Hussman Interview Q3

Please see solutions to Q1 & Q2.

3) A vehicle is filled with fuel. One gallon of fuel can propel every 10 pounds of vehicle weight a
distance of 2 miles (5 pounds of weight 4 miles, etc). A gallon of fuel weighs 8 pounds.

a) Write a general expression for the distance traveled by the vehicle

b) If f0 is the starting amount of fuel, write an analytic equation for the total distance traveled

c) If the vehicle weighs 100 pounds when empty, and has traveled 1 mile after expending half
of its starting amount of fuel, what is the starting amount of fuel, and how far does it
ultimately take the vehicle (round to 4 decimal places)?

d) Suppose you didn’t discover that an analytic solution to this problem requires calculus (now
there’s a hint!). Write a short program in C++, Basic, Matlab or pseudocode to approximate
the amount of fuel required to move the vehicle (100 pounds when empty) 5 miles.

Weight of Vehicle = V; Weight of Fuel/Gallon = F; Gallons of Fuel = G; Distance Travelled = D; Weight of Vehicle and Fuel = W

I am not sure the difference between part a and b but here is my attempt.

a) D is directly proportional to G and inversely proportional to W
D ~ G/W => D = KG/W, where K is proportionality constant
In this case when D = 2; G = 1; W = 10. Therefore, K = 20.
Thus, when W = 5; D = 20*1/5 = 4

b) W = V + F*G => D = KG/(V+FG)
Total distance travelled is tricky because as soon as you burn a little fuel the weight reduces and the further you can travel.
Therefore, to find distance travelled we first need to know what happens when fuel changes by small amount i.e. differentiate the above equation with respect to G

Now to find total distance travelled we have to integrate when fuel G ranges from f₀ to 0
Let us first solve indefinite integral

Now adding limits total distance travelled is

 
c) Weight of Vehicle = V = 100lbs; Weight of Fuel/Gallon = F = 8lbs/g; Gallons of Fuel = G = f₀ –> f₀/2; Distance Travelled = D = 1 and K = 20

Substituting,

10000+1200f₀+32f₀² = 250*(-4f₀)
32 f₀² + 2200f₀ + 100 = 0
f₀ ~ –63.8562 & –4.8938
The equation has 2 roots and both negative. So, it is clear I have made a mistake. However, my calculus is a little rusty and I will leave the solution so that the internet can explain the mistake
 
d) This will only make sense if I have the correct equation. So, I will not attempt to solve it.

Hussman Advisors Financial Analyst Pre Interview Question 2

I am solving question 2 of Hussman Advisors Financial Analyst pre interview question. Question 1 was solved earlier.

Q2 Let the value of a firm’s equity be equal to the discounted stream of deliverable cash flows
(“DCF”) and assume that firm‐wide DCF grows at rate g, so Vt = DCFt+1 / (k‐g). Let Nt be the
number of shares of stock at time t, so the per‐share stock price Pt is simply Vt/Nt. Each year, the
company pays a proportion d of DCF to existing shareholders as dividends, and then uses the
remaining proportion (1‐d) to repurchase shares. Assume that shares are repurchased at price
Pt, which simultaneously determines Nt.


a) Derive a simple expression for the per‐share dividend growth rate g*, in terms of k, d and g
only, and give a simple interpretation (Hint: Write down expressions involving DCF, don’t be
afraid to substitute equivalent expressions, and notice that all growth rates will be constant
given the assumptions above. In particular, Nt/Nt+1 = Nt‐1/Nt).


b) Show algebraically that if the per‐share dividend growth rate g* is used, the standard
dividend discount model holds even in the presence of share repurchases, so that adding in
repurchases as if they were a separate payment to shareholders would actually represent
double‐counting. Specifically, show that Vt/Nt = Dt+1 / k‐g*.

a) From the problem statement we have the following equation:

P₀ = V₀ / N₀ = (DCF₁/N₀) / (k-g)

I will use g’ instead of g* as * is generally used as multiplication sign and may confusion i.e. g’ = g*
We can also drive the price per share using dividend discount method i.e. if Dividends D_t grow at constant rate g’ then

P₀ = D₁/k-g’

Also, from the problem statement D_t = d*DCF_t
Therefore, P₀ = D₁/k-g’ = (d*DCF₁/N₀) / (k-g’)

The price per share obtained from both the methods should be same.

Therefore, k-g’ = d*(k-g) or g’ = g* = k - d*(k-g)

I am not sure if the question expected me to prove that P₀ obtained from two methods is same or not.
However, there is an entire paper that goes on to show the equivalence of Cash Flow Method and Dividend Discount Method.


b) From the result above: d * (k-g) = (k-g’)

V_t/N_t = DCF_t+1/(k-g) = D_t+1 / d*(k-g) = D_t+1 / (k-g’)

This makes sense since we assumed P_t (V_t/N_t) is equivalent to derive g’ in the first case.

Hussman Strategic Advisors Financial Analyst Pre Interview Questions

Hussman Strategic Advisors has a bunch of pre-interview questions for Financial Analyst / Financial Engineer position. I will try my best to answer some of those questions. I will tackle the 1st question in this post.

What is the payment on a 12‐year, $400,000 mortgage that compounds on the payment date every 73 days at 6%? (show your formula and calculations, and ignore leap year)

I do not really understand the question so here is my best guess. I am assuming I do not have to worry about individual days in a month. Also, I am assuming 6% is the annual rate.


Step 1: Calculate relationship between annual rate, 73 days rate and monthly rate.

365 Days = 12 months => 73 Days = 12*(73/365) months = 2.4 months

Let R be the annual rate in decimal (0.06 in this case); r be the monthly rate in decimal and R* is the rate every 73 days (or every 2.4 months)

To convert rates lets start with something familiar.
If a loan compounds every 6 months (half yearly) then 6 months rate is R/2 i.e. R/(12/6)
If a loan compounds every 3 months (quarterly) the 3 months rate is R/4 i.e. R/(12/3)
If a loan compounds every month then monthly rate is R / 12 i.e. R/(12/1)

So, if a loan compounds ever 2.4 months (73 days) the 73 day rate is R/(12/2.4) = R/5 = R*

Now, monthly rate r compounded for 2.4 months = R* i.e. (1+r)^2.4 = 1 + R*

r = (1+R*)^(1/2.4) – 1 = (1+(R/5))^(1/2.4) – 1 = (1+(0.06/5))^(1/2.4) – 1 = 0.004983 = 0.4983%

If the loan had compounded every month instead of 73 days then r = R/12 = 0.5%


Step 2: Derive Payment Formula

Let L be the loan amount ($400,000) and P be the payment every month

At the end of the 1st month:
Outstanding Loan Amount, L₁ = L*(1+r) – P
Let (1+r) = X
L₁ = L*X – P

At the end of 2nd month:
Outstanding Loan Amount, L₂ = L₁*X – P = L*X² – P*X – P

At the end of 3rd month:
Outstanding Loan Amount, L₃ = L₂*X – P = L*X³ – P*(X² – X – 1)

At the end of nth month:
Outstanding Loan Amount, L_n = L_n-1*X – P = L*Xⁿ – P*(X^n-1 – X^n-2 – … 1)

Sum of Geometric Progression (GP) 1+x+x²+x³+…x^n-1 = (xⁿ – 1)/(x – 1)
Using the GP formula and replacing X by (1+r)

L_n = L*(1+r)ⁿ – P*(((1+r)ⁿ – 1) / r)

At the end of loan term (n = 144 months in this case) outstanding amount = 0
When L_n = 0 then P = L*r*(1+r)ⁿ / ((1+r)ⁿ – 1) = L*r / (1 – 1/(1+r)ⁿ)
Therefore, P = 400,000*0.004983 / (1-1/(1+0.004983)¹⁴⁴) = 3899.39