Part 1 : Understanding Money And Its Management (Ch1,2,3,4) Part 2: Evaluating Business and Engineering Assets (5,6,7)

CHAPTER 5 Present Worth Analysis

Payback period

Chapter Opening Story – Calculating Costs in the Clouds Nature of Project: Fuel savings through flight route-mapping software Investment cost: $100 million Expected savings: $20 million per year Is it worth investing in the project? United Airlines

Ultimate Questions Is it worth investing in a new aviation software to save $20 million per year, say over 10 years? How long does it take to recover the initial investment? What kind of interest rate should be used in evaluating business investment opportunities?

Loan versus Project Cash Flows

Describing Project Cash Flows Year (n) Cash Inflows (Benefits) (B) Cash Outflows (Costs) (C) Net Cash Flows (B – C) $1,800,000 -$1,800,000 1 908,000 2 … 7 1,268,000

Payback Period Principle: How fast can I recover my initial investment? Method: Based on cumulative cash flow (or accounting profit) Screening Guideline: If the payback period is less than or equal to some specified payback period, the project would be considered for further analysis. Weakness: Does not consider the time value of money

Example 5.1 Payback Period Current cost (% saved) Savings Setup $ 335,000 70% $ 234,500 Scrap/rework $ 58,530 85% $ 49,751 Operators $ 220,000 100% Fixturing $ 185,000 $ 157,250 Programming time $ 80,000 60% $ 48,000 Floor space $ 35,000 65% $ 22,750 Maintenace $ 45,000 $ 27,000 Coolant $ 15,000 50% $ 7,500 Inspection $ 120,000 Documentation $ 5,000 $ 2,500 Expediting $ 25,000 75% $ 18,750 Total Annual Saving $908,000

Example 5.1 Payback Period N Cash Flow Cum. Flow 1 2 3 4 5 6 7 -$1,800,000 $454,000 $681,000 $908,000 $1,268,000 -$1,800,000 -$1,346,000 -$665,000 $243,000 $1,151,000 $2,059,000 $2,967,000 $4,235,000 50% 75% Payback period should occurs somewhere between N = 2 and N = 3 or 2.73 years

Figure 5-2 Cumulative Project Cash Flows over Time

Discounted Payback Period Principle: How soon can I recover my initial investment plus interest? Method: Based on cumulative discounted cash flows Screening Guideline: If the discounted payback period (DPP) is less than or equal to some specified payback period, the project would be considered for further analysis. Weakness: Cash flows occurring after DPP are ignored

Example 5.2 Discounted Payback Period 3.53 years * Cost of funds = (Unrecovered beginning balance) X (interest rate)

Key Points Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability.

Present Worth Analysis

Net Present Worth Decision Rule

Evaluation of a single project Determine the interest rate (required interest rate, minimum attractive interest rate MARR) Estimate the service life of the project Estimate the cash inflow for each period of the service life Determine the cash outflow for each period over the service life Determine the net cash flows for each period Find the present worth of each net cash flow at MARR NPW?

Example 5.3 EoY Cash flow $-1,800,000 1 $454,000 2 $681,000 3 $908,000 $-1,800,000 1 $454,000 2 $681,000 3 $908,000 4 5 6 7 $1,268,000

Excel Solution using NPV Function A B C 1 Period Cash Flow 2 ($1,800,000) 3 $454,000 4 $681,000 5 $908,000 6 7 8 9 $1,268,000 10 PW(15%)= $1,546,571 =NPV(15%,B3:B9)+B2

Present Worth Amounts at Varying Interest Rates

Figure 5-5 – Net Present Worth Profile

An Excel Worksheet to Illustrate the Process of Computing the NPW

Figure 5-6 Elements of Required Return (MARR)

Can you explain what a typical NPW really means? Suppose the company has $1,800,000. It has two options. Option 1:Take the money out and invest it in the project or Option 2: Leave the money in the firm’s treasury that can be placed in investments that yield a return equal to the MARR. These funds may be viewed as an investment pool. Let’s see what the consequences are for each option.

Figure 5-7 The Concept of Investment Pool

Project Balances

Net Future Worth and Project Balance Diagram

Key Parameters in Project Balance Diagram and Their Economic Interpretations The area of negative project balance -The exposure to financial risk The discounted payback period – The time period when the project recovers the initial investment plus the cost of the funds The area of positive project balance -The profit potential after recovering the initial investment The net future worth – The net surplus at the time of project completion

Capitalized Equivalent Worth

Example 5.4 Given: A = $2M, i = 8%, and N = ∞ Find: CE(8%) Solution: ∞ 1 2 3 ∞ CE(8%) = $25,000,000

Comparing Mutually Exclusive Alternatives

Common Terminologies Used Mutually Exclusive Projects Alternative vs. Project Do-Nothing Alternative

Revenue versus Service Projects Revenue Projects Projects whose revenues depend on the choice of alternatives Service Projects Projects whose revenues do not depend on the choice of alternative but must product the same amount of output (revenue)

Analysis Period versus Required Service period The time span over which the economic effects of an investment will be evaluated (study period or planning horizon). Required Service Period The time span over which the service of an equipment (or investment) will be needed.

Fundamental Principle in Comparing Mutually Exclusive Alternatives Principle: Projects must be compared over an equal time span (or analysis period). Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period.

Example 5.5: Analysis Period Equals Project Lives The net cost of not replacing the old system is now $71,384( = $222,937 - $151,533) Since the new system costs only $28,570, the replacement should be made now.

Analysis period differs from Project Lives Case 1: Project lives longer than the analysis period (Example 5.6) Case 2: Project lives shorter than the analysis period (Example 5.7)

Case 1: Project Lives Longer than Analysis Period Estimate the salvage value at the end of the required service period. Compute the PW for each project over the required service period. Select the least cost alternative.

Example 5.6 Present Worth Comparison Case 1: Project lives longer than the analysis period Allan company got the permission to harvest southern pines from one of the timberland area. They are considering purchasing a machine, which has the ability to hold, saw, and place trees in bunches to go down to the landing area. The operation on this timberland area must be completed in three years. Allan company could speed up the operation, but doing so is not desirable, as the market demand of the timber does not justify such speed. There are two possible models of machines that Allan could purchase for this job: Model A is a two year old used equipment, where as Model B is a brand new machine. Model A costs $205,000 and has a life of 10,000 hours before it will require any major overhaul. The operating cost will run $50,000 per year, for 2,000 hours of operation. At this operational rate, the unit will be operable for five years, and at the end of that time it is estimated that the salvage value will be $75,000. The more efficient Model B costs $275,000, has a life of 14,000 hours before requiring any major overhaul, and cost $32,500 to operate, for 2,000 hours per year in order to complete the job within three years. The estimated salvage value of Model B at the end of seven years is $95,000.

Since the life of either model exceeds the required service period of three years, Allan company has to assume some things about the used equipment at the end of that time. Therefore, the engineers at Allan estimate that, after three years, the Model A unit could be sold for $130,000 and Model B unit for $180,000. Allan summarized the resulting cash flows (in thousands of dollars) for the project as follows: Here, red color figures represent the estimated salvage values at the end of the analysis period (end of year 3). Assume the firm’s MARR is 15%, which option is more acceptable? N MODEL A MODEL B -$205,000 -$275,000 1 -50,000 -32,500 2 3 130,000 -50,000 180,000 -32,500 4 5 75,000 -50,000 6 7   95,000 - 32,500

Example 5.6: Project’s Life is Longer than Analysis Period PW(15%) = -$230,852 PW(15%) = -$233,684

Case 2: Project’s Life is shorter than Analysis Period Come up with replacement projects that match or exceed the required service period. Compute the PW for each project over the required service period. Select the least cost project.

Example 5.7 Present Worth Comparison Case 2: Project lives shorter than the analysis period Phoenix Manufacturing Company is planning to modernize one of its distributions located outside Denver, Colorado. Two options to move goods in the distribution center have been under consideration: A conveyor system and forklift trucks. The firm expects that the distribution center will be operational for the next 10 years, and then it will be converted into factory outlet. The conveyor system would last eight years, whereas the forklift trucks would last only six years. The two options will be designed differently, but will have identical capacities and will do exactly the same job. The expected cash flows for the two options, including maintenance costs, salvage values are as follows. N CONVEYOR SYSTEM LIFT TRUCK -$68,000 -$40,000 1 -13,000 -15,000 2 3 4 5 6 -15,000 + 4,000 7 8 -13,000 + 5,000  

SOLUTION Since each option has a shorter life than the required service period (10 years), we need to make explicit assumption of how the service requirement is to be met. If the company goes with conveyor system, it will spend $18,000 to overhaul the system to extend its service life beyond eight years. The expected salvage value of the system at the end of the required service period (10years) will be $6,000. The annual operating and maintenance costs will be $13,000. If the company goes with the lift truck option, the company will consider leasing a comparable lift truck that has annual lease payment of $8,000, payable at the beginning of each year, with an annual operating cost of $15,000 for the remaining required service period. The anticipated cash flows for both models under this scenario are shown in the table describes the cash flows associated with each option. Note that both alternatives now have the same required service period of 10 years. N CONVEYOR SYSTEM LIFT TRUCK -$68,000 -$40,000 1 -13,000 -15,000 2 3 4 5 6 -15,000 + 4,000 - 8,000 7 -15,000 - 8,000 8 -13,000 - 18,000 9 10 -13,000 + 6,000

N CONVEYOR SYSTEM LIFT TRUCK -$68,000 -$40,000 1 -13,000 -15,000 2 3 4 5 6 -15,000 + 4,000 7 8 -13,000 + 5,000   N CONVEYOR SYSTEM LIFT TRUCK -$68,000 -$40,000 1 -13,000 -15,000 2 3 4 5 6 -15,000 + 4,000 - 8,000 7 -15,000 - 8,000 8 -13,000 - 18,000 9 10 -13,000 + 6,000 Figure 5-12 Comparison for mutually exclusive service projects with unequal lives when the required service period is longer than the individual project life

PW (12%) Conveyor = - $68,000 - $13,000 (P/A, 12%, 10) - $18,000 (P/F, 12%, 8) CONVEYOR SYSTEM LIFT TRUCK -$68,000 -$40,000 1 -13,000 -15,000 2 3 4 5 6 -15,000 + 4,000 - 8,000 7 -15,000 - 8,000 8 -13,000 – 18,000 9 10 -13,000 + 6,000 Figure 5-12 Comparison for mutually exclusive service projects with unequal lives when the required service period is longer than the individual project life

PW (12%) Lift Trucks = - $40,000 - $15,000 (P/A, 12%, 10) - $8,000 (P/A, 12%, 4) (P/F, 12%, 5) Since these projects are service projects, the lift truck option is better choice. N CONVEYOR SYSTEM LIFT TRUCK -$68,000 -$40,000 1 -13,000 -15,000 2 3 4 5 6 -15,000 + 4,000 - 8,000 7 -15,000 - 8,000 8 -13,000 – 18,000 9 10 -13,000 + 6,000

SOLVED PROBLEMS 7; 13; 14; 32; 42

HOMEWORK PROBLEMS DUE DATE IS Monday, April 25th 2011 class time 11; 31; 35; 36