Portfolio Management puzzles, solved step by step

About Rs 6.3 crore over 10 days and Rs 9.2 crore over 21 days. With independent daily returns, variance adds across days, so volatility and value at risk scale with the square root of time: 2 x the root of 10 and 2 x the root of 21. The rule fails when returns trend or mean revert, when tails are fat, and when the book changes during the period.

Why the square root of time and not time itself?

Think of a person taking random steps left or right. After a hundred steps they are rarely a hundred steps from the start, because the left steps cancel the right ones; the typical distance is about ten, the square root of a hundred. Daily returns behave the same way when each day is independent. Variances add across independent days, so the spread of a ten-day return is the daily spread times the square root of ten, not times ten. Value at risk at a fixed confidence level is a multiple of that spread, so it scales the same way: Rs 2 crore becomes Rs 6.32 crore over ten days and Rs 9.17 crore over 21.

When does the rule give the wrong answer, and in which direction?

The rule rests on three assumptions, and each one breaks in real markets. If returns trend, so a bad day tends to follow a bad day, the true ten-day loss is larger than Rs 6.3 crore; if they mean revert, it is smaller. Fat tails make the 99% point further out than a normal curve suggests, and the ratio between the tail and the spread need not hold across horizons. And a book is not frozen: over a month a desk cuts losing positions, which the scaling ignores. Say which way each one pushes the number and the interviewer knows you understand the rule rather than having memorised it.

One more thing worth saying: the scaling also assumes the expected daily return is zero. Over a day that is harmless. Over a year, drift matters and a simple square root rule starts to overstate the loss for a portfolio with a positive expected return.

200 miles. The car needs 100 miles at 50 miles an hour, which is two hours. The bird flies without stopping for those two hours at 100 miles an hour, so it covers 200 miles, however many times it turns. Summing the legs gives the same answer: 66.7 there and back, then 22.2 there and back, each pair a third of the one before, which adds to 200.

Why is summing the legs the slow way?

Picture a dog on a walk that runs ahead to the gate and back to you, over and over, until you reach the gate. Nobody counts the dog's sprints; you just ask how long the walk took and how fast the dog runs. When something moves at a constant speed for a known time, distance is speed times time, whatever path it traces. The bird's zig-zag looks like the hard part of the question. It is a distraction. The only thing that matters is when the flying stops, and that is when the car arrives.

How do you check 200 the long way, in case the interviewer asks?

The bird and the car close at 150 miles an hour, so they first meet after 100 / 150 of an hour, 40 minutes, when the car is 33.3 miles from A and the bird has flown 66.7 miles. The bird flies 66.7 miles back to B. By then the car is 66.7 miles along, 33.3 miles from B, and the same geometry repeats on a gap a third as large. Each round trip is a third of the one before, so the legs form a geometric series: 133.3 times one over one minus a third, which is 200.

Why ask this on a quantitative research desk? Because the same move, stepping back from the path to the total, is how you price anything path-dependent in your head: ask what is conserved or fixed before tracing every step.

Macaulay duration about 2.83 years and modified duration about 2.67. At par the yield equals the 6% coupon, so the cash flows discount to 5.66, 5.34 and 89.00, which add to 100. Weight each year by its share of the price: 1 x 0.0566 + 2 x 0.0534 + 3 x 0.8900 gives 2.83. Divide by 1.06 for modified duration: a one point rise in yield cuts the price by roughly 2.67%.

What is duration, if not the time to maturity?

Picture a seesaw with three children sitting at the one, two and three metre marks. If the child at three metres is much heavier, the pivot that balances the seesaw sits close to three, not at the middle. Macaulay duration is that pivot: the average time you wait for your money, with each payment weighted by its present value. Here the three weights are the discounted coupons of 5.66 and 5.34 and the discounted final payment of 89.00. The last payment is so heavy that the balance point, 2.83 years, sits only two months short of maturity.

Why divide by 1.06 to get modified duration?

Macaulay duration is a time. Modified duration is a price sensitivity: the percentage change in price for a one point change in yield. With annual compounding the two differ by a factor of one plus the yield, so 2.833 divided by 1.06 is 2.673. Say what it means: if yields rise from 6% to 7%, the bond loses about 2.67% of its price, a little less in reality because the price-yield curve bends. That bend is convexity, and it is the natural next question.

About 800 tonnes a year, on stated assumptions. Weddings: about 1 crore a year at 35 grams each is 350 tonnes. Festivals and gifting: 30 crore households, a quarter buying 4 grams, is 300 tonnes. Investment coins and bars: 5% of households buying 10 grams is 150 tonnes. The total is 800 tonnes. Check it against published industry demand figures before using it anywhere.

How do you start when you have no idea of the answer?

Ask when a family actually walks into a jeweller. It is not random: a wedding, Dhanteras or Akshaya Tritiya, a birth, or a decision to save in coins. Splitting a total into the occasions that drive it turns one number nobody knows into several that anyone can picture. State the base first: about 140 crore people at four to five per household is roughly 30 crore households. That figure and every weight below are assumptions for the estimate, said out loud so the interviewer can challenge them.

How do you sanity check it, and where is it weakest?

Divide back: 800 tonnes over 30 crore households is about 2.7 grams per household per year, a small coin's worth on average, which feels plausible given that most households buy nothing in a typical year and a few buy a lot at a wedding. The weakest input is grams per wedding, because the spread is huge and the average is dragged up by a few large weddings. Moving it by 10 grams moves the total by 100 tonnes. Say that, and say you would check the result against published demand data from the industry before quoting it: an estimate is a structure plus assumptions, not a statistic.

On a multi-asset or wealth desk, the follow-up is usually why it matters: household gold is a large part of Indian savings, so its demand affects imports, the rupee and how much money is left for financial assets.

5 with one trade and 7 with any number. With one trade, buy at the lowest price that comes before a higher one: buy at 1 on day 2 and sell at 6 on day 5, for 5. With unlimited trades, collect every day-to-day rise and sit out every fall: 1 to 5 earns 4, 3 to 6 earns 3, a total of 7. The general answer is the sum of the positive daily changes.

How do you find the best single trade without checking every pair?

Walk through the prices once, like a shopper who notes the cheapest price seen so far and asks each day how much they would make selling today. The best single trade is the largest gap between today's price and the lowest price seen before today, found in one pass. At day 2 the low is 1. Day 3 offers 5 minus 1, which is 4; day 5 offers 6 minus 1, which is 5; no later day beats it. The trap is subtracting the lowest price from the highest overall: the high of 7 comes before the low of 1, so it cannot be sold after buying.

Why is the unlimited answer just the sum of the rises?

Any profitable trade from a low to a later high can be split into daily steps, and it gains only on the up days inside it while paying for every down day it sits through. With no limit on trades and no shorting, the best strategy holds the stock on every day it rises and nothing on every day it falls, so profit equals the sum of positive daily changes. Here that is 4 plus 3, which is 7. One pass through the prices gives the answer, which is the point of asking a coding-flavoured candidate.

Say the limitation as a portfolio manager would: this is perfect hindsight with no costs. Add a transaction cost per trade and the two answers move toward each other, because catching a small rise is no longer worth paying for.

About Rs 80.5 crore. Vacancy of 8% takes Rs 10 crore of gross potential rent down to Rs 9.2 crore collected. Operating costs of 30% of that are Rs 2.76 crore, leaving net operating income of Rs 6.44 crore. Divide by the 8% cap rate: 6.44 over 0.08 is Rs 80.5 crore, or 12.5 times the income.

Why does the cap rate price net operating income and not the rent?

Think of buying a shop that a friend runs. You would not pay for the sales the shop could make if every shelf sold out; you pay for what is left after empty days and the electricity bill. A cap rate is net operating income divided by value, so the value is only ever as good as the income that actually reaches the owner after vacancy and running costs. Gross potential rent is the ceiling, not the income.

What are the lines between gross rent and value, in order?

Say them as a ladder. Gross potential rent is what a full building earns at the rent roll. Take off vacancy and bad debt to get effective gross incomeThe rent a building actually collects after vacancy and unpaid rent, before any running costs.. Take off operating expenses, such as maintenance, insurance, property tax and management, to get net operating income. Every line above NOI moves the price by 12.5 times its size at an 8% cap rate, so a small leak near the top is a large leak in value. One extra point of vacancy costs Rs 0.1 crore of rent, Rs 0.07 crore of NOI, and Rs 0.875 crore of value.

For an exit value, the same arithmetic runs on the NOI expected in the sale year and an exit cap rate, which analysts often set a little above the entry cap rate to allow for an older building. Say which cap rate you are using and why, and say that capital spending such as a new roof sits below NOI and is not captured by this shortcut.

Only 20%. Take 1,000 applicants. 40 will default and the model flags 36 of them. 960 are good and the model wrongly flags 15% of them, 144 people. The flagged pile holds 180, and 36 of those default: one in five. The model is good at catching defaulters, but defaulters are so rare that false alarms outnumber them four to one.

Why is 90% the wrong answer when the model catches 90% of defaulters?

Picture a smoke alarm that always sounds when there is a fire and also sounds now and then for burnt toast. Because toast burns far more often than houses do, most alarms in a year are toast. A flag's meaning depends on how common the thing it looks for is: when defaulters are 4% of applicants, even a modest false alarm rate on the other 96% produces more wrong flags than right ones. The 90% is the chance a defaulter gets flagged; the question asks the reverse, the chance a flag is a defaulter.

Why count people instead of using the formula?

Bayes' rule gives the same answer, but natural frequencies are faster to say and harder to get wrong under pressure. Turn every percentage into a count of people out of a round number, and the answer is simply the flagged defaulters over everyone flagged. The formula version is 0.04 x 0.9 over (0.04 x 0.9 plus 0.96 x 0.15), which is 0.036 over 0.18, or 20%. Offer it as the check after the counts.

Then say what a lender does with it. A flag at 20% is a reason for a closer look, not a rejection. The cleared pile, by contrast, holds only 4 defaulters in 820, about 0.5%, against 4% before the model, so the model is most useful for waving through the safe majority. Cutting the false alarm rate from 15% to 5% would lift the flagged default share to about 43%.

Roughly 12 crore 5G phones a year, on stated assumptions. Take about 60 crore smartphone users out of 140 crore people. If they replace their phone every 4 years, that is 15 crore replacements, plus about 1 crore first-time buyers, so 16 crore phones. If three quarters of new phones sold are 5G, the answer is about 12 crore. A household check, one phone every two years across 30 crore households, gives 15 crore, close enough.

Why start from users and replacement, not from the population?

Think of how many school shoes a town buys in a year. The number of children matters, but the answer is set by how often each child outgrows a pair. In a market where most people already own the product, yearly sales are the user base divided by the replacement cycle, plus a smaller flow of first-time buyers. Population only tells you the ceiling; the cycle turns the stock of users into a yearly flow.

Which assumptions should you say out loud, and how do you check them?

Every number here is an assumption to be stated, not a fact to be quoted: the user base, the cycle, the first-time flow and the 5G share. The replacement cycle deserves the most care because a one-year change in it swings the answer by between a fifth and a third; the population barely matters by comparison. A three-year cycle gives 15.75 crore and a five-year cycle 9.75 crore. The 5G share is the other moving part, because it changes fast from one year to the next; ask which year the interviewer means.

Then check from a different direction. With about 30 crore households, one new phone per household every two years gives 15 crore phones, close to the 16 crore from the user build. Two methods that land near each other are more convincing than one precise-looking number. For a real estimate, replace every assumption with published industry shipment data and confirm the current figures.

If the interviewer wants value rather than units, multiply by an assumed average selling price and say that 5G phones are skewed to the middle and upper price bands, so the price assumption needs as much care as the cycle.

Durations of 12.5 and 33.3 years; the value stock falls about 6% and the growth stock about 14%. With price equal to D over (r minus g), duration is 1 over (r minus g): 1 over 0.08 and 1 over 0.03. A 50 basis point rise takes the spreads to 8.5 and 3.5 points, so prices fall by 1 minus 8/8.5, which is 5.9%, and 1 minus 3/3.5, which is 14.3%.

Why does a stock have a duration at all?

Think of two people valuing a lottery ticket: one pays out a small sum every year starting now, the other pays little now and a lot decades from now. If the interest rate rises, the second ticket loses far more value, because its money is further away and gets discounted for longer. An equity is a stream of future cash flows, so like a bond it has a duration: the weighted distance to its cash, and the growth stock's cash sits much further out. In the constant-growth model that distance comes out as 1 over (r minus g).

How do you get the price changes exactly, and why is the duration estimate too big?

The price is proportional to 1 over (r minus g), so compare spreads before and after. For the value stock the spread moves from 8 to 8.5 points, a 5.9% price fall; for the growth stock it moves from 3 to 3.5, a 14.3% fall. Duration times the rate change gives 6.25% and 16.7%, a little larger, because price is a curved function of the rate: like a bond with convexity, the stock loses less than the straight-line estimate. The curvature matters more the longer the duration.

This is the arithmetic behind a familiar market pattern: when real yields rise sharply, long-duration growth stocks usually fall more than value stocks. Say the limitation too. The model assumes growth is fixed while the rate moves; in practice rates often rise because growth is strong, which can offset part of the fall. And no real company grows at 8% forever, so the growth stock's duration is a stylised upper figure.

1,000 x (1 + 1/2 + 1/3 + ... + 1/n) miles, which grows without limit but only like the logarithm of n. Drive all n cars 1,000/n miles; together they have burned one full tank, so one car can refill the rest and be abandoned. Repeat with n minus 1 cars for 1,000/(n minus 1) miles, and so on. Four cars reach 2,083 miles, 100 cars about 5,187, and the distance tracks 1,000 x (ln n + 0.577).

When should a car drop out of the convoy?

Think of friends sharing water on a long walk, where every bottle is full at the start and nobody can carry more than one. The moment the group has drunk exactly one bottle's worth, one friend can pour the rest of theirs into everyone else's bottles and head home. A car should drop out the instant the convoy has burned exactly one tank in total, because that is the first moment its remaining fuel exactly fills the others. With n cars that happens after 1,000/n miles, since n cars burn fuel n times as fast as one.

Why does the distance grow only like a logarithm?

The total is 1,000 times the harmonic sum 1 + 1/2 + ... + 1/n. Each extra car adds the shortest leg of the journey, 1,000/n miles, so the gains shrink as the convoy grows, and the harmonic sum rises like ln n plus about 0.577. It never stops growing, so any distance is reachable in principle, but slowly: 10 cars reach about 2,929 miles, and getting past 5,000 miles takes 83 cars. Doubling the fleet adds only about 1,000 x ln 2, roughly 693 miles.

Why a hedge fund asks it: the structure is the same as scaling a strategy. Each extra unit of capital or effort buys a smaller increment, and a candidate who sees the diminishing returns and names the rate of decay is showing the instinct the desk wants. Also be ready to argue optimality in one sentence: any plan that drops a car earlier wastes fuel it cannot hand over, and dropping later wastes the fuel spent carrying a car that is no longer needed.