Compare financial model assumptions and outcomes

Which statement identifies the assumptions in a simple-interest model and a compound-interest model?

Reuse the Year 10 experiment with a separate $2,000 deposit, fixed 6% nominal annual rate and two-year term. Try annual, half-yearly, quarterly and monthly compounding. Predict how changing frequency affects both the period rate and number of updates; the rate and balance readouts are rounded only for display

What must stay fixed for this experiment to isolate compounding frequency?

Compare assumptions before outputs

Financial models can use different assumptions for the change in a quantity. State what each model holds constant, calculate both outcomes at the same horizon, and interpret the comparison for the stated role

1. 1Identify the fixed quantity Simple interest adds a fixed amount based on the initial value, while compound interest applies a fixed percentage to the current value \(\displaystyle A_{n+1}=A_n+c,\qquad B_{n+1}=RB_n\)
2. 2Use the same starting value To isolate the effect of the update rules on raw future balances, use the same starting value. If initial values differ, choose and justify a suitable normalised comparison rather than simply comparing balances \(\displaystyle A_0=B_0=1000\)
3. 3Use the same horizon Compare at the same elapsed date. Convert each model to its own matching period count: three years is twelve quarters but three annual updates \(\displaystyle A_3=1000+3c,\qquad B_3=1000R^3\)
4. 4Interpret the comparison for the role A larger future value can suit an investor, while a lower cost or balance can suit a borrower or purchaser \(\displaystyle 1331>1300\quad\Rightarrow\quad B\text{ has the larger value}\)

Construct a recurrence investigation

A fictional fund starts with $8,000. Model A pays simple interest at 4% per year on the original principal. Model B retains interest and compounds at a fixed nominal 4% per year, quarterly. No fees, deposits, withdrawals or rate changes occur. Choose a spreadsheet or recurrence tool, justify the choice and define every index and balance

1. 1Construct and connect Write initial values, recurrences and explicit rules. Construct all Model B rows from quarter 0 through quarter 12, keeping full precision internally, and a discrete plot with a chosen window. Construct the Model A values at years 0,1,2,3 and compare only matching dates. Explain the fixed coefficients in the first-order form \(\displaystyle V_{n+1}=aV_n+b\)
2. 2Verify and distinguish Record actual formulas or commands. Independently verify the last value with the explicit rule and a substitution into the recurrence. Calculate the interest in quarter 4 and total interest over the first year; explain why these are different quantities. Explain why a displayed cents-rounded row must not become the next exact input
3. 3Save independent work first Save your model choices and reasons, complete tables, discrete plots, formulas or commands, checks and written conclusions before opening the model response. Complete construction, explanations and actual external-tool operation are not automatically marked; review them yourself or with a teacher

Model response: Construct a recurrence investigation

Compare your saved work with the values, assumptions and checks. Other suitable tools and windows are acceptable when the initial index, units and model conventions are preserved

1. 1Rules, indices and tool Use A(0)=8000 and A(n+1)=A(n)+320 for n completed years, so A(n)=8000+320n. Use B(0)=8000 and B(j+1)=1.01B(j) for j completed quarters, so B(j)=8000(1.01)^j. The first model has a=1,b=320; the second has a=1.01,b=0. A spreadsheet can keep each preceding full value and generate all rows. For example, put 8000 in B2, enter =B2*1.01 in B3 and fill down to quarter 12.
2. 2Complete table and graph The Model B balances at quarter indices 0 through 12, displayed to cents, are 8000.00,8080.00,8160.80,8242.41,8324.83,8408.08,8492.16,8577.08,8662.85,8749.48,8836.98,8925.35,9014.60. The exact value in every row is given by the rule below. Plot thirteen separate integer-index points, with quarter index horizontal and dollars vertical; a suitable window is 0 to 13 and 7800 to 9200. Do not infer a balance between modelled update times from a connecting line \(\displaystyle B_j=8000\left(\frac{101}{100}\right)^j\)
3. 3Matching dates and checks Model A gives 8000,8320,8640,8960 at years 0,1,2,3. Compare those with B at quarters 0,4,8,12, not at quarters 0,1,2,3. At quarter 12, B is about $9014.60 and its three-year interest is about $1014.60. The full quarter 3 value is 8242.408; quarter 4 interest is 82.42408, about $82.42. First-year interest is 8324.83208−8000, about $324.83. Retain the exact row or full stored value between updates \(\displaystyle B_{12}=1.01B_{11}=1.01\times8925.346773322532=9014.600241055758=8000(1.01)^{12}\)

Given information

\(\displaystyle R=4\%\text{ nominal per year},\quad k=4\)

Find the percentage rate applied in one quarter. Give the percentage number without a percent sign

Given information

\(\displaystyle t=3\text{ years},\quad k=4\)

How many quarterly updates occur over three years?

Given information

\(\displaystyle B_0=8000,\quad B_{j+1}=1.01B_j,\quad j=12\)

Find Model B’s three-year balance to two decimal places

Given information

\(\displaystyle B_3=8242.408,\quad q=1.01\)

Find interest earned during quarter 4 only, to two decimal places

Why are there thirteen plotted values but only twelve updates?

Compare rates and verify a financial target

Keep the fictional $8,000 starting amount and three-year horizon. Compare Model B with Model C, which compounds annually at 4.05% per year. Neither has fees or other transactions. Choose and justify the comparison quantity before calculating. Do not select from the quoted annual percentages alone

1. 1Rates, horizon and inverse Calculate each effective annual rate and three-year balance. State which is larger under these assumptions, and how the choice criterion would differ for an amount owed rather than an investment. Under Model B, find the minimum starting amount in whole cents needed to reach at least $9,000 after three years. Check the adjacent cent amounts instead of assuming ordinary rounding preserves the target
2. 2Threshold and counterexample Find the first complete quarter when Model B’s $8,000 balance is strictly above $9,000. Show the previous quarter and qualifying quarter. Then replace the target with the exact value 8000(1.01)^12: distinguish the first quarter at least equal to this target from the first quarter strictly above it. Explain the model’s exclusions and what could change your recommendation
3. 3Save independent work first Save your model choices and reasons, complete tables, discrete plots, formulas or commands, checks and written conclusions before opening the model response. Complete construction, explanations and actual external-tool operation are not automatically marked; review them yourself or with a teacher

Model response: Compare rates and verify a financial target

Compare your saved work with the values, assumptions and checks. Other suitable tools and windows are acceptable when the initial index, units and model conventions are preserved

1. 1Effective comparison Model B’s effective annual rate is 100[(1.01)^4−1]=4.060401%, about 4.0604%. Model C’s annual compounding makes its effective rate 4.05%. Model C has C(0)=8000, C(n+1)=1.0405C(n), with n in years; its values at years 0,1,2,3 are 8000,8324,8661.122,9011.897441. Model B finishes about $2.70 higher after three years even though its quoted nominal rate is lower. The comparison is conditional, not evidence about real products. For otherwise comparable amounts owed, a lower balance would be preferable
2. 2Minimum principal to whole cents The exact initial requirement is 9000/(1.01)^12, approximately 7987.043027. A minimum whole-cent amount must be rounded upward to $7987.05. Verify that 7987.04(1.01)^12 is below 9000 and 7987.05(1.01)^12 is above 9000. An ordinary nearest-cent approximation is not a sufficient proof of meeting an at-least target
3. 3Whole-period boundary B(11) is about $8925.35 and B(12) is about $9014.60, so quarter 12 is the first above $9000; the positive multiplier greater than one establishes monotonic growth. For the new exact target B(12), the first quarter at least equal is 12, but the first strictly above is 13. Do not compare rounded displayed balances to decide exact equality
4. 4Qualified conclusion State the principal, period/rate conventions, complete-year horizon and final-only rounding assumption. Fees, access restrictions, changing rates, repayments or deposits could change the result or require a different recurrence. A model-generated value is not a guaranteed transaction or a complete financial recommendation. Cash-flow timing and progressively revised payment models are extended in the next chapter

Given information

\(\displaystyle q=1.01,\quad k=4\)

Give Model B’s effective annual percentage rate to four decimal places. Enter the percentage number

Given information

\(\displaystyle B=8000(1.01)^{12},\quad C=8000(1.0405)^3\)

Find Model B minus Model C after three years, in dollars to two decimal places

Given information

\(\displaystyle B_j=8000(1.01)^j,\quad B_j>9000\)

Find the first whole quarter when Model B is strictly above $9,000

Given information

\(\displaystyle P(1.01)^{12}\ge9000\)

Find the minimum starting amount in whole cents that reaches at least $9,000 after twelve updates

For the exact target 8000(1.01)^12, which threshold statement is correct?

Choose and test depreciation models

A fictional asset starts at $24,000. The available hypotheses are a fixed $3,000 annual loss, a 15% annual loss on the current value, or a cost of $2 per unit of use. Usage in years 1 through 4 is 1000,1500,2000,2500 units. Choose and justify a model for each stated mechanism; a larger estimated value alone is not evidence that a hypothesis is valid

1. 1Construct all three models Write initial values and recurrences with the correct year or usage index. Build complete year-end tables from year 0 through year 4, show the cumulative usage and construct a discrete graph of the unit-cost values. Compare value, total positive depreciation loss and signed change in value without interchanging them
2. 2Boundary and validity Compare the fixed annual-loss and usage-based values in every year. Find the first strict reversal of their year-1 ordering, explicitly allowing an equality year. Explain why changing annual usage does not produce a constant annual subtraction even though the cost per unit is fixed. State valid index ranges and evidence needed before using any model as an estimate of resale value
3. 3Save independent work first Save your model choices and reasons, complete tables, discrete plots, formulas or commands, checks and written conclusions before opening the model response. Complete construction, explanations and actual external-tool operation are not automatically marked; review them yourself or with a teacher

Model response: Choose and test depreciation models

Compare your saved work with the values, assumptions and checks. Other suitable tools and windows are acceptable when the initial index, units and model conventions are preserved

1. 1Rules and complete records Let F(n) and R(n) use completed years: F(0)=R(0)=24000, F(n+1)=F(n)−3000 and R(n+1)=0.85R(n). Their values for years 0 through 4 are F: 24000,21000,18000,15000,12000 and R: 24000,20400,17340,14739,12528.15. For the use index j, U(0)=24000 and U(j+1)=U(j)−2. Cumulative usage after years 0 through 4 is 0,1000,2500,4500,7000, giving year-end values 24000,22000,19000,15000,10000.
2. 2Representations and strict comparison Plot those five unit-cost values at year indices 0 through 4; a suitable window is years 0 to 5 and dollars 0 to 26000. Use separate points. Fixed annual-loss values are lower than unit-cost values in years 1 and 2, equal in year 3, and higher in year 4. The first strict reversal is year 4, not the equality year. Annual usage losses are 2000,3000,4000,5000; the constant recurrence coefficient belongs to the per-use index, not an assumed fixed annual subtraction
3. 3Quantities and limits At year 4, the unit-cost future value is $10,000, positive depreciation loss is $14,000 and signed value change is −$14,000. The fixed-loss rule reaches zero at year 8 and should not be extended to negative asset values without a different interpretation. The unit-cost rule remains nonnegative only up to 12,000 units; reducing-balance values remain positive at finite indices. These algebraic limits do not prove real valuations. Compare condition, actual use, market evidence and changes in the asset before selecting a practical model

Given information

\(\displaystyle R_0=24000,\quad R_{n+1}=0.85R_n\)

Find the reducing-balance value after four years to two decimal places

Given information

\(\displaystyle U_0=24000,\quad c=2,\quad u=1000+1500+2000+2500\)

Find the unit-cost value after the four years, in dollars

Given information

\(\displaystyle F=(24000,21000,18000,15000,12000),\quad U=(24000,22000,19000,15000,10000)\)

The entries in each sequence correspond to years 0, 1, 2, 3 and 4, respectively. At which first whole year is the fixed annual-loss value strictly higher than the unit-cost value?

At year 4 in the unit-cost model, which statement keeps value and loss distinct?