# Feasibility Study

9.1 INTRODUCTION

A feasibility study is an analysis to determine whether the project is technically and financially feasible that takes all of a project’s relevant factors into account—including economic, technical, legal, and scheduling considerations. This is done in order to determine if the project under consideration would be completed successfully.

In other words, the study tries to determine whether the project is technically and financially feasible. Financially feasible means whether the project is feasible within the estimated cost. A feasibility study also determines whether a project makes good business sense, i.e., will it be profitable?

9.2 PROPOSED METHODOLOGY

The following is a proposed outline to follow for solving a feasibility study aiming for financial viability of a project:

9.3 APPLICATIONS

SELECTED CASE NO 1

The following example illustrates the application of a feasibility study in order to provide an assessment of the practicality of a proposed project. The project involves the installment of a waste-heat-recovery system.

A waste heat recovery unit (WHRU) is an energy recovery heat exchanger that transfers heat from process outputs at high temperature to another part of the process for some purpose, usually increased efficiency. Waste heat found in the exhaust gas of various processes or even from the exhaust stream of a conditioning unit can be used to preheat the incoming gas. This is one of the basic methods for recovery of waste heat, as shown in Figure 9.1.

Consider the unit shown in the diagram. Figure 9.2, a two-pass water tube waste heat boiler. It was suggested to buy such unit and install it in an oil field. The following information is provided in order to carry out a feasibility study:

• • Cost of unit = \$250,000 with lifetime 10 years
• • Value of waste-heat to be recovered by the unit is anticipated to be = \$40,000 annually
• • Maintenance fees for the unit = \$10,000 annually
• • Money could be invested by the management at 10% annually

What is your recommendation using the feasibility study?

FIGURE 9.1 Proposed outline to solve feasibility study problems.

SOLUTION

The feasibility analysis indicates a profit of \$5,000. This is equivalent to ARR =[5,000/250,000] x 100 = 2%.

If money would be invested by the management at 10% annual rate of return:

CONCLUSION

The feasibility study shows that under the given conditions, it is not feasible to install the waste heat boiler (WHB).

FIGURE 9.2 A two-pass waste heat boiler.

SELECTED CASE NO 2

A feasibility study carried out for an oil company indicated that it is possible to invest \$1 million in either one of two projects. Anticipated cash flows generated by the two projects over the useful lifetime are given next: Table 9.1.

Give your recommendations of which project you choose based on the net present value (N.P.V.). Use selected values for the discount interest rate (more than one).

Compute the discounted cash flow rate of return (D.C.F.R.) for each project.

SOLUTION

For (1), calculation is clone for three different discount interest rates: 8%, 10%, and 12%. In addition, a graphic plot is presented (Figure 9.1 for the change of the discounted value (present value) of the cash flows for both projects with the discount rate.

Next example is fordistillaion column, shown in Table 9.2 (Table 6.7)

TABLE 9.2

N.P.V. of a Distillation

 Distillation Unit Initial Cost \$200,000 Useful Life / Year 10 Salvage Value \$10,000 Net Cash Flow each Year \$50,000 D.C.F.R \$10 Present value of cash flow of 50000 annally, for 10 years at 10 % \$307,250 The Peresnt Value of Cash Flows for 10 years, Minus original investment of \$200,000 \$107,250 Persent Value of \$10,000 salavage value to be recived at the end of years at 10% \$3,860 Total value of net cash receipts plus persent value \$111,110

Therefore, as the example shows, the choice between the two projects depends on the discount rate used. Usually, the oil company’s cost of capital for investing in the project will determine which project is selected.

From the data, it can be seen that at a discount of 12%, the present value of cash flow' from project 2 is \$1 million; and at a discount of over 12%, the present value of cash flow gives us the discount rated amount of under \$1 million. This analysis of the present value of cash flow gives us the discount rate at which anticipated cash flow' equals the initial investment, which is the D.C.F.R. For project 2, it is about 12%; for project 1, it is about 13%.

# Model Solved Examples

PROBLEM 10.1

A sum of \$10,000 is borrowed by a refining oil company. Propose four different equivalent plans of money payments for this capital over a period of 10 years assuming the interest rate is 6%.

SOLUTION

As shown in Table 10.1, plan I involves the annual payment of the interest only (\$600) until the end. Plans II and III involve systematic reduction of the principal of the debt (\$10,000). For plan II this is done by uniform repayment of principal (\$1,000/ yr) along with diminishing interest, while for plan III a scheme is devised to allow for uniform annual payment for both capital and interest all the way through until the end (\$1,359). For plan IV, on the other hand, payment is done only once at the end of the 10th year. The equivalence of the four payments is further illustrated in Figure 10.1

PROBLEM 10.2

Assume a petroleum company investment of \$10 million for an expansion to a current refinery, allocated \$1,000,000 for land and \$7,000,000 for fixed and other physical properties subject to depreciation. Additional capital of \$2,000,000 is available for operation purposes, but this sum is not subject to depreciation. Investors want a 15% interest rate (or earning rate to investors) on their money for a 10-year period. The sinking-fund method will be used, with depreciation figured at 15% per year. No income taxes are involved in order to simplify the example.

SOLUTION

First-year profit before deducting the sinking-fund depreciation charge made at the earning rate of 15% interest, and assuming no salvage value for the physical properties, is 0.15 x \$1,000,000, or \$150,000 per year.

But the oil company must earn enough additional money annually to pay for the depreciation occurring on the depreciable capital of \$700,000.

Using sinking-fund depreciation and a 15% interest rate for the sinking fund, the annual deposit in the fund is given by:

TABLE 10.1

Summary for the Four Plans for Solving Example 10.1

 Year Investment I (*) П <*) Ш 9) IV w 0 S 10.000 1 «00 l.«00 1.359 2 «00 1.540 1.359 3 «00 1.480 1.359 4 «00 1.420 1.359 5 «00 13«0 1.359 6 «00 1.300 1.359 7 «00 1.240 U«9 £ «00 1.180 1,359 9 «00 1.120 1.359 10 10,600 1.0«0 1.359 17,908

Thus, company profits before depreciation must total \$184,440 (\$150,000 + \$34,440) and not merely \$150,000 in the first year. Actually, the \$184,440 in the first year represents:

\$34,440 = the sum of annual depreciation charge

\$105,000 = the 15% interest on the un-depreciated part of the depreciable capital which is, in the first year or before any deductions, 0.15 x \$700,000 \$45,000 = the 15% interest on the non depreciable capital, or 0.15 x \$300,000 \$184,440 = the total for the first year

Thus \$139,440 (\$105,000 + \$34,440) is needed to cover (1) the depreciation deposit in the sinking fund, and (2) the interest on the depreciable capital for that year. This is also calculated by using:

FIGURE 10.1 Equivalence of the four payments.

In each succeeding year, the book value of the depreciable capital decreases, but the depreciation reserve increases in such a manner that the sum of the two always equals to \$700,000 and the total annual interest remains constant at \$105,000 even though the interest charges on each component vary.

The biggest drawback to the actual use of the sinking-fund method in business is the fact that businesses rarely maintain an actual depreciation sinking fund. The interest rate which could be obtained on such deposits would be small, probably not over 6% in the petroleum business, according to financial experts of the oil industry. An active business, such as an oil company operation, is constantly in need of working capital. This capital will usually earn much more than 6%.

A reasonable rule is that all values should be kept invested in the oil business and not remain idle. As a result, a fictitious depreciation fund is often used: The amounts which have been charged to.

PROBLEM 10.3

With reference to the investment made to procure boilers for surface facilities in an oil field, as shown in Table 10.2, calculate the payback period (P.P.) for each alternative and give reasons for selecting one and not the other.

SOLUTION

P.P. is readily calculated as follows:

As far as the P.P. as a criterion for choice, the number of years to recover the depreciable capital is the same for both types of boilers. However, the recovery of investment for boiler 1 is faster than for boiler 2 (e.g.„ compare \$20,000 to \$5,000 for the 1st year), as shown in Figure 10.2. Therefore, from the standpoint of cost of money (time value of money), investment in boiler 1 is preferable to investment in boiler 2.

TABLE 10.2

Comparison of Two Boiler Investment (Example 10.3)

 Cash Flow Year Boiler 1 Boiler 2 0 50,000 50,000 1 20,000 5,000 2 15,000 10,000 3 10,000 15,000 4 5,000 20,000 Total Cash Flow 50,000 50,000 Payback period P.P 1 = 4 Years P.P 2 = 4 Years

FIGURE 10.2 Solution of Problem I0.3.

This example points out that, when using the payout period method, oil management should also observe the rapidity of cash flows between alternatives. The alternatives may have the same number of years-to-pay-back, as they do here, but one may be more favorable than the other because the largest amount of cash flow comes in the first few years. This could be an excellent point in favor of investment in one alternative over another when both have approximately the same payout periods. It could be a strong factor in selection of one especially if a greater amount of cash “back” is needed early in the investment (Figure 10.3).

FIGURE 10.3 Change of return on investment (R.O.I.) percentage saving versus different types.

10.1 MODEL EXAMPLES Example 10.1

If an oil company expects a cash flow of \$800,000 by the end of 10 years, and 10% is the current interest rate on money, calculate the net present value (N.P.V.) of this venture.

SOLUTION

No capital investment is involved here, so the problem is simply a discounting procedure.

The present value of the cash flow:

Example 10.2

Assume that a distillation unit with an initial cost of \$200,000 is expected to have a useful life of 10 years, with a salvage value of \$10,000 at the end of its life. Also, it is expected to generate a net cash flow above maintenance and expenses amounting to \$50,000 each year. Assuming a selected discount rate of 10%, calculate the N.P.V.

where this factor is readily obtained from tables found in Appendix A. Calculations are given in the Table 10.3.

TABLE 10.3

N.P.V. for Distillation Column

 Distillation Unit Initial Cost \$200,000 Useful Life / Year 10 Salvage Value \$10,000 Net Cash Flow each Year \$50,000 D.C.F.R \$10 Present value of cash flow of 50000 annally, for 10 years at 10 % \$307,250 The Peresnt Value of Cash Flows for 10 years, Minus original investment of \$200,000 \$107,250 Persent Value of \$10,000 salavage value to be recived at the end of years at 10% \$3,860 Total value of net cash receipts plus percent value \$111,110

TABLE 10.4

Summary for the Five Types

 Capital Investment (C.l.) Type I 10,000 Type II 16,000 Type III 20,000 Type IV 26,000 n, years 5 5 5 5 A. dep. = C.I./n 2,000 3,200 4,000 5,200 A. oper. cost 100 100 100 100 Total annual cost = A. dep. + A. oper. cost 2,100 3,300 4,100 5,300 Revenue (income) \$/yr 4,100 6,000 6,900 8,850 Annual profit 2,000 2,700 2,800 3,550 R.O.I. 20.0% 16.9% 14.0% 13.7%

Example 10.3

Instead of flaring the associated natural gas separated along with crude oil, it was decided to recover the lost heat by using the waste-heat recovery system (W.H.R.S.). For pilot test runs, four designs were offered; each has a lifetime of 5 years. The savings and costs associated with each are as follows (Table 10.4).

All four designs seem to be acceptable as far as the minimum annual R.O.I., exceeding 10% (required by management). Which design is to be recommended?

SOLUTION

Using incremental comparison:

 1 2 3 4 1st comparing 1 to 2 acceptable as a basis 11.7% — — 2nd comparing 2 to 3 — -basis- 2.5% — 3rd comparing 2 to 4 — -basis- — 8.5%

Conclusion: Design 2 is recommended; it gives more profit than design 1, while return on extra investment (R.O.E.I.) is 11.7%, which is >10% (minimum).

Figures 10.3 and 10.4 are bar charts to illustrate the solution of the problem.

FIGURE 10.4 Incremental comparison versus different types.

Example 10.4

Given: Consider two possibilities related to the purchase of a heat exchanger for an oil refinery to replace an older model for which annual costs are running around \$20,950. Other details are as follows:

Purchase possibility A is a heat exchanger constructed with materials of steel and copper. Its investment cost is \$15,000. Its economic service life is estimated to be 10 years, and salvage value at the end of the 10th year is estimated at \$500. Annual labor, maintenance, repairs, and operational expenses are estimated at \$11,500; other annual direct costs are 4% of the investment cost of \$15,000, or \$600, when operating under optimum conditions.

Purchase possibility В is a stainless steel heat exchanger with an investment value of \$40,000. Its economic life is also regarded as 10 years, with a scrap value of \$1,000 at end of the 10th year. Annual labor, maintenance, repairs, and operational expenses are estimated at \$4,000; other annual direct costs are 7% of the investment cost of \$40,000, or \$2,800, when operating under optimum conditions.

The current cost of capital is 8%.

Find: Using the annual cost method, determine which purchase possibility would be more economical with respect to annual costs.

SOLUTION

 Purchase possibility A with capital recovery, formula "Find A, Given P" Purchase possibility В with capital recovery, formula "Find A, Given P" (Original cost - salvage value) (recovery factor) + (salvage value) (interest rate) (Original cost - salvage value) (recovery factor) + (salvage value) (interest rate) (\$15,000 - \$500) (0.1490) + (\$500) (0.08) = \$2,201 capital recovery of original cost and salvage value (\$40,000-51,000) (0.1490) + (\$1,000) (0.08) = \$5,891 capital recovery of original cost and salvage value Summary of annual costs with capital recovery Summary of annual costs with capital recovery Recovery of capital \$2,201 Recovery of capital \$5,891 Annual costs, maint., repairs 11,500 Annual costs, maint., repairs 4,000 Annual costs, optimum conditions 600 Annual costs, optimum conditions 2,800 Total annual costs \$14,301 Total annual costs \$12,691

With a potential savings in annual cost of \$1,610 (\$14,301 - \$12,691) in favor of the stainless steel heat exchanger, purchase possibility В appears to be the more feasible "buy" according to the annual cost method. (Only differences in costs, with cost items common to both purchase possibilities, were used.) Furthermore, the salvage value of the stainless steel exchanger (\$1,000) is \$500 more than for the steel-copper exchanger (Figure 10.5).

FIGURE 10.5 Comparison between types A and B.

The annual cost method is used where the same costs for each alternative recur annually almost in the same manner. For a series of costs which are nonuniform, an average annual cost equivalent might be calculated. For alternatives with different lifetimes, the time period for comparison might be that of the alternative with the shortest life.

Whereas the annual cost method does not give the relative amounts of capital, but the present value method does. The present value method reduces all costs to equivalent capital at a given date.

Example 10.4 (summary):

 A: (Steel and Copper) Heat Exchanger B: (Stainless Steel) Heat Exchanger Capital Investment cost (C.l.) 15,000 40,000 Lifetime (yr) 10 10 Salvage value 500 1,000 Annual labor, maintenance, repairs, operational expenses costs (\$) and 11,500 4,000 Annual direct costs (\$) 600 2,800 The current cost of capital is 8% 1,200 3,200 Capital recovery 2,201 5,891 Total annual costs 14,301 12,691

Purchase possibility В appears to be the more feasible "buy" according to the annual cost method. (Only differences in costs, with cost items common to both purchase possibilities, were used.) Furthermore, the salvage value of the stainless steel exchanger (\$1,000) is \$500 more than for the steel-copper exchanger.

Example 10.5

A feasibility study carried out for an oil company indicated that it is possible to invest \$1 million in either one of two projects. Anticipated cash flows generated by the two projects over the useful lifetime are given in Table 10.4.

• 1. Give your recommendations of which project you choose based on the N.P.V. Use selected values for the discount interest rate (more than one).
• 2. Compute the discounted cash flow rate of return (D.C.F.R.) for each project.

SOLUTION

For (1), calculation is done for three different discount interest rates: 8%, 10%, and 12%, as shown in Table 10.5. In addition, a graphic plot is presented (Figure 6.8) for the change of the discounted value (present value) of the cash flows for both projects with the discount rate.

In summarizing the results of Table 10.5, if the cash flows of project 1 and project 2 are discounted at 8%, project 2 is preferable; if the cash flows are discounted at 10%, project 2 is preferred to project 1 because the present value of project 2 is almost \$14,000 more; and if the cash flows are discounted at 12%, project 1 is slightly preferable to project 2, and will continue to be preferable to project 2 as discount rates go higher than 12%.

Therefore, as the example shows, the choice between the two projects depends on the discount rate used. Usually, the oil company's cost of capital for investing in the project will determine which project is selected.

From the data, it can be seen that at a discount of 12%, the present value of cash flow from project 2 is \$1 million; and at a discount of over 12%, the present value of cash flow gives us the discount rated amount of under \$1 million. This analysis of the present value of cash flow gives us the discount rate at which anticipated cash flow equals the initial investment, which is the D.C.F.R. For project 2, it is about 12%; for project 1, it is about 13%.

Example 10.6

During field operations, the manager in charge is considering the purchase and the installation of a new pump that will deliver crude oil at a faster rate than the existing one.

The purchase and the installation of the new pump will require an immediate layout of \$15,000. This pump, however, will recover the costs by the end of 1 year. The relevant cash flows for the case.

TABLE 10.5

 Solution of Example 10.6 Year Protect 1 Project 2 1 400,000 100,000 2 320,000 200,000 3 200,000 300,000 4 300,000 400,000 5 100,000 500,000 Total anticipated cash Flow 1.320.000 1.500.000 At 8% Discount Year Discount factor for 8% Project 1 Cash How Discounted Value (col. 1*col. 2) Project 2 Cash Flow Discounted Value (col. 1*col. 4) 1 2 3 4 5 1 0.926 400,000 370,400 100,000 92,600 2 0.856 320,000 273,920 200,000 171,200 3 0.794 200,000 158,800 300,000 238,200 4 0735 300,000 220,500 400,000 294,000 s 0.681 100,000 68,100 500,000 340,500 I_ \$1,091,723 \$1,136,505 At 10% Discount Year Discount factor for 10% Project 1 Cash How Discounted Value (col. 1*с<й. 2) Project 2 Cash Flow Discounted Value (col. 1*cof. 4) 1 2 3 4 5 1 0.909 400,000 363,600 100,000 90,900 2 0.826 320,000 264,320 200,000 165,200 3 0.751 200,000 150,200 300,000 225,300 4 0 683 300,000 204,900 400,000 273,200 5 0621 100,000 62,100 500,000 310,500 I \$1,045,123 \$1,065,105 At 12% Discount Year Discount factor for 12% Project 1 Cash Row Discounted Value (col. 1*coj. 2) Project 2 Cash Row Discounted Value (col. 1*coi. 4) 1 2 3 4 5 1 0 893 400,000 357,200 100,000 89,300 2 0797 320,000 255,040 200,000 159,400 3 0.712 200,000 142,400 300,000 213,600 4 0636 300,000 190,800 400,000 254,400 s 0.567 100,000 56,700 500,000 283,500 I_ \$1,002,143 \$1,000,205
 Year 0 1 2 Install new (larger pump) -15,000 19,000 0 Operate existing (old pump) 0 95,000 95,000

If the oil company requires 10% minimum annual rate of return on money invested, which alternative should be chosen?

SOLUTION

The present worth method is applied in solving this problem (see Chapter 6). Calculate the present worth for both alternatives, where:

Present worth = Present values of cash flows, discounted at 10% - Initial capital investment

a. For the new pump: P.V. = (190,000j/1.1 = \$172,727

b. For the old pump: P.V. = (95,000)/1.1 + (95,000)/(1.1)2

Based on the above results, keep the old pump. It gives higher present value. Example 10.7

The XVZ oil production company was offered a lease deal for oil wells on which the primary reserves are close to exhaustion. The major condition of the deal is to carry out secondary recovery operation using water-flood at the end of the 5 years. No immediate payment by the XYZ Company is required. The relevant cash flows are estimated as given in Table 10.6.

TABLE 10.6 Data for Example

 Year Net Present Worth @ 10% 0 1-4 5 6-20 0 550,000 -\$650,000 S 100,000 5227,000

The decision to be made: should the lease and the secondary flood proposal be accepted?

justify your answer, and check the present worth (P.W.) value.

SOLUTION

The fact that the proposal at hand gives a positive P.W., makes it a viable one. The project should be undertaken.

Next, calculation is carried out to check the P.W. reported in Table 10.2.

The cash flows are discounted to present values, at 10%. Using the compound interest factors listed in Appendix B, the following results are obtained: