# 经济代写 | 金融代写 | project代写 | econ代写 – Practice

### Practice

project代写 | econ代写 | 经济代写 | 金融代写 – , 该题目是值得借鉴的project代写的题目

The following are selected answers for some of the copies of the Final exams posted on the Queens Exam Bank . Please note the answers may not be correct in all cases. When the TAs are marking the Final examination papers they will occasionally find mistakes with a solution set. That original solution set may, or may not, get updated. The following is based on the solution sets I could find.

### Fall 2014

Q2: a) 42.58% ; b) 20.02% ; c) 17.18% ; d) 6.61% ; e) 29.10% Q3: a) 9% ; b) ~ 13% ; c) PW ~ 704,270 +/- 4 Q4: a) 4.97% ; b) ~ 53,087 ; c) 24.5 +/- 0.5 years Q5: AW New ~ 15,015 and AW Used ~ 16, Q6: b) EAC(1) ~ 117,400 and EAC(2) ~ 104,869 and EAC(3) ~ 107, Q7: EV = 80,500 (Construct new) Q8: a) 7 million/year ; b) 3.4 million / year ; c) Breakeven level of production = 14,166.67 metric tons Q9: Cost300,000 m^2 = 199.2 million and Cost150,000 m^2 = 114.4 million Q10: a) AW ~ 109,974 ; b) ~ 44,685 units ; c) Vendor B is preferred after 44,685 units as it has the lower Annual Cost ; d) i) ~ 90,770 ; ii) First Cost ; iii) Salvage Value

### Winter 2015

Q1: a) 37,470 ; b) 40,370 ; c) 33,665 +/- 1 ; d) 39,865 +/- 1 Q2: Cost of new plant ~ 4,788, Q3: AW Steel ~ 737.62 and AW Plastic ~ 391. Q4: i* A ~ 12% and iB ~ 14% Q5: 3.876% and BV DB(30) ~ 74, Q6: The Economic Life is 2 years and EAC ~ 30, Q7: PW ~ 35, Q8: MARR Actual = 10%, PW ~ 485, Q9: b) Choose 120 minute parking permit because EV 3 < EV 2. Therefore, EV 1 = 5.

### Fall 2015

Q1: The number of units that the company must manufacture in order to breakeven = 121,212. Q2: The estimated cost of the new laboratory ~ 1,274,524 +/- 14 Q3: Choose project B because it has the highest AW \$2,655 versus \$1, Q4: a) PW = 6,680 ; b) IRR ~ 9% ; d) Perform Incremental IRR analysis. Q5: a) Taxes = 153,000 ; b) Taxes = 224,775 ; c) PW = Purchase price X CTF

Q6: a) EAC* ~ 12,896 ; b) Recommend keeping the existing machines for two more years and then replace with the challenger in year 3 when the EACTotal of the defender is larger than the EAC* of the challenger. Q7: b) PW ~ 1.25 to 1.26 million Q8: b) Expected value = 447,

### Winter 2016

Q1: Breakeven number of units = 1, Q2: Total cost for the new solar array, with optional features ~ 354,879 and the unknown cost capacity factor ~ 1. Q3: AW SP240 ~\$13,362 and AW PL647 ~\$10, Q4: a) i* B= 15% ; b) Do not choose project as IRR (15%) < MARR(15.6%) ; c) Chose Project A as IRR (>20%) > MARR(15.6%) Q5: AW(van) ~ \$6,423 and AW(lease) = \$6, Q6: Economic Life = 2 years and EAC* ~ \$6, Q7: a) Actual MARR ~ 15% ; b) PW ~ 576,445 or ~ 576,341 (depending on approach) Q8: Expected value = \$4,

### Fall 2016

Q1: a) Either machine will produce the required 30,000 non-defective units/3-months : b) Machine #1 3,290/day and Machine #2 3,630/day c) Machine #1 6.46/unit and Machine #2 6.90/unit Q2: a) ~ 10.6 million ; b) i) 0.86 hours or about 52 minutes, ii) ~9.43 hours Q3: a) AW A ~ 1,574 and AW B ~2,654 ; b) Simple payback with Option A is 8 years and with Option B is 7 years Q4: This is a mutually exclusive problem, so rank by lowest first cost. PW XJ13 – Y19 i* ~ 18%, PW Z98 XJ3 i* ~8% Q5: AW Excavator = 102,004 and AW Lease = 90,000 (Fuel of 20,000 can be ignored) Q6: The grader has two remaining years in its economic life and the EAC* is 185, Q7: The present worth of the contract is ~ 13,222 and you should proceed Q8: EV1 = 400,000 and EV2 = 230,

### Winter 2017

Q1: a) Fuel cost savings = \$99 ; b) Yes purchase Dynolube as fuel savings = \$79.11, which is greater than purchase price Q2: a) Z 50 ~ 0.235 hours ; b) Estimated per unit selling price ~ 17.19 +/- 0. Q3: a) This is a 6-year \$10,000 annuity that starts at year 2 (first arrow end of year 3) and a 6-year \$1,000 gradient that starts a year 2 (first arrow end of year 4). ; b) PW ~ 39,895 ; c) the difference is ~ 4, Q4: Equivalent monthly MARR = 0.5%, ERR ~ 1%, On an equivalent annual basis i* ~ 12.68% Q5: a) S 8 = 14,412 ; b) CTF = 0.7494, CSF = 0.7375 ; c) needs to take into account the CTF at start and 1-t for the other amounts, payback is 6 years. ; d) PW ~ \$248,809 +/- \$ Q6: a) EAC* = 7,600 at Year 1 ; b) EAC(3)~ 8,255 and because EAC(3) is above EAC*OptionD we can keep the defender for two more years and then replace it with Option D in the third year.

Q7: a) 525,000 ; b) PW ~ 418,680 ; c) PW ~ 353, Q8: a) EV = 150,000 therefore she should undertake the small scale project. ; b) Most sensitive to rent revenues ; c) 126,

### Fall 2017

Q1: a) 312,500 parts ; b) 3,125 hours ; c) ~10.3 months ; d) 90,000 ; e) 64,063 ; f) ~89,286 ; g) 133,036 ; h) 320 ; i) 373,645 +/- 10 Q2: b) PW ~ 32,193 ; c) AW ~ 4, Q3: AW Tires ~873 and AW Tar ~ 690 Q4: a) PW L > 17% > iL > 20% ; PW M > i M ~ 15% ; b) MUST sort in order of increasing initial cost. iLM > MARR, therefore chose Project L. Q5: a) PW A ~ 78,051 and PW B ~ 146,441; b) After Tax PW A ~ 44,280 and After Tax PW B ~ 86,897 ; c) ieff-A ~ 42.6% and ieff-B ~ 20% Q6: a) MARR A ~ 9% ; b) PW ~ 704,270 +/- 4 Q7: b) EAC ~ 8,598 in Year 2 Q8: EV = 1,725,000 with the Do Nothing option.