Unit 27 · Financial Math and Capstone

Interest is the price of using someone else’s money. When you deposit money in a bank, the bank pays you interest. When you borrow, you pay it. The amount you start with is the principal. The rate is the percent charged per year. With simple interest, the rate is applied to the original principal every year, and nothing else. The formula is I = P × r × t, where I is the interest, P the principal, r the yearly rate as a decimal, and t the time in years.

Suppose you lend a cousin $2,000 at 4% simple interest for 3 years. Write 4% as 0.04. Then I = 2,000 × 0.04 × 3. Multiply in steps: 2,000 × 0.04 = 80, and 80 × 3 = 240. The interest is $240, so your cousin pays back $2,000 + $240 = $2,240 at the end. Notice that the interest is the same $80 every year, because it is always figured on the same $2,000.

Time does not have to be a whole number of years. A loan of $800 at 5% for 6 months uses t = 0.5, because 6 months is half a year. I = 800 × 0.05 × 0.5 = 40 × 0.5 = $20. A common mistake is to plug in 6 for t, which gives $240, the interest for six years, not six months. Another mistake is writing 5% as 0.5 instead of 0.05, which makes the answer ten times too large. Always convert the percent first and the time second, then multiply.

Simple interest is the right model for short loans between people, for some savings bonds, and for the interest that builds up on many student loans while the student is still in school. It is also the starting point for everything else in this chapter, because compound interest is what you get when simple interest is charged over and over on a growing balance.

Most bank accounts and most loans use compound interest: at the end of each period the interest is added to the balance, and the next period’s interest is figured on the new, larger balance. Start with $1,000 at 6% compounded once a year. After year one the balance is 1,000 × 1.06 = $1,060. After year two it is 1,060 × 1.06 = $1,123.60. After year three it is 1,123.60 × 1.06 = $1,191.02. Simple interest would have given only 1,000 + 3 × 60 = $1,180. The extra $11.02 is interest earned on interest.

Instead of multiplying year by year, use the formula A = P × (1 + r/n)^(n × t). Here A is the final amount, P the principal, r the yearly rate as a decimal, n the number of compounding periods per year, and t the number of years. For the account above, n = 1, so A = 1,000 × 1.06^3 = 1,000 × 1.191016 = $1,191.02, the same answer. If the same 6% is compounded monthly, n = 12, the rate per month is 0.06/12 = 0.005, and there are 36 months: A = 1,000 × 1.005^36 = 1,000 × 1.19668 = $1,196.68. More frequent compounding gives a little more.

Watch for two common errors. First, the rate per period must be the yearly rate divided by n; using the full 6% each month would give a wild answer. Second, the exponent is the total number of periods, n × t, not just t. For $2,000 at 4% compounded quarterly for 1 year, the rate per quarter is 0.01 and there are 4 quarters: A = 2,000 × 1.01^4 = 2,000 × 1.04060 = $2,081.21. A student who uses 1.04^4 gets $2,339.72, which would be 4% every quarter, or 16% a year.

A handy estimate is the rule of 72: divide 72 by the yearly percent rate to estimate how many years money takes to double. At 6%, 72 ÷ 6 = 12 years. Check: 1.06^12 = 2.012, very close to 2. At 8%, doubling takes about 72 ÷ 8 = 9 years. The rule also works for debt: a balance growing at 24% doubles in about 3 years if nothing is paid.

Two banks both advertise 6%. One compounds yearly, the other monthly. Which is the better deal? The advertised 6% is the nominal rate, sometimes printed as APR, the annual percentage rate. It does not tell you how much a dollar actually grows in a year when interest is compounded more than once. The number that does is the effective annual rate, which banks call the APY, annual percentage yield.

To find the APY, grow $1 for one year and see how much it gained. At 6% compounded monthly the monthly rate is 0.005, and there are 12 months: 1.005^12 = 1.061678. So $1 becomes $1.0617, a gain of 6.17%. The formula is APY = (1 + r/n)^n − 1. For 6% compounded quarterly, (1.015)^4 − 1 = 0.0614, or 6.14%. Compounded daily, (1 + 0.06/365)^365 − 1 = 0.0618, or 6.18%. Notice that the gain from compounding more often shrinks quickly: monthly to daily adds only about 0.01%.

Now compare offers correctly. Bank A pays 4% compounded monthly; Bank B pays 4.05% compounded once a year. Bank B’s number looks bigger. But Bank A’s APY is (1 + 0.04/12)^12 − 1 = 1.0407 − 1 = 4.07%, which beats 4.05%. A common mistake is to compare nominal rates without checking how often each one compounds. Another is to forget the minus 1, which gives 1.0407 instead of 0.0407.

The same idea works against you on a credit card. A card that says 24% APR charges 2% a month, and if you paid nothing all year the balance would grow by 1.02^12 − 1 = 0.268, or 26.8%, not 24%. In the United States, lenders must state the APR in a loan offer, and savings accounts advertise the APY. Knowing which number you are looking at is the first step in any comparison.

Jada’s $12,000 car loan at 6% for 48 months is an amortized loan: the same payment every month, and at the last payment the balance is exactly zero. Each month the lender first charges interest on the current balance. The monthly rate is 0.06 ÷ 12 = 0.005. In month one, interest is 12,000 × 0.005 = $60. Jada’s payment is $281.82, so 281.82 − 60 = $221.82 goes to principal, and her new balance is 12,000 − 221.82 = $11,778.18.

In month two the interest is figured on the smaller balance: 11,778.18 × 0.005 = $58.89. Now 281.82 − 58.89 = $222.93 goes to principal. Every month the interest part shrinks and the principal part grows, because the balance keeps falling. A table that lists this month by month is an amortization schedule. After 12 payments Jada owes about $9,264; after 24, about $6,359; after 36, about $3,274; after 48, zero.

Where did $281.82 come from? The lender uses the payment formula M = P × i ÷ (1 − (1 + i)^(−n)), where P is the amount borrowed, i is the monthly rate, and n is the number of payments. For Jada, P = 12,000, i = 0.005, n = 48. First find 1.005^48 = 1.2705, so (1.005)^(−48) = 1 ÷ 1.2705 = 0.7871. Then 1 − 0.7871 = 0.2129. The top is 12,000 × 0.005 = 60. Finally 60 ÷ 0.2129 = $281.82. You do not need to memorize the formula, but you should be able to check a payment with it.

Marco’s 12% loan has a monthly rate of 0.01, and the same formula gives $316.01. His first-month interest is $120, double Jada’s. Over 48 months Jada pays 13,527.36 − 12,000 = about $1,527 in interest; Marco pays about $3,168. Two things cut interest: a lower rate, and a shorter loan. A 36-month loan has a higher payment but far less total interest, because the balance is gone sooner.

A credit card is a loan with no fixed end date. Each month the card company adds interest to whatever balance you carried over, then asks for at least a minimum payment, often a small percent of the balance or a flat amount like $25, whichever is larger. If you pay the full balance every month by the due date, most cards charge no interest at all. If you do not, interest starts and keeps going.

Take a $1,000 balance on a card with a 24% APR. The monthly rate is 24% ÷ 12 = 2%. In month one the interest is 1,000 × 0.02 = $20. Suppose you pay $50. The new balance is 1,000 + 20 − 50 = $970. Month two: interest is 970 × 0.02 = $19.40, and after another $50 the balance is $939.40. Paying $50 a month, it takes 26 months to clear the $1,000, and you pay about $1,290 in all, so about $290 of interest. That is with no new purchases at all.

Now suppose the minimum is 3% of the balance or $25, whichever is larger. The first minimum is $30, of which $20 is interest, so the balance drops by only $10. As the balance falls the minimum falls too, until it hits $25. Paying only the minimum on that $1,000, with the same 24% rate, takes about six years and costs about $1,870 in all. That is why the CARD Act of 2009 requires statements to show how long minimum payments would take.

The two mistakes that cost the most are carrying a balance you could have paid off, and paying only the minimum. The fix is arithmetic: figure the monthly interest, then choose a payment large enough that most of it hits the principal. On a $600 balance at 18% APR the monthly interest is 600 × 0.015 = $9, so a $60 payment sends $51 to principal, while a $25 payment sends only $16.

A mortgage is a loan used to buy a home, with the home as the lender’s security. Buyers usually pay part of the price up front, the down payment, and borrow the rest. On a $250,000 home, a 20% down payment is 0.20 × 250,000 = $50,000, and the mortgage is $200,000. Mortgages commonly run 30 or 15 years, and the payment comes from the same formula as a car loan, with n = 360 or n = 180 months.

Take $200,000 at 6% for 30 years. The monthly rate is 0.005 and n = 360. Then 1.005^360 = 6.0226, so (1.005)^(−360) = 0.1660, and 1 − 0.1660 = 0.8340. The payment is 200,000 × 0.005 ÷ 0.8340 = 1,000 ÷ 0.8340 = $1,199.10. Over 360 months the borrower pays 360 × 1,199.10 = $431,676, so the interest alone is about $231,676, more than the amount borrowed. The same loan over 15 years has a payment of $1,687.71, and total interest of about $103,788. The 15-year payment is about $489 higher, but it saves about $128,000.

The rate matters just as much as the length. At 4% the 30-year payment is $954.83 and total interest about $143,739. At 7% the payment is $1,330.60 and total interest about $279,018. One percentage point on a mortgage is worth tens of thousands of dollars. Property taxes and insurance are added on top of the payment, so the monthly cost of owning is higher than the loan payment alone.

Student loans follow the same rules with one twist. On many loans, interest builds up while the student is still in school, usually as simple interest, and is added to the balance when repayment begins. A $10,000 loan at 6% left alone for 4 years of school gains 10,000 × 0.06 × 4 = $2,400 before the first payment is due. After that, a $20,000 balance at 5% repaid over 10 years costs $212.13 a month and about $5,456 in interest. Paying the interest during school, even in small amounts, keeps the balance from growing.

Would you rather have $500 today or $520 a year from now? If your savings account pays 5%, take the $500: in a year it grows to 500 × 1.05 = $525, which beats $520. This is the time value of money. Money now can be put to work, so a future amount has to be larger to be worth the same. To compare amounts at different times, move them to the same date.

Moving money forward is compound growth. Moving it backward is finding the present value: divide by the growth factor instead of multiplying. What is $1,000 five years from now worth today at 5%? The growth factor is 1.05^5 = 1.2763, so the present value is 1,000 ÷ 1.2763 = $783.53. Check: 783.53 × 1.2763 = $1,000. A common mistake is subtracting 5 × 5% = 25% to get $750; that ignores compounding and gives too little.

Regular saving uses the same idea many times. If you put $200 into an account every month at 6%, the monthly rate is 0.005, and after 10 years (120 deposits) the account holds 200 × (1.005^120 − 1) ÷ 0.005 = 200 × (1.8194 − 1) ÷ 0.005 = 200 × 163.88 = $32,776. You deposited only 120 × 200 = $24,000; the other $8,776 is interest. This is the future value of a stream of equal deposits, and the formula is FV = PMT × ((1 + i)^n − 1) ÷ i.

Time is the most powerful input. At 7%, $100 a month for 10 years grows to about $17,300, but the same $100 a month for 40 years grows to about $262,500, even though you deposit only four times as much. A saver who puts in $100 a month from age 20 to 30 and then stops, letting the money grow for 30 more years, ends with more than a saver who starts at 30 and deposits every month until 60. Starting early beats depositing more.

A job offer says $16 an hour. The number that lands in your bank account is smaller. Gross pay is what you earn before anything is taken out. Working 30 hours a week for 4 weeks, gross pay is 16 × 30 × 4 = $1,920. From that, the employer takes deductions: federal income tax withholding, Social Security tax, Medicare tax, and in Illinois a flat state income tax. What is left is net pay, or take-home pay.

Here is an example paycheck for someone earning $3,200 a month. In this example the federal withholding is 10%: 0.10 × 3,200 = $320. Social Security is 6.2%: 0.062 × 3,200 = $198.40. Medicare is 1.45%: 0.0145 × 3,200 = $46.40. Illinois state income tax is 4.95%: 0.0495 × 3,200 = $158.40. Add the deductions: 320 + 198.40 + 46.40 + 158.40 = $723.20. Net pay is 3,200 − 723.20 = $2,476.80. About 22.6% of the gross pay never reaches the worker. Actual federal withholding depends on income and the form you fill out, so treat the 10% as an example.

A budget is a plan for the net pay, made before the month starts. One simple plan is the 50/30/20 rule: about 50% for needs like rent, food, transportation and utilities, about 30% for wants, and about 20% for savings and paying down debt. On $2,476.80 of net pay, needs get 0.50 × 2,476.80 = $1,238.40, wants get 0.30 × 2,476.80 = $743.04, and savings get 0.20 × 2,476.80 = $495.36. The exact split is a choice; the point is that every dollar is assigned.

The most common budget mistake is planning with the gross number. A person who thinks of themselves as making $3,200 and rents a $1,500 apartment believes rent is 47% of pay. Against net pay it is 1,500 ÷ 2,476.80 = 61%, leaving very little room. Always budget from the number on the pay stub, and check the stub itself: a percentage that does not match its line is worth a question to the employer.

Inflation is a general rise in prices over time. If inflation runs at 3% a year, something that costs $100 today costs about $103 next year, and 103 × 1.03 = $106.09 the year after. Inflation compounds just like interest. Over 10 years at 3%, prices multiply by 1.03^10 = 1.3439, so the $100 item costs about $134.39. A $2.50 bus fare would be about 2.50 × 1.03^20 = 2.50 × 1.806 = $4.52 after 20 years at that rate.

The flip side is purchasing power, what a fixed amount of money can actually buy. Cash under a mattress does not grow, so after 10 years of 3% inflation, $100 buys only what 100 ÷ 1.3439 = $74.41 buys today. A savings account paying 1% grows the dollars but not the buying power: the money grows 1% while prices grow 3%, so the saver falls about 2% behind each year. The real rate of return is roughly the nominal rate minus the inflation rate. An account paying 5% during 3% inflation earns a real return of about 2%.

This is why a fixed income is risky over a long time. A retiree with a pension of $2,000 a month that never changes keeps receiving $2,000, but after 20 years of 3% inflation that money buys what about 2,000 ÷ 1.806 = $1,107 buys today. It is also why long-term savings usually go into investments that historically outpace inflation, rather than into cash, and why many pensions and Social Security include a yearly cost-of-living adjustment.

The U.S. government tracks inflation with the Consumer Price Index, a measure of the cost of a fixed basket of goods and services. Inflation rates change from year to year, so any long projection should be labeled with the rate it assumes. A common mistake is to compare a price from decades ago with a price today without adjusting; a $5,000 car in a grandparent’s story was not a cheap car in that year’s dollars.

Offers are written to look good. The math job is to make them comparable. A gym charges $40 a month or $400 for a year paid up front. Over 12 months the monthly plan costs 12 × 40 = $480, so the yearly plan saves $80 if you use it all year. The break-even point is 400 ÷ 40 = 10 months: if you quit before 10 months, the monthly plan was cheaper. The decision depends on an honest guess about yourself, but the math tells you what that guess has to be.

A phone store offers plan A, $30 a month plus an $800 phone paid today, or plan B, $60 a month with the phone included, both for 24 months. Plan A costs 24 × 30 + 800 = 720 + 800 = $1,520. Plan B costs 24 × 60 = $1,440. Plan B is $80 cheaper over the contract, even though it looks more expensive each month. Now change one input: if you would keep the phone for 36 months on plan A, that plan costs 36 × 30 + 800 = $1,880 while plan B for 36 months costs $2,160. The longer you keep the phone, the more plan A wins. Always state the time span you assumed.

Car dealers often offer a choice: 0% financing, or a cash rebate with a normal loan. On a $20,000 car over 60 months, 0% financing means 20,000 ÷ 60 = $333.33 a month and $20,000 in all. The other choice is a $2,000 rebate, so you borrow $18,000 at 6%: the payment formula gives $347.99 a month, and 60 × 347.99 = $20,879.40 in all. The 0% deal is cheaper by about $879. But if the loan rate were 2% instead of 6%, the rebate would win. The answer changes with the rate, so compute it; do not guess.

The method is the same every time. First, choose one time span and use it for both offers. Second, list every cost, including up-front fees, monthly charges and interest. Third, add each side up and compare. Fourth, ask what would have to change for the other choice to win, and decide whether that change is likely. The fourth step is what separates a calculation from a decision.

Some choices last years: rent or buy, take a job with a lower salary and free transit, pay off a loan early or invest. The same four steps apply, but the numbers are bigger and the assumptions matter more. Take renting versus owning. A family in a Chicago suburb can rent a home for $1,600 a month or buy a similar one with a mortgage payment of $1,199, plus about $350 a month in property tax and insurance and about $200 a month in upkeep. Owning costs 1,199 + 350 + 200 = $1,749 a month, $149 more than renting.

But part of the mortgage payment lowers the loan balance, and that part is not gone; it becomes equity, the share of the home the family actually owns. In the first year of the $200,000 loan at 6%, about $2,456 of the payments goes to principal, about $205 a month. Counting that as saving rather than spending, the true monthly cost of owning is about 1,749 − 205 = $1,544, now $56 less than renting. Add the $50,000 down payment, which could have earned interest, and the comparison shifts again. There is no single right answer; there is an honest calculation with its assumptions written down.

A second kind of choice is between paying debt and saving. Suppose you have an extra $100 a month and a card balance at 24% APR. Paying the card is like earning 24% with no risk, because every dollar paid stops 2% a month of interest. No savings account comes close. But if the debt is a 3% student loan and a retirement account offers an employer match, the match may be worth more. Compare rates on both sides, and remember that a guaranteed 24% saving beats an uncertain 7% return.

Finally, test the answer. Change one assumption at a time: rent rises 3% a year, the home does not gain value, the family moves after 5 years, the loan rate is 7%. If the choice stays the same under every reasonable change, you can act with confidence. If it flips, you have found the assumption that actually decides the question, and that is the one to investigate. This habit of wiggling the inputs is the heart of the modeling capstone in the next chapter.

A mathematical model is a simplified description of a real situation, built from numbers, variables and rules, that lets you predict or explain something. Every model begins with a question, and vague questions make useless models. Is the light badly timed is a feeling. Does the 30-second green on Main Street clear the cars that arrive during the 45-second red at 5:15 pm is a question with numbers in it. The second one can be answered; the first cannot.

Test a question with three checks. Is it measurable: does it name quantities you can count or time, like cars per minute or seconds of green? Is it bounded: does it say where and when, so the data has a clear scope? Is it decidable: can you say in advance what result would mean yes and what would mean no? For the traffic light, yes means the clearing time is less than the green, and no means it is more.

Strong capstone questions usually come from something you already care about and can reach. Does the cafeteria line move faster with two registers than one? How much of a phone battery does one hour of video use? Does the Red Line really run every 6 minutes at rush hour, and how much does it vary? How many more books does the school library lend in October than in March? Each of these names a quantity, a place and a time, and each can be answered with a few days of careful counting.

Weak questions fail one of the checks. Why is traffic so bad in our town is not bounded. Is our school the best in Illinois is not decidable, because best has no agreed measure. Would a new stadium help the economy is measurable in principle but far beyond a student’s data. Before going further, write your question in one sentence and make sure someone else can tell exactly what data would settle it.

A model is simpler than the world, and that is its strength. The traffic model assumes cars arrive at a steady rate, that every car needs the same 2 seconds to cross, that no one turns, and that the red is always 45 seconds. None of these is exactly true. They are assumptions, choices to ignore some details so the model can be built. The rule is not to avoid assumptions; it is to write every one of them down before you start.

Next, name the variables, the quantities that can change. In the traffic model: the red time R in seconds, the arrival rate r in cars per minute, the green time G in seconds, and the queue Q in cars. Some are inputs you measure or set, like R, r and G. Some are outputs the model computes, like Q and the clearing time C. Say what each one means and what its unit is. A model with a variable in minutes multiplied by a rate in cars per second will be off by a factor of 60.

Now write the rules. The queue at the end of a red is Q = r × R ÷ 60, because r is per minute and R is in seconds. With r = 10 and R = 45, Q = 10 × 45 ÷ 60 = 7.5 cars. The clearing time is C = 2 + 2 × Q seconds, so C = 2 + 15 = 17 seconds. The light clears if C ≤ G. With G = 30, it does. Each rule came from an assumption, and if an assumption is wrong the rule is wrong with it.

Sort the assumptions by how much they matter. Cars arriving in bursts instead of steadily would change Q a little. A truck that takes 5 seconds to move would change C a little. A left-turn lane that blocks through traffic could change everything. The big assumptions are the ones you should plan to check with data, or at least flag in your report as the limits of the model.

Data collection is where most student projects go wrong, not in the math. Before anyone stands at the corner, make a data plan: what will be recorded, in what unit, over what interval, who records it, and where the sheet lives. For the traffic project the plan was: count cars that stop for the red, one count per red cycle, recorded as a whole number with the clock time, for 10 cycles each morning and evening, by two students who compare sheets afterward.

Use an interval long enough to smooth out luck. Counting cars for 10 seconds gives numbers like 1 or 3 that swing wildly. Counting for 2 minutes gives numbers like 18 or 22 that settle down. Five morning counts of 18, 22, 21, 19 and 20 cars in 2-minute intervals total 100 cars in 10 minutes, an arrival rate of 100 ÷ 10 = 10 cars per minute. Each single count is noisy; the average is steady.

Record more than you think you need, in the same layout every time. A row of the traffic sheet held: date, time, red length in seconds, cars queued, seconds to clear, weather, notes. The weather column seemed pointless until a rainy day showed clearing times of 3 seconds per car instead of 2. Keep raw numbers, not rounded ones, and never erase: cross out and write beside, so a strange value can be checked later.

Think about sampling. Ten cycles on one Tuesday morning describe that Tuesday. To say something about weekday mornings, spread the counts across several days. Counting only when a friend is free, or only when the line looks long, builds bias into the data before any math happens. Choose the times in advance and stick to them, even when the corner looks boring.

Raw data is never clean. The traffic sheets, when typed into a spreadsheet, contained a count of 210 cars in one 2-minute interval. Every other count was between 15 and 25. A single-lane road cannot deliver 210 cars in 2 minutes, so 210 is almost surely a typo for 21. The right move is to check the paper sheet, fix it if the sheet says 21, and write in a cleaning log: row 14, changed 210 to 21, matched paper sheet.

The second problem was units. One student recorded clearing times in seconds, another in minutes and seconds, and a third wrote 0.5 for half a minute. The 0.5 sat next to 17 and 19 and would have crushed the average. Convert everything to one unit, and label the column with it: clearing time (seconds). A quick check that catches most unit mistakes is to sort the column and look at the smallest and largest values.

Third, missing values. A row with a blank because a phone died is not a zero, and typing 0 would tell the model that no cars came. Leave the cell blank or mark it missing, and count how many rows are missing. If 2 of 40 rows are missing, drop them and say so. If 20 of 40 are missing, the data plan failed and the counts need redoing. Never fill a blank with a guess and then treat it as measured.

Finally, outliers, real values that sit far from the rest. On the rainy day, clearing took 3 seconds per car instead of 2. That is not an error; it is information. Keep it, flag it, and decide whether the question is about all days or about dry days. Removing an outlier is allowed only when you can say why it is not a genuine observation, and the report must state that it was removed.

The first model should be the simplest one that could be right. For the traffic question, that is a linear model: cars in the queue grow in a straight line with red time. The class timed reds of different lengths at several lights and recorded the queue: 30 seconds gave 5 cars, 45 gave 8, 60 gave 10, 90 gave 14, and 120 gave 21. Plot red time on the horizontal axis and queue on the vertical axis. The points rise roughly along a line through the origin, which makes sense: a red of 0 seconds should have 0 cars.

Fit the line. A rate of 10 cars per minute means 10 ÷ 60 = 1/6 car per second, so the model is Q = R ÷ 6. Check it against the data: 30 ÷ 6 = 5, matching exactly; 45 ÷ 6 = 7.5, close to 8; 60 ÷ 6 = 10, exact; 90 ÷ 6 = 15, one more than 14; 120 ÷ 6 = 20, one less than 21. The largest miss is 1 car. A spreadsheet’s least-squares line gives about Q = 0.17R − 0.14, nearly the same, which is reassuring.

A model earns trust by predicting values it has not seen. The class held back one measurement: a 75-second red that queued 12 cars. The model predicts 75 ÷ 6 = 12.5, a miss of half a car. That is a prediction check, and passing it means more than fitting the original points, because a model can be bent to fit points it was built from. Always keep some data aside for this.

Then combine the pieces. Queue Q = R ÷ 6 in the morning; clearing time C = 2 + 2Q. For R = 45, Q = 7.5 and C = 17 seconds, under the 30-second green. In the evening the rate doubles, so Q = R ÷ 3 = 15 and C = 32, over the green. The first model already answers the question. Later models will refine it, but a simple model that explains the main effect is worth more than a complicated one nobody can check.

Two students proposed different lines for the queue data. Model 1 is Q = R ÷ 6. Model 2 is Q = 0.2R − 2, which someone drew by eye. To compare them, compute each model’s residuals, the measured value minus the prediction, and then the mean absolute error, the average size of the residuals. Model 1 predicts 5, 7.5, 10, 15, 20 for the five reds; the residuals are 0, 0.5, 0, −1, 1; the sizes average to (0 + 0.5 + 0 + 1 + 1) ÷ 5 = 0.5 car. Model 2 predicts 4, 7, 10, 16, 22; residuals 1, 1, 0, −2, −1; average size (1 + 1 + 0 + 2 + 1) ÷ 5 = 1.0 car. Model 1 misses by half as much.

Error is not the only test. A model should also make sense at the edges. Model 2 predicts −2 cars for a red of 0 seconds, which is impossible; Model 1 predicts 0. A model whose formula gives nonsense at natural boundary values, such as negative cars or a queue that keeps growing after the light turns green, has a wrong shape even if its error on the data looks acceptable.

Sometimes a curved model fits better. Suppose someone proposes Q = 0.0015R^2, a quadratic. At R = 120 it predicts 21.6, close to 21, but at R = 30 it predicts only 1.35, far from 5, so its error is worse overall. Beware of the opposite trap too: a model with enough adjustable numbers can pass through every data point and still be useless, because it is fitting the noise. Prefer the simplest model whose error is small and whose shape makes sense, and add complexity only when the data clearly demands it.

Write the comparison as a small table: model, formula, mean absolute error, behavior at R = 0, number of adjustable parameters. A reader can then see why you chose what you chose. In the traffic project the table made the decision obvious: the linear model through the origin had the lowest error, one parameter, and a sensible zero.

A model’s answer depends on numbers you measured roughly. Sensitivity analysis asks: if an input were a bit different, would the conclusion change? Take the evening case: rate 20 cars per minute, red 45 seconds, 2 seconds per car, green 30 seconds. Base result: Q = 15, C = 32 seconds, the light fails by 2 seconds. Now change one input at a time by about 20% and recompute.

Arrival rate 16 instead of 20: Q = 16 × 45 ÷ 60 = 12, C = 2 + 24 = 26 seconds, the light passes. Rate 24: Q = 18, C = 38, fails badly. Seconds per car 2.4 instead of 2: C = 2 + 2.4 × 15 = 38, fails. Seconds per car 1.6: C = 2 + 24 = 26, passes. Startup delay 3 seconds instead of 2: C = 33, one second worse, barely matters. So the conclusion is sensitive to the arrival rate and to the seconds per car, and nearly insensitive to the startup delay.

That tells you where to spend effort. Measuring the startup delay more carefully is a waste; measuring the evening arrival rate on more days is not. The class went back and counted five more evenings, getting rates from 18 to 22. Even at the lowest, 18, Q = 13.5 and C = 29 seconds, only one second inside the green, so on most evenings the line does not clear. The conclusion survived, and now it came with a range instead of a single number.

Sensitivity also answers what-if questions for the people who will read the report. What green would clear 20 cars per minute? Solve 2 + 2 × 15 ≤ G, so G ≥ 32 seconds; a 36-second green leaves margin. What rate could a 30-second green handle? Solve 2 + 2Q ≤ 30, so Q ≤ 14, and Q = r × 45 ÷ 60 ≤ 14 gives r ≤ 18.7 cars per minute. A village engineer can use those two numbers directly.

A graph can tell the truth or bend it with the same numbers. Suppose the average evening wait rose from 40 seconds to 44 seconds, an increase of 4 ÷ 40 = 10%. On a bar chart whose vertical axis starts at 0, the second bar is 10% taller, which looks like what it is. On a bar chart whose axis starts at 38, the first bar has height 2 and the second has height 6, so it looks three times taller. Same data, a very different impression. Bar charts should start at zero, because the eye reads bar height as size.

Line graphs and scatter plots can start above zero when the data lives in a narrow band, but then the axis must be clearly labeled and the reader warned. Every axis needs a label with a unit: red time (seconds), cars queued. Show the actual data points, not only the fitted line, so a reader can judge how well the model fits. When you show the model, draw it only over the range where you have data; extending Q = R ÷ 6 out to a 10-minute red claims something you never measured.

Choose the chart for the question. A scatter plot with a fitted line shows whether queue length grows with red time. A bar chart compares the morning and evening clearing times against the 30-second green, with the green drawn as a horizontal line. A table is best when the reader needs exact numbers, such as the sensitivity results. A pie chart almost never fits a modeling project, because it hides the sizes it claims to show.

Before you present, do the squint test. Look at each graph from across the room and say aloud the one thing it shows. If what you say is not what the data says, fix the graph. The most common fixes are starting the axis at zero, adding units, removing a 3-D effect that distorts the heights, and using the same scale on two graphs that will be compared.

A modeling report is an argument, and a good argument has four parts. The claim is one sentence: at 5:15 pm on weekdays, the 30-second green on Main Street does not clear the cars that arrive during the 45-second red. The evidence is the data: on five evenings, cars arrived at 18 to 22 per minute, and cars cleared at about 2 seconds each after a 2-second start. The reasoning is the model: Q = r × 45 ÷ 60 gives 13.5 to 16.5 cars, and C = 2 + 2Q gives 29 to 35 seconds, so on most evenings the clearing time exceeds 30 seconds.

The fourth part is the limits. Say what the model assumed and what would change the answer: arrivals were counted on five evenings in one month; turning cars were counted as through cars; rain slowed clearing to 3 seconds per car and was excluded; the green was measured as 30 seconds on every cycle observed. Then say what you recommend and what it would take: a 36-second green clears 20 cars per minute with margin, but lengthening one green shortens another, which this model did not study.

Write for a reader who is smart but busy. Put the claim first, not last. Give one clear graph and one small table. Show the key calculation with the actual numbers, so the reader can check it in a minute: Q = 20 × 45 ÷ 60 = 15, C = 2 + 2 × 15 = 32 > 30. Define each variable the first time. Avoid words like obviously and clearly; if it were obvious, you would not need the model.

Then read it as a skeptic. Where could the reader say so what, or how do you know that? Every place you find one, add the number or the reason that answers it. A common weakness is a claim stronger than the evidence: the light is broken becomes, with the same data, the light’s green is too short for evening traffic on most weekdays we measured. The second claim is smaller and much harder to argue with.

A presentation is the argument spoken, with the same four parts in the same order. Start with the claim and the key number: the evening green needs about 32 seconds and has 30. Show the one graph. Then give the evidence and the reasoning, using the actual calculation on a slide so the audience can follow it. End with the limits and the recommendation. Practice until the whole thing fits in five minutes, because the questions afterward are where the model is really tested.

Prepare for the challenges you would make yourself. What if you counted on unusually busy days? Answer with the range: five evenings, 18 to 22 per minute, and the light fails at 19 or more. What if drivers speed up when the line is long? Answer with sensitivity: even at 1.6 seconds per car the clearing time is 26 seconds, so that would flip the conclusion, and it is worth measuring. What about the cross street? Answer honestly: not modeled, and it is the first thing to study next.

When you do not know, say so, and say what you would do to find out. A defense is not a debate to win; it is a check that the model holds up. If a question reveals a real flaw, thank the questioner and write it down. The traffic class was asked whether the counts included cars turning right on red, which never join the queue. They had counted them. Removing an estimated 2 per minute lowered the evening rate to about 18 and the clearing time to 29 seconds, one second under the green. The conclusion weakened to on the busier evenings, and the report was better for it.

Speak the numbers with their units, point at the graph when you cite it, and keep the model’s formula visible so the audience can test any what-if on the spot. A listener who can plug in their own number and get your answer is a listener who believes you.

A modeling project fails most often on time, not on math. Plan it in checkpoints, each with a date and a thing you can show. For a four-week capstone: week 1, a one-sentence question that passes the three checks, a list of assumptions and variables, and a data plan. Week 2, the raw data, cleaned, with a cleaning log. Week 3, a first model with its error, a comparison to at least one alternative, and a sensitivity table. Week 4, the graph, the written argument, and a practiced presentation.

Each checkpoint is a chance to catch trouble early. If the week 1 question is not measurable, you find out before wasting a week of counting. If the week 2 data has 20 missing rows, you can recount before building a model on holes. The traffic class discovered at the week 2 checkpoint that nobody had recorded the green time; they had assumed it, and a single evening of timing fixed it. A plan without checkpoints would have found that gap on presentation day.

Budget the hours honestly. Counting cars for 10 cycles takes about 15 minutes per session, and five sessions morning and evening is about 2.5 hours across the week. Cleaning and entering the data takes another hour. Fitting and comparing models in a spreadsheet takes two hours the first time, and sensitivity tests one more. Writing takes longer than anyone expects; plan three hours and a day away from the draft before rereading it as a skeptic.

Keep a project log with one line per work session: date, what was done, what is next, what is worrying. The worrying column is the most valuable one, because it becomes the limits section of the report. Finally, decide in week 1 what done means: a claim, one graph, one table, a page of argument and a five-minute talk. A project with a finish line gets finished.