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Created:
 29th of August 2025
12:35:27 AM

Modified:
 30th of August 2025
08:21:37 AM

Breakeven: Simple vs Compound Interest

Breakeven analysis — what it is and why it matters

Breakeven analysis identifies the point where two return formulas deliver the same maturity. Here we compare a simple-interest offer (linear growth with time) against a compound-interest offer (exponential growth with compounding). At the crossing, an investor is indifferent; on one side of the crossing simple interest pays more, and on the other side compound interest does.

Breakeven condition is given here for practice; in tests, derive it.

Engineers are often faced with financial decisions, each of which hold good under specific conditions. This problem should introduce you to a scenario where you will be required to have different constraints, and you ease one while keeping the other constraints constant. Ofcourse, with improved computing power, scenarios with more than one constraint being modified at the same time is possible, however we will use such scenarios for later.

Phrasing of the question.

This question was phrased thus.

Task 1: Rs* ≈ XX.XX% ; A* ≈ ₹Y,YYY
Task 2: N* ≈ ZZ.ZZ years ; A* ≈ ₹W,WWW
Task 3: <SI higher by|CI higher by> ₹V,VVV

Scenario

You are responsible for negotiating the best return for some extra money that you have lying with you from your business. You wish to have the information on what rates or duration that you should keep as the levels below which you should not go below when negotiationg with different financial institutions.

Principal: ₹1,50,000. Two offers:

• Option A — Simple Interest (SI): rate R_s is negotiable.
• Option B — Compound Interest (CI): R_c = 8.4% p.a., compounded quarterly (m = 4).

Formulas

$$ \text{SI}=\frac{P\,N\,R_s}{100}, \qquad A_{\mathrm{SI}}=P+\text{SI}=P\!\left(1+\frac{N\,R_s}{100}\right) $$ $$ A_{\mathrm{CI}}=P\!\left(1+\frac{R_c}{100\,m}\right)^{mN} $$ $$ \textbf{Breakeven:}\quad P\!\left(1+\frac{N\,R_s}{100}\right)=P\!\left(1+\frac{R_c}{100\,m}\right)^{mN} $$

What to find

1. Unknown rate. With N = 6.75 years, find R_s* (% p.a., two decimals) and the common maturity A*.
2. Unknown years. With R_s = 11.2%, find N* (years, two decimals) and A*.
3. Quick check. With N = 6.75 and R_s = 10.5%, which option is higher and by how many rupees?

As an exercise, we are restricting to fixed values of input. However, try entering different values to find the ideal solution for the scenario that you are envisioning. In practice, these values become constraints.

Option 1: Python (Using the while-loop iteration only)

We use the while loop to still work out a solution, however the algorithm uses a bisection method which will aim to achieve covergence with fewer iterations.

# si_ci_breakeven_while.py
# Breakeven (SI vs CI) using only while loops (no function defs)

# Inputs
P  = 150000.0   # rupees
Rc = 8.4        # % p.a. (compound)
m  = 4          # compounding per year
N0 = 6.75       # years for Task 1 and Task 3

# ---------- Task 1: find Rs* for fixed N0 (bisection) ----------
target = (1.0 + Rc/(100.0*m))**(m*N0)   # = A_CI/P at N0
lo, hi = 0.0, 50.0                      # search Rs in [0%, 50%]
eps = 1e-12
while (hi - lo) > eps:
    mid = 0.5*(lo + hi)
    lhs = 1.0 + N0*mid/100.0            # = A_SI/P at N0
    if lhs >= target:
        hi = mid
    else:
        lo = mid
Rs_star = 0.5*(lo + hi)
A_star1 = P*(1.0 + N0*Rs_star/100.0)

print(f"Task 1: Rs* ≈ {Rs_star:.2f}% ; A* ≈ ₹{round(A_star1):,}")

# ---------- Task 2: find N* for fixed Rs (bisection) ----------
Rs_fixed = 11.2                          # % p.a. (simple)
loN, hiN = 0.0, 50.0                     # years bracket
while (hiN - loN) > eps:
    midN = 0.5*(loN + hiN)
    lhs  = 1.0 + midN*Rs_fixed/100.0
    rhs  = (1.0 + Rc/(100.0*m))**(m*midN)
    if lhs >= rhs:
        hiN = midN
    else:
        loN = midN
N_star  = 0.5*(loN + hiN)
A_star2 = P*(1.0 + Rs_fixed*N_star/100.0)

print(f"Task 2: N* ≈ {N_star:.2f} years ; A* ≈ ₹{round(A_star2):,}")

# ---------- Task 3: single-horizon comparison ----------
Rs_offer = 10.5
A_si = P*(1.0 + Rs_offer*N0/100.0)
A_ci = P*((1.0 + Rc/(100.0*m))**(m*N0))
diff = A_si - A_ci
print(f"Task 3: {'SI higher by' if diff>=0 else 'CI higher by'} ₹{abs(round(diff)):,}")
    

Option 2: Python (Defining functions to create more reusable code)

A similar approach, but functions are defined to achieve modularity.

# Inputs
P  = 150000.0   # rupees
Rc = 8.4        # % p.a. (compound)
m  = 4          # compounding per year
N0 = 6.75       # years for Task 1 and Task 3

import math

def comp_factor(N, Rc, m):
    # This function returns the Maturity Amount at the
    # end of the period of investment for a given rate
    # of interest, using Compound Interest.
    return (1.0 + Rc/(100.0*m))**(m*N)

def A_SI(P, N, Rs):
    # This function returns the Maturity Amount at the
    # end of the period of investment for a given rate
    # of interest, using Simple Interest.
    return P*(1.0 + N*Rs/100.0)   # uses PNR/100

def A_CI(P, N, Rc, m):
    return P*comp_factor(N, Rc, m)

# ---------- Task 1: solve for Rs* with bisection (while loop) ----------
target = comp_factor(N0, Rc, m)     # equals A_CI/P at N0
lo, hi = 0.0, 50.0                  # search Rs in [0%, 50%]
eps = 1e-10
while hi - lo > 1e-10:
    mid = 0.5*(lo + hi)
    lhs = 1.0 + N0*mid/100.0        # equals A_SI/P at N0
    if lhs >= target:
        hi = mid
    else:
        lo = mid
Rs_star = 0.5*(lo + hi)
A_star1 = A_SI(P, N0, Rs_star)      # common maturity

print(f"Task 1: Rs* ≈ {Rs_star:.2f}% ; A* ≈ ₹{round(A_star1):,}")

# ---------- Task 2: solve for N* with bisection (while loop) ----------
Rs_fixed = 11.2
loN, hiN = 0.0, 50.0                 # years bracket
while hiN - loN > 1e-10:
    midN = 0.5*(loN + hiN)
    lhs  = 1.0 + midN*Rs_fixed/100.0
    rhs  = comp_factor(midN, Rc, m)
    if lhs >= rhs:
        hiN = midN
    else:
        loN = midN
N_star  = 0.5*(loN + hiN)
A_star2 = A_SI(P, N_star, Rs_fixed)

print(f"Task 2: N* ≈ {N_star:.2f} years ; A* ≈ ₹{round(A_star2):,}")

# ---------- Task 3: quick comparison ----------
Rs_offer = 10.5
diff = A_SI(P, N0, Rs_offer) - A_CI(P, N0, Rc, m)
who  = "SI higher by" if diff >= 0 else "CI higher by"
print(f"Task 3: {who} ₹{abs(round(diff)):,}")

Five short practice items

1. Given P = ₹2,25,000, N = 5.80, CI R_c = 9.0% monthly (m=12): find R_s* and A*.
2. P = ₹3,00,000, R_s = 10.1%, CI R_c = 7.4% semi-annual (m=2): find N* and A*.
3. P = ₹4,00,000, cap N = 6.25, CI R_c = 8.6% quarterly (m=4): find R_s* and A*.
4. P = ₹1,20,000, R_s = 10.8%, CI R_c = 8.4% with m = 1, 4, 12: report each N*.
5. P = ₹2,75,000, R_s = 13.0%, CI R_c = 9.2% monthly (m=12): find minimal

Some

With principal ₹1,50,000, years N, simple rate R_s% p.a., compound rate R_c% p.a., and compounding frequency m per year, the formulas are:

$$ \text{SI}=\frac{P\,N\,R_s}{100}, \qquad A_{\mathrm{SI}}=P+\text{SI}=P\!\left(1+\frac{N\,R_s}{100}\right) $$ $$ A_{\mathrm{CI}}=P\!\left(1+\frac{R_c}{100\,m}\right)^{mN} $$

Breakeven (given here for practice):

$$ P\!\left(1+\frac{N\,R_s}{100}\right) = P\!\left(1+\frac{R_c}{100\,m}\right)^{mN} $$

The principal P cancels from the breakeven equation, so the crossing depends on rates, compounding frequency, and time. This gives two natural “negotiation levers”: (i) fix the tenure N and solve for the matching rate R_s^*; or (ii) fix the quoted simple rate R_s and solve for the matching tenure N^*.

Monotonicity guarantees (existence and uniqueness at a glance)

• Unknown rate, time fixed. For fixed N>0, the left-hand side \(1+\frac{N\,R_s}{100}\) increases strictly and smoothly as R_s increases, whereas the right-hand side is a positive constant. There is exactly one solution, and it has the closed form $$ R_s^{*}(N)=\frac{100}{N}\left[\left(1+\frac{R_c}{100\,m}\right)^{mN}-1\right]. $$
• Unknown time, rate fixed. With R_s fixed, \(g(N)=1+\frac{N\,R_s}{100}\) and \(h(N)=\left(1+\frac{R_c}{100\,m}\right)^{mN}\) are both continuous and strictly increasing in N. For small N the linear term can exceed the compound term; for large N the exponential dominates. Continuity plus opposing behaviours ensures a single crossing \(N^*>0\). Convexity of the exponential rules out multiple positive roots.

How the Python while-loop code works

The script computes three items using only while loops and simple comparisons: (1) the matching simple rate R_s^* for a fixed tenure, (2) the matching tenure N^* for a fixed simple rate, and (3) a one-shot comparison at a given rate and tenure. The common technique is bisection: keep a lower and upper bound that straddle the solution, test the midpoint, and halve the bracket until it is sufficiently tight.

1. Task 1 — find R_s^* for N=N0. Compute the compound growth factor at the given tenure: $$ \text{target}=\left(1+\frac{R_c}{100\,m}\right)^{mN_0}. $$ Start with a safe rate bracket, e.g. [0, 50] (% p.a.). At each step set mid=(lo+hi)/2, form the linear left side \(1+\frac{N_0\,\text{mid}}{100}\), and compare it to target. If the left side is too large, move the upper bound down; otherwise move the lower bound up. Stop once hi − lo is below a tiny tolerance. The midpoint is R_s^*. The common maturity is \(A^*=P\!\left(1+\frac{N_0\,R_s^*}{100}\right)\).
2. Task 2 — find N^* for fixed R_s. Use the same bracket-and-halve logic on years, e.g. [0, 50]. Compare the linear left side \(1+\frac{N\,R_s}{100}\) with the exponential right side \(\left(1+\frac{R_c}{100\,m}\right)^{mN}\). Move the bound that keeps the (unique) root inside the bracket and stop when the interval is tight. Report \(N^*\) and \(A^*=P\!\left(1+\frac{N^*\,R_s}{100}\right)\).
3. Task 3 — straight comparison. Evaluate \(A_{\mathrm{SI}}=P\!\left(1+\frac{N_0\,R_s}{100}\right)\) and \(A_{\mathrm{CI}}=P\!\left(1+\frac{R_c}{100\,m}\right)^{mN_0}\), then print which is larger and by how many rupees.

Why this works smoothly: the quantities you are solving for are monotone in their unknowns. That guarantees a single solution in any bracket that encloses it, and bisection will reach it with steady, predictable convergence—no derivatives, no fragile steps.