### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Engr Econ SocialGlobal Impltn EGN 3615

USF

GPA 3.76

### View Full Document

## 60

## 0

## Popular in Course

## Popular in General Engineering

This 10 page Class Notes was uploaded by April Prosacco on Wednesday September 23, 2015. The Class Notes belongs to EGN 3615 at University of South Florida taught by Michael Weng in Fall. Since its upload, it has received 60 views. For similar materials see /class/212654/egn-3615-university-of-south-florida in General Engineering at University of South Florida.

## Reviews for Engr Econ SocialGlobal Impltn

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/23/15

Chapter 7 Solutions 74 The rate of return exceeds 60 so the interest tables are not useful F P 1 in 25000 5000 1 13 1 i 250005000 3 171 i 071 Rate of Return 71 78 FA i 20 10725105 40 and interpolating i 6 136786403678640996 676 exact value 6774 711 3000 11967 PA i 30 PA i 30 300011967 25069 Performing Linear Interpolation PA i 30 i 25808 1 24889 125 i 1 02525808 2506925808 24889 1201 a Nominal Interest RateYear 1201 x 12 1441 b Effective Annual Interest Rate 1 00120112 1 0154 154 713 Solve for the rate that yields a NPW of zero NPW 0 14000 PA i 30 200000 PF i 30 250000 T i 5 NPW 14000 15372 200000 02314 250000 11488 Try i 0 NPW 14000 13765 200000 01741 250000 22470 Performing Linear Interpolation NPW 1 11488 5 22 470 6 i 5 10 11488 22470 11488 534 731 PW of Cost PW of Bene ts 925 40 PA i 10 1000 PF i 10 Try i 5 925 40 7722 1 000 06139 92278 i too high Try i 45 925 40 7913 1 000 06439 96042 i too low i 497 736 Interest paid monthly 50000 01212 500 NPW 0 500 PA i 240 50000 PF i 240 45000 Try i 100 NPW 500 90819 50000 00918 45000 499950 Try i 125 NPW 500 75942 50000 00507 45000 449400 Interpolating we get i 11317month Nominal rate 11317 12 1358 Annual effective rate 1 00113212 1 1446 744 Set PW of Cost PW of Benefits 1845 50 PA i 4 2242 PF i 4 Try i 7 450 3387 2242 07629 1879 gt 1845 T y i 8 450 3312 2242 07350 1813 lt 1845 Rate of Return 7 1 1879 1 8451879 1 813 752 for 6 months Nominal annual rate of return 2 752 150 Equivalent annual rate of return 1 007522 1 156 745 a The monthly payments are600036 16667 over 3 years NPW 0 6000 250 16667 PA i 36 so PA i 36 3450 The tablesdon t go to a low enough interest rate so must solve 1i351i1i35 3450by trial and error or Excel using the IRR function Excelyields i 000232 so la 1 000232 1 00282 or 282 b The fact that the dealer would accept 5200 cash for the car indicates its true worth so the extra 800 is a hidden finance charge Your payments are still based on the original 6000 cost but you only receive a car worth only 5200 NPW 0 5200 250 16667PA i 36 so PA i 36 2970 andinterpolating i 1 0253040729703010728847 1081 exact value 1079 so la 1 001081 1 1377 exact value 1375 747 First determine the monthly payments forthe loan where i 412 03333 so A 6000 AP 03333 36 000333310003333351000333336 1 17714 a NPW 0 6000 250 17714 PA i 36 so PA i 36 3246 andinterpolating i 050 0253287132463287131447 0572 so r 12 0572 686 and la 1 00057212 1 00709 or 709 b Worth ofthe car 6000 800 5200 but the payments are determined by the actual cost to buyer here 6000 Thus the payments are the same as above NPW 0 5200 250 17714 PA i 36 so PA i 36 27944 and interpolating i 125 02528847 2794428847 27661 1440 so r 12 1440 1728 and la 1 00144012 1 01872 or 1872 c The actual value ofthe car seems to be the most important factor 752 PW of Benefits PW of Cost 0 15000 PF i 4 9000 80 PA i 4 0 Try i 12 15000 06355 9000 80 3037 28954 Try i 15 15000 05718 9000 80 2855 65140 Performing Linear Interpolation i 12 3 2895428954 651 40 1292 754 a Total Annual Revenues 500 12 months 4 apt 24000 Annual Revenues Expenses 24000 8000 16000 To find lnternal Rate of Return the Net Present Worth must be 0 NPW 16000 PA i 5 160000 PF i 5 140000 At i 12 NPW 8464 At i 15 NPW 6816 lRR 12 3 84648464 6816 137 b At 137 the apartment building is more attractive than the other options 758 Year Cash Flow 0 9000 1 4 800 5 8 400 9 6 000 PW of Cost PW of Bene ts 9000 400 PA i 8 400 PA i 4 6000 PF i 9 Try i 3 400 7020 400 3717 6000 07664 8893 lt 9000 Try i 2 2 400 7170 400 3762 6000 08007 9177 gt 9000 Rate of Return 2 2 12 9177 90009177 8893 281 763 The number of months between August 15 and January 15 is 5 Month Annual Permit Semester Permit 0 1 00 65 1 O 2 O O 3 O O 4 O O 5 65 To solve for the monthly interest rate set the two PWs equal to each other so 100 65 65 PF i 5 Thus 1 i395100 6565 053846 Solving get i 01318 or 1318 and la 1 01318 1 3418 or 342 Unless the student is graduating in January or just doesn t have the 100 it is clearly better to buy the permit a year at a time 769 Year A B B A 0 2000 2800 800 1 3 800 1 100 300 Comguted ROR 97 87 61 The rate of return on the increment B A exceeds the Minimum Attractive Rate of Return MARR 5 therefore the higher cost alternative B should be selected 771 Year A B B A 0 10000 5000 5000 1 10 1993 1193 800 Comguted ROR 15 20 961 AROR 961 gt MARR 8 Select A 774 Using incremental analysis computed the internal rate of return for the difference between the two alternatives Year A B 0 9000 17 3000 8 1 200 Note lnternal Rate of Return IRR equals the interest rate that makes the PW of costs minus the PW of Benefits equal to zero 9000 3000 PA i 7 1200 PF i 8 0 Try i 25 9000 3000 3161 1200 01678 68436 lt 0 Try i 30 9000 3000 2802 1200 01226 44688 gt 0 i 25 5 6843644688 68436 280 actual value is 279 The contractor should choose Alternative A and lease because 28 gt 15 MARR 780 a 2000 150 100 PA i 20 PA i 20 1 850100 185 i per month The alternatives are equivalent at a nominal 9 annual interest b Take Alt 1 the 2000 and invest the money at a higher interest rate 781 a Salvage 015 X 380000 57000 and firm s interest rate 12 Year Purchase Lease Purchase Lease 0 380000 60000 320000 1 0 60000 60000 2 0 60000 60000 3 0 60000 60000 4 0 60000 60000 5 0 60000 60000 6 57 000 0 57000 NPW 0 320000 60000 PA IRR 5 57000 PA IRR 6 and interpolating IRR 3 0525382538 2730 324 also 324 from Excel The IRR is well below the firm s interest rate on the borrowed amount 320000 from leasing so lease the bulldozer b The rm receives 65000 more than it spends on operating and maintenance costs Year Purchase Lease Purchase Lease 0 380000 60000 320000 15 65000 60000 125000 6 65000 0 122000 57 000 NPW 0 320000 125000PA IRR 5 122000 PF IRR 6 and interpolating IRR 30 597789778 22346 315 3142 from Excel Clearly the situation has changed The interest rate on the borrowed amount is now well above the firm s interest rate so buy the bulldozer The rate of return forthe bulldozer will clearly be largest for this cash ow and is given by PW 0 380000 65000 PA ROR 6 57000 PF ROR 6 and interpolating ROR 4 0557775777 960 443 443 from Excel Note that the author has failed to give a practical scenario for how the 65000 benefit can be realized if the bulldozer is purchased instead of leased Year A B A B NPW at 7 NPW at 9 0 9200 5000 4200 4200 4200 1 1850 1750 100 93 92 2 1850 1750 100 87 84 3 1850 1750 100 82 77 4 1850 1750 5100 3891 3613 5000 5 1850 1750 100 71 65 6 1850 1750 100 67 60 7 1850 1750 100 62 55 8 1850 1750 100 58 50 Sum 211 104 AROR 83Choose Alternative A 784 Year Zappo Kicko Kicko Zappo 0 56 90 3 1 56 0 56 2 0 0 0 Compute the incremental rate of return on Kicko Zappo PW of Cost PW of Bene t 34 56 PF i 1 PF i 1 3456 06071 From interest tables incremental rate of return gt 60 AROR 647 hence the increment of investment is desirableBuy Kicko 787 This is an unusual problem with an extremely high rate of return Available interest tables obviously are useless One may write PW of Cost PW of Bene ts 05 35 1 i391 09 1 i392 39 1 i393 86 1 i394 For high interest rates only the rst few terms ofthe series are significant Try i 650 PW of Benefits 351 65 091 652 391 653 861 654 0467 0016 0009 0003 0495 Try i 640 PW of Benefits 351 64 091 642 391 643 861 644 0473 0016 0010 0003 0502 i 642Calculator Solution i 6429 788 5200000 Income 300 Income gradient 1000 Deposit A 01 Horizon years 400 Savings rate Cumulative Year Salarv Deposit Savings 1 5200000 520000 520000 2 5356000 535600 1076400 3 5516680 551668 1671124 4 5682180 568218 2306187 5 5852646 585265 2983699 6 6028225 602823 3705870 7 6209072 620907 4475012 8 6395344 639534 5293546 9 6587204 658720 6164009 10 6784821 678482 7089051 11 6988365 698837 8071450 12 7198016 719802 9114109 13 7413957 741396 10220069 14 7636375 763638 11392510 15 7865467 786547 12634757 For any row Salary 1 003 Previous year s Salary Deposit Percent Deposit Current year s Salary Savings 1 004 Previous year s Savings Current year s Deposit Amount saved is 12634757 in 15 years Chapter 6 Solutions 64 D 100 FP 6 2 200 FP 6 4 NF 6 6 100 1124 200 1 262 01434 5231 69 500 D FA 12 3 05D D PA 12 2 D 3374 05 1690 D 5005564 8986 613 Equate the value of annual deposits in year 16 with the value ofthe annual college expenses first deposit at end of year 3 A FA 6 14 25000 PA 6 4 A 25000 3465 21 015 4122 624 Compute the equivalent future sum for the 26000 and the four 440 payments F 26000 FP 1 4 440 FA 1 4 26000 1041 440 4060 2527960 This is the amount of money still owed at the end of the four months Now solve for the unknown n 2527960 840 PA 1 n PA 1 n 2572960840 3009 From the 1 interest table n is almost exactly 36 Thus 36 payments of 840 will be required 629 The 11 deposit are beginningof period deposits so the compound interest factors must be adjusted for this situation Pnow1 500000 PF 1 12 500000 08874 443700 A Pnow1AP 111 443700 00951 42196 Quarterly beginningof period deposit 42196 637 a EUAC 5000 35000 NP 6 20 5000 35000 00872 8052 b Since the EUAC ofthe new pipeline is less than the 10000 annual cost of the existing pipeline it should be constructed 641 EUAC Comparison Gravity Plan Initial Investment 28 million AP 10 40 28 million 01023 286400 Annual Operation and maintenance 10000 Annual Cost 296400 Pumping Plan Initial Investment 14 million AP 10 20 14 million 01023 143200 Additional investment in 10thyear 200000 PF 10 10 NP 10 40 200000 03855 01023 7890 Annual Operation and maintenance 25000 Power Cost 50000 for 40 years 50000 Additional Power Cost in last 30 years 50000 FA 10 30 NF 10 40 50000 164494 000226 18590 Annual Cost 244680 Select the Pumping Plan 649 Machine X EUAC 5000 NP 8 5 5000 02505 1252 Machine Y EUAC 8000 2000 NP 8 12 2000 008 150 1106 Select Machine Y 653 Annual Cost of Diesel Fuel 50000 km35 kmliter x 068liter 97143 Annual Cost of Gasoline 50000 km28 kmliter x 072liter 128571 EUACdiesel 24000 4000 NP 6 5 4000 006 97143 fuel 900 repairs 1000 insurance 20000 02374 240 287143 785943 EUACgasoline 19000 6000 NP 6 4 6000 006 1 28571fuel 700 repairs 1000 insurance 709751 The gasoline taxi is more economical 655 Alternative A EUAB EUAC 1000 10000 NP 8 1000 10000 008 200 Alternative B EUAB EUAC 1762 15000 NP 8 20 1762 15000 01019 234 Alternative C EUAB EUAC 5548 20000 NP 8 5 5548 20000 02505 538 Select C 657 Choose alternative with minimum EUAC a 12month tire EUAC 3995 AP 10 1 4395 b 24month tire EUAC 5995 AP 10 2 3454 c 36month tire EUAC 6995 AP 10 3 2813 d 48month tire EUAC 9000 AP 10 4 2840 Buy the 36month tire 673 a Payment PMT 075 360 92000 74025 b Owedyr1 PV 075 348 74025 9137111 so paid 62992000 068 C Owedyr 10 PV 075 240 74025 8227505 d nterest25 IPMT 075 25360 92000 68013 Principal25 PPMTO75 25360 92000 6012 675 a Payment PMT 05 360 145000 86935 b NPER 05 1000 145000 2588 months 2157 years c NPER 05 286935 145000 2381 months 1984 years

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.