662.436

678910

5.7956.7287.6528.5669.471

5.6016.4727.3258.1628.983

5.2426.0026.7337.4358.111

4.9175.5826.2106.8027.360

4.6235.2065.7476.2476.710

4.3554.8685.3355.7596.145

4.1114.5644.9685.3285.650

3.8894.2884.6394.9465.216

3.7844.1604.4874.7725.019

3.6854.0394.3444.6074.833

3.4983.8124.0784.3034.494

3.3263.6053.8374.0314.192

3.1673.4163.6193.7863.923

3.0203.2423.4213.5663.682

2.9513.1613.3293.4633.571

2.8853.0833.2413.3663.465

2.7592.9373.0763.1843.269

2.6432.8022.9253.0193.092

1112131415

10.36811.25512.13413.00413.865

9.78710.57511.34312.10612.849

8.7609.3859.98610.56311.118

7.8878.3848.8539.2959.712

7.1397.5367.9048.2448.559

6.4956.8147.1037.3677.606

5.9376.1946.4246.6286.811

5.4535.6605.8426.0026.142

5.2345.4215.5835.7245.847

5.0295.1975.3425.4685.575

4.6564.7934.9105.0085.092

4.3274.4394.5334.6114.675

4.0354.1274.2034.2654.315

3.7763.8513.9123.9624.001

3.6563.7253.7803.8243.859

3.5443.6063.6563.6953.726

3.3353.3873.4273.4593.483

3.1473.1903.2233.2493.268

1617181920

14.71815.56216.39817.22618.046

13.57814.29214.99215.67816.351

11.65212.16612.65913.13413.590

10.10610.47710.82811.15811.470

8.8519.1229.3729.6049.818

7.8248.0228.2018.3658.514

6.9747.1207.2507.3667.469

6.2656.3736.4676.5506.623

5.9546.0476.1286.1986.259

5.6695.7495.8185.8775.929

5.1625.2225.2735.3165.353

4.7304.7754.8124.8444.870

4.3574.3914.4194.4424.460

4.0334.0594.0804.0974.110

3.8873.9103.9283.9423.954

3.7513.7713.7863.7993.808

3.5033.5183.5293.5393.546

3.2833.2953.3043.3113.316

2122232425

18.85719.66020.45621.24322.023

17.01117.65818.29218.91419.523

14.02914.45114.85715.24715.622

11.76412.04212.30312.55012.783

10.01710.20110.37110.52910.675

8.6498.7728.8838.9859.077

7.5627.6457.7187.7847.843

6.6876.7436.7926.8356.873

6.3126.3596.3996.4346.464

5.9736.0116.0446.0736.097

5.3845.4105.4325.4515.467

4.8914.9094.9254.9374.948

4.4764.4884.4994.5074.514

4.1214.1304.1374.1434.147

3.9633.9703.9763.9813.985

3.8163.8223.8273.8313.834

3.5513.5563.5593.5623.564

3.3203.3233.3253.3273.329

2627282930

22.79523.56024.31625.06625.808

20.12120.70721.28121.84422.396

15.98316.33016.66316.98417.292

13.00313.21113.40613.59113.765

10.81010.93511.05111.15811.258

9.1619.2379.3079.3709.427

7.8967.9437.9848.0228.055

6.9066.9356.9616.9837.003

6.4916.5146.5346.5516.566

6.1186.1366.1526.1666.177

5.4805.4925.5025.5105.517

4.9564.9644.9704.9754.979

4.5204.5244.5284.5314.534

4.1514.1544.1574.1594.160

3.9883.9903.9923.9943.995

3.8373.8393.8403.8413.842

3.5663.5673.5683.5693.569

3.3303.3313.3313.3323.332