GENTECH 3EE3 MID-TERM REVIEW QUESTIONS Review Problem Questions GEN TECH 3EE3 WEEK 6 LECTURE Administrative Items Mid Term Test: o Scheduled for Wednesday, February 16th, 7:00 – 9:00 PM (Online) o A2L Midterm Study Groups – 10 Discussion Forum Groups (self-enrol) o Chapters 1-4 including bonds and mortgage questions o This Exam contains 10 multiple choice questions (A2L) and 20 FIB problems (A2L) o This is a closed-book, double sided hand-written 8.5 X 11 crib sheet is permitted, Midterm 3EE3 Formula Sheet and tables are provided at the end of the Exam paper, and only ‘McMaster Approved’ calculators are permitted 2 Capitalized Formula Problem 3 What is the PW of a cost to install= $2,500,000, infinite life, maintenance of $300,000 every 10 years, annual i=5% SOLUTION PW = 2,500,000 + 300,000(A/F,i,n)/i PW = 2,500,000 + 300,000(A/F,5%,10)/0.05 PW = 2,500,000 + 300,000(0.07950)/0.05 PW = $2,977,000 CB Electronix needs to expand its plant. It is considering two alternative plans. Both will last indefinitely. The MARR is 15%. Use an AW comparison to select the preferred alternative. Plan 1: Expand by 20 000 square feet now First cost = $2 000 000 Annual Maintenance costs = $10 000 per year Periodic maintenance costs = $15 000 every 15 years, starting in 15 years 4 Capitalized Formula Problem Plan 1: Expand by 20 000 square feet now First cost = $2 000 000 Annual maintenance costs = $10 000/yr Periodic maintenance costs = $15 000 every 15 years, starting in 15 years. 5 Capitalized Formula Problem Plan 2: Expand by 12 500 square feet now and 7 500 square feet in 10 years. First cost $1 250 000 $1 000 000 Annual maintenance cost $5 000 years 1-10 $11 000 starting in year 11 Periodic maintenance costs N/A $15 000 every 15 years starting 15 years after 2nd addition 6 Capitalized Formula Problem Solution Plan 1 Using the capitalized cost formula for long lived assets: AW = (2 000 000 i) + 10 000 + 15 000(A/F,15%,15) AW = 2 000 000(0.15) + 10 000 + 15 000(0.02102) = $310 315 P = A/i A = Pi 7 Capitalized Formula Problem Plan 2 Step AW Equation Answer 1 First cost 1 250 000(0.15) + 1 000 000 (P/F,15%,10) (0.15) $224 578.50 2 Maintenance cost: years 1 to 10 5000(P/A,15%,10)(0.15) $3 764.01 3 Maintenance cost: years 11 and on [(11 000)/(0.15)](P/F,15%,10) (0.15) $2 720.00 4 Periodic maintenance costs [15000(A/F,15%,15)/(0.15)](P/F,15 %,10)(0.15) $77.90 Total $231 140 8 Solution Capitalized Formula Problem Solution AW of plan 1: $310 315 AW of plan 2: $231 140 Select plan 2, which has the lowest AW 9 Capitalized Formula Problem Non Uniform Annuity Problem 10 Write the general mathematical expression that calculates A when PW = 0 for the cash flow diagram shown below. Calculate A if P = $50,000, X = $5,000, i = 5% and g = 2%. 0 1 2 3 P 98 10 11 12 13 38 39 40 A A A A A A X X(1+g) X(1+g)2 X(1+g)27 X(1+g)28 X(1+g)29 Non Uniform Annuity Problem, Cont’d 11 SOLUTION P +A(P/A,i,10) – x(P/A,g,i,30)(P/F,i,10) = 0 A = [x(P/A,g,i,30)(P/F,i,10) – P][1/(P/A,i,10)] (A/P,i,10) = 1/(P/A,i,10), substitute into above expression A = [x(P/A,g,i,30)(P/F,i,10) – P](A/P,i,10) or (A/P,i,10) = (F/P,i,10)(A/F,i,10), substitute into above expression A = [x(P/A,g,i,30)(P/F,i,10) – P](F/P,i,10)(A/F,i,10) A = [x(P/A,g,i,30)(P/F,i,10)(F/P,i,10) – P(F/P,i,10)](A/F,i,10) A = [x(P/A,g,i,30) – P(F/P,i,10)](A/F,i,10) io = (1+i)/(1+g) – 1 = (1+0.05)/(1+0.02) – 1 = 0.029412 A = [5,000(P/A,2.94%,30)/(1+0.02) – 50,000(F/P,5%,10)](A/F,5%,10) A = $1,222 PW Comparison Method Problem 12 A company is upgrading one of their oldest operating lines at their packaging facility. Two options are available for upgrading; option A and B. Option A upgrading costs $65,000 now and yields annual savings of $25,000 in the first year, $24,000 the second year, $23,000 in the third year, and so on. Option B costs $15,000 and saves $6,500 in the first year, with the savings decreasing by 15% each year thereafter. If the upgraded packaging line will last for eight years, which upgrading option is better using the present worth method and 8% MARR Draw a cash flow diagram for both options. PW Comparison Method Problem, Cont’d 13 SOLUTION OPTION A: PWA = -65,000 + [(25,000 – 1,000(A/G,8%,8)](P/A,8%,8) PWA = -65,000 + [(25,000 – 1,000(3.0985)](5.7466) PWA = $60,859.16 OPTION B: io = (1+i)/(1+g) – 1 = (1+0.08)/
(1-0.15) – 1 = 0.270588 PWB = -15,000 + 6,500(P/A,g,i,n)/(1+g) = -15,000 + 6,500(P/A,i o,8)/0.85 PWB = -15,000 + 24,100.36 = $9,100.36 Since PWA > PWB, select OPTION A! Uniform Annuity Cash Flow Problem 14 You would like to save $10,000 for an upgrade to your business computer system in 2 years time. If you deposit $350 every month (starting today), what nominal interest rate is required in order for you to reach your financial goal if the investment is compounded monthly SOLUTION Given F = 10,000, A = 350 n = 25 months A = F(A/F,i,n) or F=A(F/A,i,n) A = F(A/F,i,n) = F{i/[(1+i)^n-1]} 350 = 10,000{i/[(1+i)^n-1]} i = 1%; A = 354.07 i = 2%; A = 312.20 Linear Interpolation: i = 1+(2 – 1)[(350 – 354.07)/(312.20 – 354.07)]=1.097206 Therefore, r = mi = 12(1.097206) = 13.166% compounded monthly Mortgage Problem 15 You have just bought a house for $300,000 and put $50,000 down. The rest of the cost has been obtained from a mortgage. The mortgage has a nominal interest rate of 3%, compounded monthly with a 10-year amortization period. The term of the mortgage is 5 years. What is the monthly mortgage payment How much will you owe in 5 years SOLUTION P = 300,000 – 50,000 = 250,000 i = 0.25% n = 120 months A = P(A/P,i,n) = 250,000(A/P,0.25%,120) = 250,000(0.009656) A = $2,414.02 F5 = P(F/P,i,n) – A(F/A,i,n) F5 = 250,000(F/P,0.25%,60) – 2,414.02(F/A,0.25%,60) F5 = 250,000(1.161617) – 2,414.02(64.647) F5 = $134,345.83 Bond Problem 1 16 A man buys a corporate bond from a bond brokerage house for $925. The bond has a face value $1,000 and pays 4% of the face value each year. If the bond is paid off at the end of 10 years, what rate of return will the man receive SOLUTION PW of Cost = PW of Benefits $925 = $40 (P/A, i%, 10) + $1,000 (P/F, i%, 10) Try i = 5% $925 = $40 (7.722) + $1,000 (0.6139) = $922.78 (i is too high) Try i = 4.5% $925 = $40 (7.913) + $1,000 (0.6439) = $960.42 (i is too low) i* = 4.97% Bond Problem 2 17 At what coupon interest rate will a $20,000 bond yield a nominal 12% interest compounded quarterly if the purchaser pays $18,000 and the bond becomes due in 20 years Assume the bond interest is payable quarterly. SOLUTION x = coupon rate per quarter NPW = 0 = ($20,000* x) (P/A, 3%, 80) + $20,000 (P/F, 3%, 80) – $18,000 ($20,000* x) (30.201) = $18,000 – $20,000 (0.0940) x = ($18,000 – $20,000 (0.0940))/($20,000 × 30.201) = 0.0267 = 2.67% Coupon rate = 2.67%/qtr × 4 qtr/yr = 10.68% Supplementary Problem 1 18 A corporation sold 12% bonds with a face value of $200,000. These bonds mature in in six years and interest is paid semi-annually. The bonds were yield 10%. What is the issue price SOLUTION PW=200,000(P/F,5%,12) + 200,000(0.12/2)(P/A,5%,12) PW=200,000(0.556837) + 200,000(0.06)(8.863252) PW=$217,726.50 Supplementary Problem 2 19 A company issued 5-year, $1 million bonds with a coupon rate of 6% and yield of 5%. The bonds pay interest semi-annually. What is the issue price of the bond SOLUTION PW=1,000,000(P/F,2.5%,10) + 1,000,000(0.06/2)(P/A,2.5%,10) PW=1,000,000(0.7811984) +1,000,000(0.03)(8.7520639) PW=$1,043,760.32 Supplementary Problem 3 20 How are bonds priced SOLUTION Bonds are priced at the present value of all future cash flows (i.e. interest paid out at compounding period and final face value paid out in the future) at the bond yield. Bond yield is defined as the amount of return that an investor will realize on a bond. Therefore, bonds are prices at the present value of all future cash flows at yield. Supplementary Problem 4 21 A $92,000 mortgage has an interest rate of 9% compounded semi- annually, for a 5-year term. The mortgage is amortized over 20-years. What is the monthly payment SOLUTION ie=(1+r/m)^m – 1 = (1+i)^6 – 1 = 0.09/2 = 4.5% solve for i (monthly interest rate); (1+i)^6 = 1.045; therefore, i = (1.045)^(1/6) – 1 = 0.0073631 A=92,000(A/P,0.73631%,240) A=$818.05 Supplementary Problem 4 22 On February 1, 2013, Sawa Inc. issued 5-year, $1 million bonds with a coupon rate of 6% and yield of 5%. The bonds pay interest semi- annually on August 1 and February 1 each year. What is the issue price of the bond SOLUTION P = 1000000(P/F, 2.5%, 10) + 30000(P/A, 2.5%, 10) P = $1,043,760 NOTE: Because the coupon rate is paid as a semi-annually cash flow, use 5%/2 = 2.5% semi-annually interest rate as the yield. Questions Period 23