As mentioned in earlier sections, cash flows are the amounts of money estimated for future projects or observed for project events that have taken place. All cash flows occur during specific time periods, such as 1 month, every 6 months, or 1 year. Annual is the most common time period. For example, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoing cash flows. And an estimated receipt of $500 every month for 2 years is a series of 24 incoming cash flows. Engineering economy bases its computations on the timing, size, and direction of cash flows.

Of all the steps in Figure (1–1) that outline the engineering economy study, estimating cash flows (step 3) is the most difficult, primarily because it is an attempt to predict the future. Some examples of cash flow estimates are shown here. As you scan these, consider how the cash inflow or outflow may be estimated most accurately.

Cash Inflow Estimates:

Income: -$150,000 per year from sales of solar-powered watches

Savings: -$24,500 tax savings from capital loss on equipment salvage

Receipt: -$750,000 received on large business loan plus accrued interest

Savings: -$150,000 per year saved by installing more efficient air conditioning

Revenue: -$50,000 to -$75,000 per month in sales for extended battery life iPhones

Cash Outflow Estimates:

Operating costs: -$230,000 per year annual operating costs for software services

First cost: -$800,000 next year to purchase replacement earthmoving equipment

Expense: -$20,000 per year for loan interest payment to bank

Initial cost: -$1 to -$1.2 million in capital expenditures for a water recycling unit

All of these are point estimates, that is, single-value estimates for cash flow elements of an alternative, except for the last revenue and cost estimates listed above. They provide a range estimate, because the persons estimating the revenue and cost do not have enough knowledge or experience with the systems to be more accurate. Once all cash inflows and outflows are estimated (or determined for a completed project), the net cash flow for each time period is calculated.

where NCF is net cash flow, R is receipts, and D is disbursements.

At the beginning of this section, the timing, size, and direction of cash flows were mentioned as important. Because cash flows may take place at any time during an interest period, as a matter of convention, all cash flows are assumed to occur at the end of an interest period.

Figure (1-4): A typical cash flow time scale for 5 years

Figure (1-5): Example of positive and negative cash flows

It is important to understand that future (F) and uniform annual (A) amounts are located at the end of the interest period, which is not necessarily December 31. Remember, end of the period means end of interest period, not end of calendar year.

The cash flow diagram is a very important tool in an economic analysis, especially when the cash flow series is complex. It is a graphical representation of cash flows drawn on the y axis with a time scale on the x axis. The diagram includes what is known, what is estimated, and what is needed. That is, once the cash flow diagram is complete, another person should be able to work the problem by looking at the diagram.

Cash flow diagram time t = 0 is the present, and t = 1 is the end of time period 1. We assume that the periods are in years for now. The time scale of Figure (1–4) is set up for 5 years. Since the end-of-year convention places cash flows at the ends of years, the “1” marks the end of year 1.

The direction of the arrows on the diagram is important to differentiate income from outgo.

§  A vertical arrow pointing up indicates a positive cash flow.

§  A down-pointing arrow indicates a negative cash flow.

§  We will use a bold, colored arrow to indicate what is unknown and to be determined.

For example, if a future value F is to be determined in year 5, a wide, colored arrow with F= ? is shown in year 5. The interest rate is also indicated on the diagram. Figure (1–5) illustrates a cash inflow at the end of year 1, equal cash outflows at the end of years 2 and 3, an interest rate of 4% per year, and the unknown future value F after 5 years. The arrow for the unknown value is generally drawn in the opposite direction from the other cash flows; however, the engineering economy computations will determine the actual sign on the F value.

Before the diagramming of cash flows, a perspective or vantage point must be determined so that + or – signs can be assigned and the economic analysis performed correctly. Assume you borrow $8500 from a bank today to purchase an $8000 used car for cash next week, and you plan to spend the remaining $500 on a new paint job for the car two weeks from now. There are several perspectives possible when developing the cash flow diagram—those of the borrower (that’s you), the banker, the car dealer, or the paint shop owner. The cash flow signs and amounts for these perspectives are as follows.

Figure (1-6): Cash flows from perspective of borrower for loan and purchases.

One, and only one, of the perspectives is selected to develop the diagram. For your perspective, all three cash flows are involved and the diagram appears as shown in Figure (1–6) with a time scale of weeks. Applying the end-of-period convention, you have a receipt of +$8500 now (time 0) and cash outflows of -$8000 at the end of week 1, followed by -$500 at the end of week 2.

Each year Exxon-Mobil expends large amounts of funds for mechanical safety features throughout its worldwide operations. Carla Ramos, a lead engineer for Mexico and Central American operations, plans expenditures of $1 million now and each of the next 4 years just for the improvement of field-based pressure-release valves. Construct the cash flow diagram to find the equivalent value of these expenditures at the end of year 4, using a cost of capital estimate for safety-related funds of 12% per year.

Figure (1–7) indicates the uniform and negative cash flow series (expenditures) for five periods, and the unknown F value (positive cash flow equivalent) at exactly the same time as the fifth expenditure. Since the expenditures start immediately, the first $1 million is shown at time 0, not time 1. Therefore, the last negative cash flow occurs at the end of the fourth year, when F also occurs. To make this diagram have a full 5 years on the time scale, the addition of the year -1 completes the diagram. This addition demonstrates that year 0 is the end-of-period point for the year -1.

Figure (1-7): Cash flow diagram, Example (1-8).

An electrical engineer wants to deposit an amount P now such that she can withdraw an equal annual amount of A1 = $2000 per year for the first 5 years, starting 1 year after the deposit, and a different annual withdrawal of A2 = $3000 per year for the following 3 years. How would the cash flow diagram appear if i= 8.5% per year?

The cash flows are shown in Figure (1–8). The negative cash outflow P occurs now. The withdrawals (positive cash inflow) for the A1 series occur at the end of years 1 through 5, and A2 occurs in years 6 through 8.

Figure (1-9): Cash flow diagram with two different A series, Example (1-9).

1.     Economic Equivalence

Economic equivalence is a fundamental concept upon which engineering economy computations are based. Before we delve into the economic aspects, think of the many types of equivalency we may utilize daily by transferring from one scale to another. Some example transfers between scales are as follows:

Length:

12 inches = 1 foot            3 feet = 1 yard            39.370 inches = 1 meter

100 centimeters = 1 meter            1000 meters = 1 kilometer            1 kilometer = 0.621 mile

Pressure:

1 atmosphere =1 newton/meter ² = 10 ³ pascal = 1 kilopascal

Speed:

1 mile = 1.609 kilometers            1 hour = 60 minutes

110 kilometers per hour (kph) = 68.365 miles per hour (mph)

68.365 mph = 1.139 miles per minute

As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to $106 one year from today.

If someone offered you a gift of $100 today or $106 one year from today, it would make no difference which offer you accepted from an economic perspective. In either case you have $106 one year from today. However, the two sums of money are equivalent to each other only when the interest rate is 6% per year. At a higher or lower interest rate, $100 today is not equivalent to $106 one year from today.

In addition to future equivalence, we can apply the same logic to determine equivalence for previous years. A total of $100 now is equivalent to $100/1.06 = $94.34 one year ago at an interest rate of 6% per year. From these illustrations, we can state the following: $94.34 last year, $100 now, and $106 one year from now are equivalent at an interest rate of 6% per year. The fact that these sums are equivalent can be verified by computing the two interest rates for 1-year interest periods.

The cash flow diagram in Figure (1–10) indicates the amount of interest needed each year to make these three different amounts equivalent at 6% per year.

Figure (1-10): Equivalence of money at 6% per year interest.

Manufacturers make backup batteries for computer systems available to Batteries+ dealers through privately owned distributorships. In general, batteries are stored throughout the year, and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries+ outlet. Make the calculations necessary to show which of the following statements are true and which are false about battery costs.

1)      The amount of $98 now is equivalent to a cost of $105.60 one year from now.

2)      A truck battery cost of $200 one year ago is equivalent to $205 now.

3)      A $38 cost now is equivalent to $39.90 one year from now.

4)      A $3000 cost now is equivalent to $2887.14 one year earlier.

5)      The carrying charge accumulated in 1 year on an investment of $20,000 worth of batteries is $1000.

1)      Total amount accrued = 98(1.05) = $102.90 ≠ $105.60; therefore, it is false. Another way to solve this is as follows: Required original cost is 105.60/1.05 = $100.57 ≠ $98.

2)      Equivalent cost 1 year ago is 205.00/1.05 = $195.24 ≠ $200; therefore, it is false.

3)      The cost 1 year from now is $38(1.05) = $39.90; true.

4)      Cost now is 2887.14(1.05) = $3031.50 ≠ $3000; false.

5)      The charge is 5% per year interest, or $20,000(0.05) = $1000; true.