Tutorial 7 Nov9 solution PDF

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NAME:________________________

SID:_______________

85-313 (01 & 02) Engineering Economics

Tutorial-ID: ______

Tutorial Assignment #7

Due: November 9, 2018 (Due by the end of this tutorial)

Problem #1: New windows are expected to save $600 per year in energy costs over their 30-year life for Fox Fabricating. At an initial cost of $12,000 and zero salvage value, using IRR, are they a good investment? Fox’s MARR is 8%. Solution: Setting disbursements equal to receipts (present worth) and solving for i*: 12,000 = 600(P/A, i*, 30) (P/A, i*, 30) = 20 From Interest Table (P/A, 2%, 30) = 22.396 and (P/A, 3%, 30) = 19.6 The IRR for this project is between 2% and 3%. By linear interpolation, we found i* = 2.86% This is less than Fox's MARR of 8%, so the windows are not a good investment.

Problem #2: Charlie has a project for which he had determined a Present Worth of $28,494. He now has to calculate the IRR for the project, but unfortunately he has lost complete information about the cash flow. He knows only that the project has - a five-year service-life and a first cost of $180,000, - a set of equal revenue cash flows occurred at the end of each year for 5 years, and - MARR used for calculation the Present Worth was 10%. What is the IRR for this project? Solution: We know that 28,494 = –180,000 + X(P/A, 10%, 5) where X is the unknown cash flow at the end of each of the five years. Solving for X: X = 208,494/3.7908 = 55,000 We calculate IRR from: 180 000 = X(P/A, i*, 5) Substituting in the known value for X and solving for i*: (P/A, i*, 5) = 180 000/55,000 = 3.273 ⇒ From the interest tables, we found that (P/A, 20%, 5) = 2.9906 and (P/A, 15%, 5) = 3.3522 By linear interpolation we found i* = IRR = 16.09% The IRR is approximately 16.09%.

TA on duty: Lucas Chauvin, Sijie Zhang, & Zihan Wang

NAME:________________________

SID:_______________

Tutorial-ID: ______

Problem #3: (a) New windows are expected to save $10,000 per year in energy costs over its 30-year life for XYZ Inc. At an initial cost of $94,269 and zero salvage value, find the IRR for this new windows installation project. Suppose that XYZ’s MARR is 8%, should the new windows be installed? Solution: Setting disbursements equal to receipts (present worth) and solving for i*: 94269 = 10000(P/A, i*, 30) (P/A, i*, 30) = 9.4269 From Interest Table (P/A, 10%, 30) = 9.4269 IRR=10% The IRR for this project is exactly 10% which is more than MARR of 8%. So the project should be accepted. (b) Six mutually exclusive projects A, B, C, D, E, and F, are being considered by XYZ. The new windows project discussed in part (a) is identified here as project A. They have been ordered by first costs so that project A has the lowest first cost, F the largest. The data in the table below apply to these projects. The data can be interpreted as follows: the IRR on the incremental investment from project C to D is 8%. Which project should be chosen using a MARR of 8%? Project A B C D E F

IRR on overall Investment ? 18% 17% 16% 19% 14%

A 14% 16% 10% 16% 9%

IRR on Increments of Investment Compared With Project B C D

7% 13% 6% 10%

8% 14% 12%

7% 6%

E

10%

Do-nothing to A: IRR on do-nothing to A is the same as IRR for A (from (a)) = 10% > MARR, so the current winner is A. From A to B: IRR on increment from do-nothing to B is 14% > MARR, current-winner is therefore B. From B to C: IRR on increment from B to C is 7% < MARR, current-winner is still B. From B to D: IRR on increment from B to D is 13% > MARR, current-winner is D. From D to E: IRR on increment from D to E is 7% < MARR, current-winner is still D. From D to F: IRR on increment from D to F is 6% < MARR, current-winner is still D. No more challenger, the final project selected is Project D.

TA on duty: Lucas Chauvin, Sijie Zhang, & Zihan Wang...