11+ AEV 348 2024 PDF

Fall 2024 Evaluation II Tate N. Cao Ron and Jane Graham School of Professional Development Jin Zhu Department of Chemical and Biological...

Fall 2024 Evaluation II Tate N. Cao Ron and Jane Graham School of Professional Development Jin Zhu Department of Chemical and Biological Engineering As we gather here today, we acknowledge we are on treaty 6 Territory and the Homeland of the Metis. We pay our respect to the First Nations and Metis ancestors of this place and reaffirm our relationship with one another. Announcement SLEQ (10 minutes) Paws -> SLEQ -> Access SLEQ -> GE348 01 Assignment #3: Due October 28th? #4: October 29th ~ November 6th #5-#6: after midterm 2 Case Study #1 Due November 5th (Individual submission) Midterm 1 Marking in progress 2 Today Define annual equivalent Convert any series of cash flows into its annual cash flow equivalent Apply annual equivalent value to compare alternatives with equal, common multiple or continuous lives, or a fixed study period Textbook: Chapter 4.4 , 4.5 NPV: 4.1 , 4.2 , 4.3, 4.5 3 Net Present Value (NPV) Review The alternative with the higher NPW or NPV is selected. NPW or NPV must be a POSITIVE value to make the project desirable. NEGATIVE NPV means you are going to pay for a project out-of-pocket Discount Rate The discount rate chosen is often called MARR (Minimum Acceptable/ Attractive Rate of Return) MARR must exceed the weighted average cost of capital (WACC) and the rate of return of the opportunity cost (options for investing) 4 Annual Equivalent Value Analysis Alternatives are compared based on their equivalent annual cash flows Annual Equivalent Value = Annual Equivalent Benefit - Annual Equivalent Cost We can convert the NPV into AEV 5 Benefit of AEV Comparable - When project lives are unequal Investment decisions Recourse Allocation Decision - Estimate the annual cost savings to justify the purchase of new equipment Justification of Purchase Financial Feasibility Monitoring - Compare to estimated annual cost savings to monitor the progress of the project Performance Measurement Early Detection of Issues Accountability and Reporting 6 AEV Analysis Criteria AEV shall be converted from NPV 𝑖(1+𝑖)𝑁 (1+𝑖)𝑁 −1 Decision Criteria: Single Project Evaluation (go or no go) If AE(i) > 0, accept the investment If AE(i) = 0, indifferent If AE(i) < 0, reject the investment Mutually exclusive alternatives Pick the highest AEV value if it is for profit project Pick the lowest AEC if it is non-profit/ public project 7 Example 1 - Annual Equivalent Value After a company invested in a project that costs $500,000, their yearly cash savings and maintenance costs for 8 years were as follows: At the end of year 1 the project makes a profit of $50,000, and then increases $5,000/year every year after. The maintenance costs for the project is $3,000 starting at year 4 and is the same for every year after. Was this project worth investing in? (company MARR is 9%) 8 Example 1 - Solution Cash Flow Diagram 0 1 2 3 4 5 6 7 8 Initial Cost (500,000) Annual Revenue (A) 50,000 50,000 50,000 50,000 50,000 50,000 50,000 50,000 Gradient Revenue (G) - 5,000 10,000 15,000 20,000 25,000 30,000 35,000 Maintenance (A) (3,000) (3,000) (3,000) (3,000) (3,000) - 147,831.34 AEV=-26,709.34 Not a good investment, losing $26,709.34 annually, Obvious from NPV 9 2 - Repeating Cycle The Smart Company wants to invest in solar panels to decrease their energy consumption. They want to invest in silicon cells which are relatively low in cost, but the cells degrade over time and need to be upgraded every 5 years. The initial cost of the cells is $500,000 and saves the company $250,000 the first year, but decreases by $50,000 every year after, due to the cells degrading. What is this projects AEV? Use a MARR of 10% 10 Example 2 - Solution Indefinite cycles of repeating projects using NPV is difficult mathematically Using AEV to look at one cycle, note, the reinvestment of $500K at year 5 should be included in the next cycle of cash flow. i=10%, N=5, A=$250K, G=-50K, I=-500K =−500 +250 ( / ,10%,5)−50 ( / ,10%,5) NPV=104,606.61 Then convert NPV for the first 5 years into AEV AEV=104,606.61(A/P,10%,5) AEV=$27, 594.96 11 Analysis Period vs Useful Lives of Alternatives Analysis Period: the time period covered by the analysis. The consequences of each of alternative must be considered, also known as study period, planning horizon, or project life. In this example, it is indefinite. Useful Lives of Alternatives: this is how long a capital asset can be useful and after that must be replaced. In this example, it is 5 years. 12 Example 3 – Infinite Analysis Period Now, The Smart company is looking at a different model of solar panels that need to be replaced every 4 years and costs $125,000 to implement, and saves $100,000 the first year, decreasing by $20,000 every year after. Which model should the company pick (compared to 10.2)? (MARR=10%) 13 Example 3 – Solution AEV(silicon cells )=$27, 594.96 For the alternative: i=10%, N=4, A=$100K, G=-20K, I=-125K =−125 +100 ( / ,10%,4)−20 ( / ,10%,4) NPV=104,424.22 Then convert NPV for the first 4 years into AEV AEV=104,606.61(A/P,10%,4) AEV= $ This project will save the company $32,942.79 a year, compared to the $27,594.96 for the 5 year cycle model. More economical option. 14 Example 4 - Comparing Projects A company needs to buy a new piece of equipment and use the machine for 5 years. Given the following cash flows, which model is the best option based on the AEV method? (MARR 10%) Model A Model B Initial Cost: ($50,000) Initial Cost: ($60,000) Yearly O&M: ($5,000) Yearly O&M: ($2,000) Salvage Salvage Price: $7,000 Price: $10,000 Note: Since the machines are used for the same purpose, they will generate the same revenue for the company, so revenue can be ignored for this calculation. 15 Example 4 – Solution In 5 minutes, working with another person beside you One person works on Model A The other works on Model B Which one would you pick? Is there an easier way mathematically? 16 Example 10.4 – Solution Model A: i=10%, N=5, I=-50K, A=-5K, S=7K NPVA=-50K-5K(P/A,10%,5),+7K(P/F,10%,5)=-64,607.48 AEVA=-64,607.48(A/P,10%,5) AEVA Model B: i=10%, N=5, I=-60K, A=-2K, S=10K NPVB=-60K-2K(P/A,10%,5),+10K(P/F,10%,5)=-61,372.36 AEVB=-61,372.36 (A/P,10%,5) 𝐵 17 Capital Recovery Costs Capital Recovery Costs: When a project is being analyzed, and only the costs to implement the project are being considered, the Annual Equivalent Value is often called the Annual Equivalent Cost. A typical Capital Recovery Cost is amortization of loan Depriciation Operating costs Opportunity costs Formula 𝑀𝐴𝑅𝑅 Where I is initial cost of the asset and S is future salvage value of the asset Formula assume we do not carry signs (“+” or “-”) in P or S 18 Capital Recovery Costs Formula 𝑀𝐴𝑅𝑅 Where I is initial cost of the asset and S is future salvage value of the asset Assumptions: Formula assume we do not carry signs (“+” or “-”) in P or S MARR consistency 19 Example 5 - Capital Recovery Cost A tractor trailer has an initial cost of $130,000 and an estimated salvage value of $25,000 at the end of its 5-year life What is the capital recovery cost for the tractor if the companies MARR is 15%? 20 Example 5 – Solution i=15%, N=5, I=-130K, S=25K CRC=-130K(A/P, 15%, 5)+25K(A/F, 15%, 5) CRC= The capital recovery cost is , the tractor shall create $ value each year to be cover its capital cost 21 Monitoring in Unit Cost/Profit In certain situations, it’s useful to calculate the Annual Equivalent Value into a unit profit or unit cost. Example: In the last example, the capital recovery cost is If the tractor trailer operates 250,000 km per year The unit cost per kilometer: 𝐴𝐸𝐶 𝐶𝑅𝐶 $35,073.31 𝑘𝑚 𝑘𝑚 250,000𝑘𝑚 22 Example 6 - “Buy or Make” A company has been buying a product from a supplier for 4 years. The supplier has a fixed unit price of $15/unit for the remainder of the 10-year contract. The company could make the product themselves for the remaining 6 years. The machine would cost $25,000. At the end of the 6 years, the machine will have a salvage value of $2,000. Operating costs per year is $2,000 and maintenance costs are $750 a year. The company needs 1500 units a year, and if the MARR is 20%, should they buy or make? 23 Example 6 – Solution i=20%, N=6, I=-25K, S=2K, A=O&M=2K+750=2.75K Cost to operate per year: AEV=-25K(A/P, 20%, 6)+2K(A/F, 20%, 6)-2.75K AEV = -$7,517.64 + $201.41 - $2,750 = $10,066.23 Each year 1,500 units will be produced Unit cost =10,066.23/1,500 Unit cost = $6.7108 / unit