Integer Sequence Explorer

My integer sequence database explorer is finally up and running over at my website.  Essentially, this application allows the user to choose two integer sequences from many common sequences, then runs a query which selects common members from each sequence.

One of the interesting tasks has been to write many small applications which produce the sequences to be loaded into the database relations.  I started out a little too optimistic, and wrote code which calculated the first million primes, squares, sum of divisors, and others.  I quickly found that when I ran queries on these tables, they took way to long, and my server was timing out.  I incrementally scaled things back until I was left with the first 10,000 members of most sequences.  Now my task is to re-design the database to allow for larger searches.

I will be loading up more sequences over the next few weeks, and posting explanations and links for the sequences.

Object Oriented Factorization Program in PHP

I love actuarial stuff.  But, when I am learning any kind of new programming or mathematics, I always think number theory.

On my deathbed, I will probably not be thinking about loss ratios.  But I will be thinking about divisibility.  Number theory is the thing that originally drew me, and many others, to mathematics.

So, in the interest of learning PHP, I wanted to put a factorization application up on my website. But I wanted to be able to factorize more than one number at a time, so that one can observe the relationship between properties of different integers. So, I thought, what I want is several instances of the factorizer. That means writing an object oriented application.
You will observe below that the constructor does all of the work. That is because number theory stuff is very calculation intensive. Many of the variables, like phi, can be calculated as by-products of other calculations.
 class DestructNumber { var $N; private$sumofdivisors = 0; private $divisorlist=array(); private$primedivisorlist=array(); private $primedivisorstring; private$phi = 1; // this constructor does all of the work // that allows the utility to use a more efficient // algorithm than a strictly obect oriented approach // would allow 

public function __construct($N) {$this->number = $N; // find half of divisors$counter = 1; $end = floor(sqrt($this->number)); for ($m=1;$m <= $end;$m++ ) { if ($this->number%$m == 0) {$this->divisorlist[$counter]=$m;$this->sumofdivisors=$this->sumofdivisors +$m; $counter++; } } //find other half of divisors for($m = $counter-1;$m > 0;$m--) {$quotient = $this->number/$this->divisorlist[$m]; if ($this->divisorlist[$m] !=$quotient) {$this->divisorlist[$counter] = $quotient;$this->sumofdivisors=$this->sumofdivisors +$quotient; $counter++; }}  // extract primes$bar=1; for($m = 2;$m <= $counter-1;$m++) { if (DestructNumber::primep($this->divisorlist[$m])) {$this->primedivisorlist[$bar][1]=$this->divisorlist[$m]; $bar++;} } //find powers of primes for ($m=1; $m <=$bar-1; $m++) {$power=1; $x = round(pow($this->primedivisorlist[$m][1],$power));

while ($this->number %$x == 0
&& $x <=$this->number)
{
$power++;$x = round(pow($this->primedivisorlist[$m][1], $power)); }$this->primedivisorlist[$m][2] =$power-1;
}
// create prime divisor string
$this->primedivisorstring =””; for ($m = 1; $m <=$bar-1;$m++) {$this->primedivisorstring .= $this->primedivisorlist[$m][1];
if ($this->primedivisorlist[$m][2]>1 )
{ $this->primedivisorstring .= ““.$this->primedivisorlist[$m][2].”“; } if ($m < $bar-1) {$this->primedivisorstring .= “*”;}
$this->phi =$this->phi * round(pow($this->primedivisorlist[$m][1], ($this->primedivisorlist[$m][2]-1)))*($this->primedivisorlist[$m][1]-1);
}
}

All of the work has been done by my constructor. Now, I just need some methods to return the values.
 // define methods private function primep ($N){$stop = floor(sqrt ($N)); for ($m=2; $m <=$stop; $m++) { if ($N%$m ==0) return 0; } return 1; } public function get_N(){ return$this->number; } public function get_sigma(){ return $this->sumofdivisors-$this->number; } public function get_numberdivisors(){ return count($this->divisorlist); } public function get_divisorstring(){ return$this->primedivisorstring; } public function get_phi(){ return $this->phi; } public function get_prime(){ return (count($this->divisorlist) == 2 ) ? "Yes" : "No"; } public function get_perfect(){ return ($this->sumofdivisors ==$this->number) ? "Yes" : "No"; } public function get_square(){ return ( count($this->divisorlist)%2 == 1) ? "Yes" : "No"; } } There are some fun things here. A number is prime if it has two divisors. A number is square if it has an odd number of divisors. Here is the link for my web factorizer. Next, I will be posting a database mining application to explore number properties. A Simple Windows Application for Factorizing Integers So, I spend so much time working and doing mathematics and walking the dogs that I don’t always spend much time on programming. But, over the last year, I have realized that everything that I need to make windows applications is already on my computer. I don’t need any fancy software: just a text editing program, and an open source compiler to compile from the command line. So my first code that I am posting here is a C++ application for factorizing integers. It just runs in a window with no fancy GUI. But it factorizes numbers very nicely. The link for the downloadable exe file is below the code.  #include <iostream> #include <cmath> #include <cstdlib> #include <cstdint> #include <string> // #include <NTL/ZZ.h> get this number theory stuff eventually  bool primep (int64_t N); int main() { using namespace std; std::string another ="N"; do { int64_t N; int64_t divisorlist[1000]; int64_t primedivisorlist[100][3]; int64_t sumofdivisors =0; int64_t phi=1; int counter (1); bool squarep (false); cout<< "This application factors integers of up to seventeen digits."<<endl; cout << "Enter an integer to be factored: " <<endl; cin>>N; cin.ignore(32767, '\n'); for (int m=1; m<= floor(sqrt(N));m++ ) { if (N%m == 0) {divisorlist[counter]=m; counter++;} } for(int m=counter-1;m> 0;m--) { int64_t quotient = N/divisorlist[m]; if (divisorlist[m] != quotient) {divisorlist[counter]=quotient; counter++; }} // now extract primes int bar=1; for(int m =2; m<=counter-1;m++) { if (primep(divisorlist[m])) {primedivisorlist[bar][1]=divisorlist[m]; bar++;} } // now find powers of primes for (int m=1; m<=bar-1; m++) { int power=1; int64_t x = round(pow(primedivisorlist[m][1], power)); while (N % x ==0 && x<=N) { power++; x = round(pow(primedivisorlist[m][1], power)); } primedivisorlist[m][2]=power-1; } cout<<N<<" = "; for(int m =1; m<=bar-1;m++) { cout<<primedivisorlist[m][1]; if (primedivisorlist[m][2]>1 ) cout<< "^"<< primedivisorlist[m][2]; if (m<bar-1) cout<<"*"; phi=phi*round(pow(primedivisorlist[m][1], (primedivisorlist[m][2]-1)))*(primedivisorlist[m][1]-1); } cout<<endl; cout<< "Number of divisors: "<<counter-1<<endl; cout<<"The divisors are: "<<endl; for(int m =1; m<=counter-1;m++) { cout<<divisorlist[m]<<" "; sumofdivisors = sumofdivisors+divisorlist[m]; } cout<<endl; cout<<"The sum of the aliquot divisors of "<<N<<" is "<<sumofdivisors-N<<". Hence, this number is "; if (sumofdivisors >(2*N)) {cout<<"abundant. "<<endl;} else if (sumofdivisors<(2*N)){cout<<"deficient. "<<endl;} else {cout<<"perfect. "<<endl;} cout<<"Phi, the number of integers less than and relatively prime to "<<N<<" is "<<phi<<endl; cout<<N<<" is "; if (primep (N)) cout<<" "; else cout<<"not "; cout<<"prime."<<endl; squarep = (counter-1)%2; cout<<N<<" is "; if (squarep) cout<<" "; else cout<<"not "; cout<<"a perfect square."<<endl; cout<<"Another number? "; std::getline(std::cin, another); } while ((another == "yes") || (another == "y")); return 0; } bool primep (int64_t N)  //returns t if N is prime { for (int m=2; m<= floor(sqrt (N)); m++) { if (N%m ==0) return false; } return 1; }  The code is very simple and self-explanatory. It is not the most efficient and beautiful way to factorize a number, but it works well enough within the realm of integers that can be easily handled by the base C++ variable types. Here is the link for the exe file: Factorizer Multi-Period Binomial Options Here is my latest program for calculating the prices of options using a multi-period binomial tree. It works, but represents a bunch of strange methods to solve unexpected problems. First, I decided to calculate the price at each node by finding the price at the terminal nodes, then use a formula, rather than calculate the prices recursively. This technique, as you will see, led to the problem of calculating binomial coefficients. The thing that I am most satisfied with is that I was able to get the program to output to a text file, which can be opened in Excel to visualize the data. The only thing that the program does not do is to do the same calculation with the Black-Scholes formula, and visualize how the binomial calculation approaches the Black-Scholes result. First, take a look at my output: We can see how the results approach a limit with each iteration. The pricing for periods 34 and 35 is starting to get weird because of limits of integer sizes and floating point number sizes. Here is a chart of the above data: Finally, here is the code: #include <cstdlib> #include <iostream> #include <cmath> #include <iomanip> #include <fstream> //Calculates price of European call or put option, and portfolio required to duplicate option, using binomial tree int nchoosek(int n,int k); //choice holds the binomial tree so that it doesn’t need to be recalculated unsigned long long choice [50][50]={}; int main() { using namespace std; float S0, r,div, K, h,t,sigma, u, d,pStar, uS,dS,Cu,Cd,delta, bondValue; double optionPrice; double prices [50]; int n; string callPut; int putTrue; //fill in nchoosek tree for (int m=1; m<50;m++) { choice [m][0]=1; choice [m][m]=1;} for (int m=2;m<(50);m++) { for (int l=1;l<m;l++){ choice[m][l]=choice[m-1][l-1]+choice[m-1][l]; } } //get data from user cout<<“Calculate the price of a European call or put, using a one period binary tree, then find rate with n period tree.”<<endl; cout<<“Current Stock Price?”; cin>>S0; cout<<“Continuously Compounded Risk Free Rate?”; cin>>r; cout<<“Continuous Dividend Rate?”; cin>>div; cout<<“Volatility?”; cin>>sigma; cout<<“Strike Price?”; cin>>K; cout<<“Time to Expiration?”; cin>>t; cout<<“Number of Periods? (less than 50 please)”; cin>>n; cout<<“Call or Put?”; cin>>callPut; if ((callPut == “put”)||(callPut==”p”)) putTrue=-1; else putTrue=1; //calculate likely upward and downward motions u=exp((r-div)*t+sigma*sqrt(t)); d=exp((r-div)*t-sigma*sqrt(t)); uS=u*S0; dS=d*S0; cout<<“Upward Move “<<uS<<endl; cout<<“Downward Move “<<dS<<endl; //calculate payoff of call, given motions above //the variable putTrue turns Us-K into K-Us. Cool. Cu=fmax(0, putTrue*(uS-K)); Cd=fmax(0, putTrue*(dS-K)); //calculate Delta, Beta, and call option price delta=exp(-div * t)*(Cu-Cd)/(S0*(u-d)); bondValue=exp(-r * t)*((u*Cd)-(d*Cu))/(u-d); optionPrice=delta*S0+bondValue; cout<<“To duplicate the option, buy “<<delta<<” shares=”” of=”” the=”” stock=”” for=”” “<<delta*s0<<“=”” dollars=”” and=”” borrow=”” “<<-1=”” *=”” bondvalue<<“=”” at=”” rate=”” “<<r<<“=”” a=”” total=”” price=”” “<<optionprice<<<span=”” class=”hiddenSpellError” pre=”” data-mce-bogus=”1″>endl; cout<<“Price of Option “<<optionPrice<<endl; cout<<“Prices above 30 levels deep may accumulate errors wildly”<<endl; //Multi Period tree int j=1; while(j<(n+1)){ h=t/j; u=exp((r-div)*h+sigma*sqrt(h)); d=exp((r-div)*h-sigma*sqrt(h)); pStar=(exp((r-div)*h)-d)/(u-d); optionPrice=0; for (int m=0; m<(j+1); m++) { //This is the big doozy. //Instead of computing the price at each interior node, only the terminal nodes are calculated //and the price is found using the binomial theorem optionPrice=optionPrice+nchoosek(j,m)*pow(pStar,j-m)*pow((1-pStar),m)*fmax(0,(putTrue* ((S0*pow(u, j-m)*pow(d,m))-K))); } optionPrice=optionPrice*exp(-r*t); prices [j]=optionPrice; cout<<“Periods: “<<j<<” “; cout<<“Option Price: “<<optionPrice<<endl; ++j; } //output prices to text file //open the text file in excel //data is delimited by commas ofstream outf(“Results.txt”); if (!outf) { cerr << “File Error”<<endl; exit(1); } outf<<“Depth of Tree”<<“,”<<“Price”<<endl; j=1; while(j<(n+1)){ outf << j <<“,”<< prices[j]<< endl; ++j; } return 0;} int nchoosek(int n, int k) { if (n>49 ||k>49)return 0; else { return choice[n][k]; }} The heart of everything is this ugly beast: optionPrice=optionPrice+nchoosek(j,m)*pow(pStar,j-m)*pow((1-pStar),m)*fmax(0,(putTrue* ((S0*pow(u, j-m)*pow(d,m))-K))); This is better put as $C=C+\binom j m p^{*^{j-m}}(1-p*)^m*max(0,put(S_ou^{j-m}d^m-K)$ You can figure out why this works fairly easily. I like the bit where I multiply (F_{0,T}-K) by -1 to get (K-F_{0,T}. More than anything, this exercise shows how errors creep into these long calculations. Looking at the graph, you can see how our numbers are already wandering away from the correct numbers after the 20th iteration. A Simple C++ Program for Computing Binomial Option Prices I have been advised that C++ and VBA are commonly requested languages for actuaries. So, while I am studying for exam MFE, I have been writing simple programs. My native language is LISP. I like its beauty, and its adaptability. I am sure that C++ has its promoters, but I think that it is a shame that such an ugly programming language has become one of the most used. Of course, I am still a novice to the language, so my code is still extra ugly. Computes call price using binomial model: #include <cstdlib> #include <iostream> #include <cmath> #include <iomanip> //Calculates price of call option, and portfolio required to duplicate option, using binomial tree int main() { using namespace std; float S0, r,div, K, h, sigma, u, d, uS,dS,Cu,Cd,delta, bondValue, callPrice; //get data cout<<“Calculate the price of a call, using a one period binary tree”<<endl; cout<<“Current Stock Price?”; cin>>S0; cout<<“Continuously Compounded Risk Free Rate?”; cin>>r; cout<<“Continuous Dividend Rate?”; cin>>div; cout<<“Strike Price?”; cin>>K; cout<<“Time to Expiration?”; cin>>h; cout<<“Volatility?”; cin>>sigma; //calculate likely upward and downward motions u=exp((r-div)*h+sigma*sqrt(h)); d=exp((r-div)*h-sigma*sqrt(h)); uS=u*S0; dS=d*S0; cout<<“Upward Move “<<uS<<endl; cout<<“Downward Move “<<dS<<endl; //calculate payoff of call, given motions above if (uS > K) Cu = uS-K; else Cu = 0; if (dS > K) Cd = dS-K; else Cd = 0; //calculate Delta, Beta, and call option price delta=exp(-div * h)*(Cu-Cd)/(S0*(u-d)); bondValue=exp(-r * h)*((u*Cd)-(d*Cu))/(u-d); callPrice=delta*S0+bondValue; cout<<“To duplicate the call, buy “<<delta<<” shares=”” of=”” the=”” stock=”” for=”” “<<delta*s0<<“=”” dollars=”” and=”” borrow=”” “<<-1=”” *=”” bondvalue<<“=”” at=”” rate=”” “<<r<<“=”” a=”” total=”” price=”” “<<callprice<<<span=”” class=”hiddenSpellError” pre=”” data-mce-bogus=”1″>endl; cout<<“Price of Call “<<callPrice<<endl; //find the price of some other related calls cout<<“Some other Call prices: “<<endl; float step; step = 0.05*K; K= K-5*step; for (int z=0;z<11;z++){ if (uS > K) Cu = uS-K; else Cu = 0; if (dS > K) Cd = dS-K; else Cd = 0; delta=exp(-div * h)*(Cu-Cd)/(S0*(u-d)); bondValue=exp(-r * h)*((u*Cd)-(d*Cu))/(u-d); callPrice=delta*S0+bondValue; cout<<setw(8)<<k<<” “<<callprice<<<span=”” class=”hiddenSpellError” pre=”” data-mce-bogus=”1″>endl; K=K+step; } return 0; } //Wow what an ugly program Here is a sample run: New Study Materials for Exam FM You may have noticed that I have been adding some study materials to my blog, as a side project. If you click on the “Easy FM Quizes” tab, you will find some interactive quiz problems. These are dumped over from my easy practice problems I made when I was studying for the exam. I have more than 1000 more exercises that I will transfer over in the coming weeks. I am even more excited about the exercises that I am putting into nice PDF problem sets. If you have heard me talk about math, you know that I learned algebra and calculus from early 20th century text books. I have a sentimental attachment to the problem sets at the ends of chapters. It has given me great satisfaction to write at simple LaTeX class that duplicates the feel of these textbook problems. When I am done, they will be a single document. The links for these problem sets will be posted on the right side bar. Enjoy! Some Study Materials For Exam MFE It looks like I will take MFE, Models For Financial Economics, in March, 2015. The best source for the information is the SOA page for the test. It has links for the current Syllabus, the sample problems, and video solutions from the Wisconsin School of Business. The great thing about this exam is that the only recommended study resource is the MacDonald book, which we all still have from exam FM. G. Stolyarov II has a very nice free study guide. At first glance I like it, and I like his studying hints. Beyond that, I will be creating study cards again, reviewing FM material, and probably using Coaching Actuaries. Once again I will be studying to instrumental jazz music. My goal is to get to the point of taking timed exams a few weeks before the actual exam. So, my countdown on the sidebar will be to February 20, but I will be taking the actual exam somewhere around the ides of March. Wish me luck. Eight Secret Tricks to Passing (Exam FM) 1. Be over-prepared: The syllabus is so extensive, and there are so many twists that may be put on standard material, that there are always going to be questions that look foreign. The more that you are prepared, the more that you will be able to handle some of these. Being over prepared also helps to deal with emotions. There were points during the exam when a little voice in my head said “There is no way that you are going to pass this.” Yet, that same voice was in my head during many sample exams which I passed with flying colors. Experience and over-preparation reveals that you can succeed even when you are not at your best. If you walk into the exam center with worries, or a cold,or the flu, or pain in your back, you know that you can still succeed. 2. Fill in the tough spots: It can hurt your head to learn new things. Identify the areas with which you struggle, and try to get your head around them. Two weeks before the exam, I still had some head-pain topics. Things like duration matching, convexity, interest rate swaps, and put-call parity. But, they were on my daily study schedule, and I forced myself to confront them each day. It paid off. 3. Fill in the fundamentals: Once you gain proficiency in a topic, the fundamentals look different. In the month before the test, I went back and reviewed some of the basics of interest theory. At this point, the basics were easy, and I picked up on some of the finer points that I missed the first time around. Amortization schedules, for instance. After I learned the basic formulas, it was easy to ignore the schedules they are based on. Yet many problems are made much easier by using schedules, rather than relying on formulas. 4. Memorization is easy: Once you understand things thoroughly, you don’t really need to do much memory work for these exams. After months of working problems, most things will be anchored in your head. And yet suppose that you find yourself trapped in a dark alley by two annuities, one of which is twice as long as the other. Then you will be glad that you memorized $\frac{a_{\overline{2n}\lvert }}{a_{\overline{n}\lvert }}=1-v^n$ I have never encountered this fact in practice, but if I did, recognizing this identity might save me a couple of minutes. Spend a little time on memory work, it is easy. 5. Test symbolic solutions with numbers: If you have a problem with a purely symbolic solution, and you can narrow it down to a couple of possible solutions, you can frequently replace the variables with reasonable numbers, and see if the result is true. This is my favorite new solution technique I learned while studying for this exam. 6. Read MacDonald: Actually read it. After you learn the math, go back and read it. If you look at the notes from the sample problems, you will notice that the exam writers have made it pretty clear that the exam will be based on the book. Those non-numeric multiple choice questions can be very challenging. Read the book. 7. Number your Scrap Paper: I learned this trick when taking practice exams. If you have time to review problems at the end of the exam, you will need to find the scratch-work easily. 8. Practice with dull number two pencils: Two weeks before the exam, I put all of my mechanical pencils away, and worked only with old fangled sharpenable pencils. Two hours into the exam, you will be working with dull stubs, so you need to be prepared. Problems in the Queue A week left until the financial mathematics exam. The best thing that I can say is that I can honestly not have worked any harder at studying. In the last three months, I have studied at least three hours a day, and frequently I have done eight to ten hours a day. Much of that time has been absolutely focused. The question is: have I studied as smartly as possible? I won’t know the answer to that until after the examination. I have done about two hours of problems today. I see that there are 19 problems remaining in the queue. These are problems that I have seen before, and that I still wish to spend some time with. I spend enough time with each problem that I feel I am confident that I understand several ways to get to a solution, or until I get sick of looking at it. If I get sick of looking at it, it comes up on the queue again later today or tomorrow. If I am fluent with it, I don’t look at it for a few days. Eventually, it disappears from the queue. Earlier today, I already did some easy numerical exercises. Later today, I will probably do an exam, which is a whole other kind of problem solving, because of time pressure (5 minutes per problem.) What I am going to do here is type in problems as they come up on my screen, then write up solutions. If I don’t understand the problem, you will get to watch me struggle. Person X enters into a long forward contract. If the spot price at expiration were S, the payoff would be -20. If the spot price at expiration were 1.2S, the payoff would be X. Person Y enters into a short forward contract. If the spot price at expiration were 0.8S, the payoff would be 40. If the spot price at expiration were 1.1S, the payoff would be Y. The forward price on each contract is the same. What is X+Y? We aren’t going to do a diagram for this one. I tried, on paper, and since we don’t know the values of 0.8S, 1.1S, or 1.2S, it is hard to know where to place them in relation to the forward price. So, let’s go with straight algebra, and see if that leads us to a reasonable solution. Payoff on Long Forward = S – F Payoff on Short Forward = F – S From Givens: $S-F = -20 \\ 1.2S=S=X \\ F-0.8S=40 \\ F-1.1S = Y$ Add equations 1 and 3 to get S=100. Then solve for F and get F=120. Using equations 2 and 4: $X+Y = 1.2(100)-120+120-1.1(100) = 10$ That wasn’t too bad. These derivative problems can be a little intimidating at first. A 15 year bond with semiannual coupons has a redemption value of$100. It is purchased at a discount to yield 10% compounded semiannually. If the amount for accumulation of discount in the 27th payment is \$2.25, which of the following is closest to the total amount of discount in the original purchase price?

We have all sorts of information here, so it should be easy to get to an answer. First, since the coupons are semiannual, we can just think entirely in 6 month terms. Since the bond is sold at a discount, we know that the coupon rate is less than the interest rate. We know that we can solve this problem by finding the original purchase price, and subtracting it from the redemption price. First, we need the coupon, which we can find using the premium discount formula.

Let’s start with that. Assuming that F=C, the amount of discount in the kth payment is:

$F(i-r)v^{n-k+1}$

Here are our givens:

$n=30 \\ F=C=100 \\ i=0.1 \\ F(i-r)v^{n-1+k} = 2.25 \\ 100 (0.05-r)1.05^{-(31-27)} = 2.25 \\ r=0.02265 \\$

Now solve for the original sale price. We might as well continue with the premium discount formula:

$P = C+(Fr-Ci) a_{\overline{n}\lvert } \\ P=100-0.2735\frac{1-1.05^{-30}}{0.05} \\ P=57.96$

Discount in original price = 42.04.

Let’s just do one more.

To accumulate 8000 at the end of 3n years, deposits are made at the end of the each of the first n years and 196 at the end of each of the next 2n years.  (1+i)^n = 2.

What is n?

These problems are much simplified by visualizing them in the right way. The easiest way to think of it is payments of 98 from 1 to 3n years, plus payments of 98 in years 2n+1 through 3n.

In math:

$98s_{\overline{3n}\lvert }+98_{\overline{2n}\lvert }=8000 \\ \frac{(1+i)^{3n}}{i} + \frac{(1+i)^{2n}}{i}=81.633 \\ \frac{2^3-1}{i} + \frac{2^2-1}{i} = 81.633 \\ i= 12.25%$

I am going to post one more, because it is a real bear. To solve it, you trust that the math will lead you to the answer.

Given a k year bond with semiannual coupons, and a yield rate of 10% convertible semi-annually, sold at a price p.

If the coupon rate had been r-0.04, the price would be P-200.

Calculate the present value of a 3k year annuity immediate paying 100 at the end of each 6 month period, at a rate of 10% semiannually.

Start at the end. We need to find:

$100 \frac{1-1.05^{-6k}}{0.005}$

Which means that what we really need is k (although v^k will do).

Dive in:

$P=1000\frac r 2 \frac{1-1.05^{-2k}}{0.05}+1000(1.05)^{-2k} \\ P-200 = 1000\frac{r-0.04}{2}\times \frac{1-1.05^{-2k}}{0.05}+1000(1.05)^{-2k} \\ \text{We can see that the redemption values will not be significant}\\ 1000 \frac r 2 a_{\overline{2k}\lvert }=1000 \frac{r-0.04}{2}a_{\overline{2k}\lvert }+200 \\ 1000 \frac r 2 a_{\overline{2k}\lvert }-1000 \frac{r-0.04}{2}a_{\overline{2k}\lvert }=200 \\ 500a_{\overline{2k}\lvert }(r-r+0.04)=200 \\ a_{\overline{2k}\lvert } = 10 \\ k = 7.1 \\ 100 \frac{1-1.05^{-6\times7.1}}{0.05} =1749.75$

Forward Contracts

The thing about math at a certain level is that there are no more easy exercises.  I have learned to make a habit of creating simple exercises.  These are for the forward price, which is the contracted price to buy an asset at time T in the future; and the prepaid forward price, which is the price paid now for an asset that will be delivered at time T.  In these problems, r is the continuous interest rate, and delta is the continuous dividend rate.
$F^P_{0, T} = S_o = S_o -PV(divs) = S_0 e^{-\delta} T$
$F_{0, T} = S_0 e^{rT} = S_oE^{rT} - AV(divs) = S_0e^{(r-\delta)}T$

1. $S_0 =1000, r=0.04, \delta = 0.01\quad F^P_{0, 6m}?$
2. $S_0 =800, r=0.02, \delta = 0\quad F_{0, 2m}?$
3. $S_0 =800, r=0.02, \delta = 0\quad F^P_{0, 2yr}?$
4. $S_0 =500, r=0.04, \delta = 0\quad F_{0, 2yr}?$
5. $S_0 =500, r=0.04, \delta = 0\quad F_{0, 6m}?$
6. $S_0 =500, r=0.04, \delta = 0\quad F^P_{0, 1yr}?$
7. $S_0 =100, r=0.03, \delta = 0.01\quad F^P_{0, 2yr}?$
8. $S_0 =100, r=0.03, \delta = 0.01\quad F^P_{0, 3m}?$
9. $S_0 =1000, r=0.04, \delta = 0.01\quad F_{0, 1yr}?$
10. $S_0 =1000, r=0.04, \delta = 0.01\quad F_{0, 6m}?$
11. $S_0 =800, r=0.02, \delta = 0\quad F^P_{0, 5m}?$
12. $S_0 =100, r=0.03, \delta = 0.01\quad F_{0, 1yr}?$
13. $S_0 =100, r=0.03, \delta = 0.01\quad F_{0, 9m}?$
14. $S_0 =500, r=0.04, \delta = 0\quad F^P_{0, 9m}?$
15. $S_0 =1000, r=0.04, \delta = 0.01\quad F^P_{0, 1yr}?$
16. $S_0 =800, r=0.02, \delta = 0\quad F_{0, 1yr}?$

Solutions:

1. $1000e^{-0.01*0.5}=995.01$
2. $800e^{0.02*(\frac 1 6)}=802.67$
3. $800$
4. $500e^{.04*2}=541.64$
5. $500e^{0.04*.5}=510.10$
6. $500$
7. $100e^{-0.01 *2}=98.02$
8. $100e^{-0.01*0.25}=99.75$
9. $1000e^{0.04-0.01}=1030.45$
10. $1000e^{(0.04-0.01)*0.5} =1015.11$
11. $800$
12. $100e^{0.03-0.01}=102.02$
13. $100e^{(0.03-0.01)*0.75}=101.51$
14. $500$
15. $1000e^{-0.01}=990.05$
16. $800e^{0.02}=816.16$

Later today, I will post some tougher ones that require a little thinking.