Retirement Equation

When can I retire from my job and live the life I want?

After playing the retirement game, it becomes clear that just having a good job is not quite enough in today’s world. By “good” I mean good pay, of course. And, “job”, unfortunately, often means “practically no family life except on days off”. Hence the popularity of the topic, “how much money does it take to retire from this job?” in company lunch breaks! I noticed this topic started coming up regularly in the IT MNC that I worked in, just after the year 2000 crash. I suppose the realization that one’s job is not secure also became more acute after that year, and seems to have remained quite acute from then onwards.

If you happened to read the analysis of the retirement game, you may have noticed that with certain investing strategies plus some good luck, one can build up a big enough corpus to take early retirement! This observation lead to the present article that will attempt to answer the following question:

How much money do I need to retire from my job right now?

I remember we used to have numerous lunch break discussions around this question, but they never yielded any proposals that everyone could agree upon. The fatalistic camp’s solution was, “You can retire today provided you live within your means”. The pessimistic camp came up with “No amount of money that we can make will allow us to retire”. The engineering camp came up with the simple formula,

Survival period = (Savings) / (expenses)

I think, in those discussions, we all missed the significant role played by the principle of compounding.  Well, this post presents an analysis that takes compounding into account.

What factors need to be taken into account to answer the retirement question? In my mind, at least the following ones:

  • K Rate of inflation
  • R Rate of return on investment
  • T Life expectancy, or time to live
  • X0 Initial Annual expenditure, or lifestyle level
  • P0 Initial net worth (at time of retirement)

These factors can be used to find solutions for these time dependent variables

  • P(t) net worth at time t
  • X(t) Expenditure level at time t

Although the first three factors vary with time and circumstances, and their values can only be guessed (since they relate to the future), we will perform our analysis assuming constant average values, with the expectation that some insights can nevertheless be gained.

Note: This post does not address the question, “how does one accumulate a high net worth?”

Basic Algorithm: We shall start with simple models and then explore the results of a more detailed analysis. In all models, however, the basic idea is the same, and in fact this same algorithm was used in the design of the Retirement Game:

  1. Start with a certain net worth, or principal.
  2. Add return on investments.
  3. Subtract expenses
  4. Repeat from #2 for the next time period

Note, however, both net worth and expenses are moving targets: they vary with time.

Reference values: To get a feel for how the models work, we shall use the following numerical values for our protagonist Sal from the previous post:

  • K Inflation multiplier, pa. 1.075 (that is, 7.5% per annum)
  • R Return on investment, pa. 1.075 to 1.15 (7.5% to 15% per annum)
  • T Time to live from day of retirement. 60 years
  • X0 4.0 Lakh Rs per year (i.e. 33,333 Rs per month)

T = 60 is a reasonable round value for a person in late 20’s or early 30’s. R is post-tax return.

NOTE: 1 Lakh = 100,000 and 1 Crore = 1,00,00,000 = 10 million.

So now let us look at some retirement models:

Return Equals Inflation (R = K) Model

Simplest model to analyze is to nullify the impact of inflation by taking return on investments to equal rate of inflation. Although return will keep pace with inflation, expenses will eat into the principal. Further, in real money terms, expenses stay constant with time. As real principal is steadily consumed to meet real expenses, we get a simple linear relation:

P0 = X0 * T

This, you will recall, is what the engineering camp had come up with. Using our reference numbers, we get:

P0 = 4.0 Lakh Rs/yr * 60 yr = 2.4 Crore Rs = 24 Million Rs

The equation for P(t) is given below. It is obtained by solving eqn 3 described below, for the special case R = K.

eqn01               k = ln(K)           (eqn 1)

Figure 1 shows a plot of P(t) with initial savings of 240 Lakh (blue curve) and 150 Lakh (red curve).


Figure 1: Net worth when return on investment equals inflation

With respect to the blue curve, it is worth noting that a person not sensitized to the dangers of inflation will feel very secure and relaxed for the first 45 years as net worth grows from 2.4 Crore to 15 Crore even after dipping into the principal for expenses. But around the 55th year panic will set in. In the next few years, he is left wondering, “What happened?”. The red curve shows what happens with initial savings of 150 Lakh: the peak is not as prominent, and of course the money doesn’t last as long.

Steady State Model

This model takes a more optimistic view, that return on investments is greater than rate of inflation, to the extent that expenses can just be met without eating into the inflation-adjusted principal. This is a steady state because real principal remains constant, so one can live indefinitely off it.

(R – K) * P0 = X0

P0 = X0 / (R – K)

Thus, if better acumen yields 15% return, i.e. 7.5% above inflation rate, Sal can live a steady state life with a roughly five times smaller P0 (net worth at retirement) than in the previous model:

P0 = 4.0/(0.075) =  53.3 Lakh


We can also turn this equation around. Instead of finding P0 , determine value of return R, for a given P0. Thus for example, given a starting principal of 100 Lakh, Sal will need

R = K + X0/P0 = 1.075 + 4/100 = 1.115, i.e. 11.5% return per annum.

Figure 2 is a plot of required return on investment for varying values of P0. X0 is set to 4 lakh and inflation to 7.5%.


Figure 2: Steady State Model, R versus P0

Some interesting observations can be made from Figure 2:

  1. Sustained returns of 25%, 30% or more are extremely difficult, if not impossible. So small sized initial savings, say 25-50 Lakh, just don’t work.
  2. Beyond a sweet spot of around 100-150 Lakh, critical value of R doesn’t ease off drastically. In other words, it is worth putting in serious effort to increase R even by a few percent, so that you don’t have to keep working longer to accumulate a larger corpus.

Big Bang Model

If return on investment is higher than that required in the steady state model, then real net worth will continue to grow without bound, exponentially. This, of course is the most desirable state! One has to generate a high R and a high P0 to attain this.

A General Solution

Now we address the more general case. Let’s ease into this by first solving for expenses, X(t). Remember, X is annual expenditure, a cash flow rate. X is measured in Rs/year, not Rs.

dX/dt = k X;  X(0) = X0

eqn02          (eqn 2)

Lowercase k in the exponent is related to annual rate of inflation K, as follows. For t = 1 year,

K = X(1) / X0 = ek

Hence k = ln(K). For example, k = 0.07232  (for 7.5% inflation).

Similarly, exponent for return on investment, r = ln(R).

To take a numerical example for eqn 2, the daily expenditure on the first day of the 10th year will be as follows.

X(10) =  X0 e^10k  = 4 * 2.061 = 8.244 Lakh/year.

Divide X(10) by 365 to get 824400 / 365 = 2258 Rs / day.

Now let’s look at P(t). Change in principal P(t) in time dt comes from two factors:

  • Increase due to return on investment in time dt
  • Decrease due to expenses in time dt

That is,

dP = r P dt – X dt

Using value of X from eqn 2, the equation can be written as:

eqn03         (eqn 3)

To take a numerical example for eqn 3, with P = 60 Lakh, X = 4 Lakh and returns of 10% pa (r = 0.0953) on a particular day, dt = 1/365,

Day’s return = r P dt = 60,00,000 * 0.0953 *(1/365) = 1566.74 Rs

Day’s expenses = X dt = 4,00,000 / 365 = 1095.89 Rs

Day’s reinvestment = dP = 1566.74 – 1095.89 = 470.85 Rs

Incidentally, this only looks okay until you realize that due to 7.5% inflation, P lost its value in one day by k P dt = 60,00,000 * 0.07232 *(1/365) = 1188.82 Rs.

Now, solving equation 3, we obtain (for r ≠ k):



Another way to examine equation 4 is to calculate P in real money terms (i.e., discounted for inflation):


Several observations can be made by inspecting equation 4; these are discussed in the next section.

Big Crunch Model

When R is too small to yield Steady State or Big Bang, net worth will eventually drop to zero. If P0 > B, P will keep rising, leading to the big bang model. P0 = B is the steady state model.

Net worth will eventually drop to zero when P0 < B. Let’s explore this case some more.

Case r > k

B  = X0 / (r-k) is positive.

Since P0 < B, P may rise initially due to second term but eventually first term will drag P down. In fact, P will become zero at T = ln(B/(B-P0)) / (r-k)

Figure 3 plots P(t) for several values of P0 and R. Only curves with P0 < B are plotted.


Figure 3: P(t) for several values of P0 (initial savings) and R (return on investment)

Notice how powerful R is compared to P0 :

  • 50 Lakh lasts for 40 years with R=16% but only 22 years with R=14%
  • 50 Lakh with R=16% lasts longer than 75 Lakh with R=12%

Increasing returns has far more impact than increasing initial net worth!

Case r < k

B = X0 / (r-k) is negative. Let B’ = -B.  Then


Here, the first term may cause P to rise but eventually second term drags P down.

P will become zero at T = ln((B’ + P0)/B’) / (k – r)

Figure 4 plots P(t) for several values of P0 and R. Compare with Figure 3. In particular observe that 50 Lakhs initial savings lasts only 10 years at 4% R (fig 4), but 40 years at 16% R (fig 3).


Figure 4: P(t) for R < K

Curtailing Expenditure

An alternative question to ask could be, “given the amount of money I have, how much can I afford to spend?” This post is already quite long, so this question is left as an exercise for the reader (just solve for X0 in eqn 4).

The Retirement Question

So finally, we are ready to answer the question,

“What is the least amount of money needed to retire today?”

The answer comes from eqn 4, which is reproduced below:



T = time to live. Set P(T) to zero and solve for P0:


Figure 5 is a plot of R vs P0 for T = 60, 7.5% inflation, X0 =  4 Lakh pa. Remember that r = ln(R) and k = ln(K).


Figure 5: How much money is needed to retire

So the most truthful answer is:

Well, it depends.

It depends on external factors such as inflation rate and investment climate. It depends on internal factors such as your TTL (time to live), your expenditure levels.

But I think the most important factor is the ability to generate a good return on investment. That, coupled with decent sized initial corpus, should do the trick.

Homework: Experiment with these Excel spreadsheets for your own set of parameters. (Download the zip file. The one with no macros computes #1 and #2. The one with macros also finds #3, but you need to enable macros).

  1. T = how many years money will last, given P0, K, R.
  2. P0 = initial funds needed, given  T, K, R
  3. R = required rate of return, given P0, K, T

Disclaimer: Arguments developed here use a simplistic economic model that certainly differs from reality. In addition, accuracy of the conclusions depends on the accuracy of the input parameters. In other words, beware of GIGO (Garbage In Garbage Out).

Further Reading



[3] How much you need to save to meet your future  retirement targets