ChE 356

Homework #4

due 2/16/07

1. Solve 3.1 but change i to 8%.

The payment for a 10 year loan of $100,000 at 8% interest is given by

The table of payments divided into payments on interest and payments on principal is given below:

Year / Amount Paid / Principal / Interest / Principal Balance
1 / $14,902.95 / $6,902.95 / $8,000.00 / $100,000.00
2 / $14,902.95 / $7,455.18 / $7,447.76 / $93,097.05
3 / $14,902.95 / $8,051.60 / $6,851.35 / $85,641.87
4 / $14,902.95 / $8,695.73 / $6,207.22 / $77,590.27
5 / $14,902.95 / $9,391.39 / $5,511.56 / $68,894.54
6 / $14,902.95 / $10,142.70 / $4,760.25 / $59,503.15
7 / $14,902.95 / $10,954.11 / $3,948.84 / $49,360.46
8 / $14,902.95 / $11,830.44 / $3,072.51 / $38,406.34
9 / $14,902.95 / $12,776.88 / $2,126.07 / $26,575.90
10 / $14,902.95 / $13,799.03 / $1,103.92 / $13,799.03

2. If you set aside $200/month in a college education fund over 10 years, what amount will be accumulated at a 6% annual interest rate (ignore income taxes)? Is that enough to pay for several years of tuition/fees (at current rate of $10,000/yr) at UT?

The future value of a series of uniform payments (with interest compounded monthly) is given by

The future value of a series of uniform payments (with interest compounded yearly) is given by

This is enough to pay for three years of tuition/fees at the current rate at UT. You will need to study hard to qualify for a scholarship for your senior year.

3. Resolve the table for insulation rate of return on p.104 for fuel cost of $6/106 Btu. Assume fuel can be converted to steam at an efficiency of 80%.

Converting the units into $/kJ we get a fuel cost of $5.69/106 kJ. Recalculating the values in the table (assuming 100% efficiency) gives:

thickness x (cm) / Insulation cost ($) / Value of fuel saved ($/year) / Payback period (years) / Return on investment (% per year) / Net present value ($) / Internal rate of return (%)
1 / $2,136.28 / $6,322.80 / 0.3379 / 295.97% / $17,167.03 / 295.67%
2 / $4,272.57 / $9,801.65 / 0.4359 / 229.41% / $25,769.55 / 228.81%
3 / $6,408.85 / $12,003.03 / 0.5339 / 187.29% / $30,525.02 / 186.31%
4 / $8,545.13 / $13,521.44 / 0.6320 / 158.24% / $33,223.74 / 156.82%
5 / $10,681.42 / $14,632.03 / 0.7300 / 136.99% / $34,694.32 / 135.08%
6 / $12,817.70 / $15,479.65 / 0.8280 / 120.77% / $35,372.98 / 118.33%
7 / $14,953.98 / $16,147.81 / 0.9261 / 107.98% / $35,511.20 / 105.00%

Should we assume 80% efficiency we get:

thickness x (cm) / Insulation cost ($) / Value of fuel saved ($/year) / Payback period (years) / Return on investment (% per year) / Net present value ($) / Internal rate of return (%)
1 / $2,136.28 / $7,903.50 / 0.2703 / 369.97% / $21,927.27 / 369.80%
2 / $4,272.57 / $12,252.06 / 0.3487 / 286.76% / $33,148.90 / 286.43%
3 / $6,408.85 / $15,003.79 / 0.4271 / 234.11% / $39,561.73 / 233.54%
4 / $8,545.13 / $16,901.80 / 0.5056 / 197.79% / $43,403.61 / 196.94%
5 / $10,681.42 / $18,290.04 / 0.5840 / 171.23% / $45,710.32 / 170.04%
6 / $12,817.70 / $19,349.56 / 0.6624 / 150.96% / $47,027.12 / 149.40%
7 / $14,953.98 / $20,184.76 / 0.7409 / 134.98% / $47,668.38 / 133.01%

Note: If you use the “r” value to calculate NPV, you get slightly higher numbers than those shown here. These values were calculated using the NPV function in Excel. Either method is fine.