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Dynamic Programming - Recursive Implementation

I'm having a bit of a problem modelling an optimization problem in matlab using a recursive algorithm (which is a requirement).

The problem is as such:

Decide the quantity of fish to catch each year considering a time window of 10 years knowing there are presently 10000 fishes in the lake, year 1, the growing rate of fish is the number of fishes present in the lake at the beginning of each year + 20%.

Let x be the quantity of fish to catch, $5 the price of each fish and the cost of catching fish:

    0.4x + 100 if x is <= 5000; 
    0.3x + 5000 if  5000 <= x <= 10000; 
    0.2x + 10000 if x > 10000;

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decide the number of fish to catch each year, for 10 years, in order to maximize the profit.

Future gains are depreciated by a factor of 0.2/year, which means that earning $1 in year 1 is the same as $0.8 in year 2 and so on.

I currently have defined the following objective function:

   x -> quantity of fish to catch
    b-> quantity of fish availavable in the beginning of year i
    c(x,b) -> cost of catching x fish with b fishes available

    f_i(b) = max {(5x - c(x,b)) + 0.8 * f_i+1((b - x) * 1.2)}

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How would i go about implementing this in matlab?

This is what i have so far:

Main file

  clear; % make sure previously defined variables are erased.
global FishBeginningYear Years FishSold MaxProfit  MatrixProfit

FishBeginningYear = 10; % In thousands (x1000)
Years = 10;
FishSold = 0;
MatrixProfit = zeros(10,52);

% Border Conditions
if FishBeginningYear <= 5
   MatrixProfit(10, FishBeginningYear) = 5 * FishSold - (0.4 * FishSold + 1);
elseif (FishBeginningYear > 5) && (FishBeginningYear <= 10)
    MatrixProfit(10, FishBeginningYear) = 5 * FishSold - (0.3 * FishSold + 5);
else
   MatrixProfit(10, FishBeginningYear) = 5 * FishSold - (0.2 * FishSold + 10);
end

MaxProfit  = pesca_rec(2, 10);

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Function

    function y = pesca_rec(Year, FishBeginningYear)
global FishSold MaxProfit  MatrixProfit Years

y = MatrixProfit(Year, FishBeginningYear);
if y ~= 0 %Memoization
    return
end %if

auxMax=-1000;
kchosen=0;

for k = 1 : Years
     
    % Costs
    if FishSold <= 5
       Cost = 0.4 * FishSold + 100;
    elseif (FishSold > 5) && (FishSold <= 10)
        Cost = 0.3 * FishSold + 5000;
    else
        Cost = 0.2 * FishSold + 10000;
    end

    MaxProfit  = 5 * FishSold - Cost + 0.8 * pesca_rec(k + 1, FishBeginningYear - FishSold); % * 1.2
	
	if auxMax < MaxProfit 
		auxMax=MaxProfit ;
		kchosen=k;
	end %if MaxProfit 

end; %

MatrixProfit(k,FishSold) = kchosen;
y=auxMax;

end %function

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This only computes fot the case where the year is 10 and b = 10 (value in thousands).
This is the same poblem describes as "Discounted Profits Problem" in book

EDIT: Can it be done in Java? Then it would be simpler to translate to matlab.

EDIT2: Code edited. Now i'm getting "Maximum recursion limit of 500 reached.". Heeelp

hypercube

The "book" appears to be in Portuguese which is more than difficult for me to comprehend....

PauloRP

ASKER

What do you mean? The book is in english. Plus, i already explained my problem in plain english.

TommySzalapski

If you change the .pt in the link to .com, then it displays the book in English (at least for me).

Look at the top of page 59. There is your recursive function, right there. See how the function f(t,b) is based on f(t+1,floor(s(b-xt)))? So the function f calls itself. Recursion.

PauloRP

ASKER

@TommySzalapski Can you show me a code example? I can't seem to grasp the concept. Evem more i don't know what the stop cndition would be. I keep getting errors with 0 indexing in matlab :(

TommySzalapski

The stop condition is if t > T. Notice how f(t,b) only depends on f if t <= T. When t = T+1, f(t,b) = 0;

So you just need an if statement something like
if t > T
f = 0
else
f = f(t+1, floor(....
end

PauloRP

ASKER

@TommySzalapski Great. Can you take a look at my implementation?

I'm getting an "Attempted to access prop(2,0); index must be a positive integer or logical." error on line 66 of my Main file.

My implementation is:

Main file

   clear;
    
    global M Kdep Cost RecursiveProfit ValorF prop
    
    Kdep=[10; 20; 30; 40; 50; 60; 70; 80; 90; 100]; %max nr of fish in the lake at the beginning of each year, 10 years, in thousands. Growth factor = 20%
    
    M=1000; 
    
    %Cost and Profit of selling i fishes given that there are j at the beginning of the year
    for i = 1:50
    	for j = 1:11
    		Cost(i,j) = 0.2 * i + 10;
            RecursiveProfit(i,j) = 5 * i - Cost(i, j);
    	end
    end
    
    
    for i = 1:10
    	for j = 1:10
    		Cost(i,j) = 0.3 * i + 5;
            RecursiveProfit(i,j) = 5 * i - Cost(i, j);
    	end
    end
    
    for i = 1:5
    	for j = 1:5
    		Cost(i,j) = 0.4 * i + 0.1;
            RecursiveProfit(i,j) = 5 * i - Cost(i, j);
    	end
    end
    
    %prop = 1 : 10;
    
    ValorF = -M * ones(10, 50);
    
    %Border conditions  
    for a = 1:5
        ValorF(10, a) = 5 * a - (0.4 * a + 1); %On Year 10, if there are <= a thousand fishes in the lake ...
    	prop(10, a) = a;
    end
    
    for b = 6:10
        ValorF(10, b) = 5 * b - (0.3 * b + 5); %On Year 10, if there are 6 <= a <= 10  thousand fishes in the lake ...
    	prop(10, b) = b;
    end
    
    for c = 10:41
        ValorF(10, c) = 5 * c - (0.2 * c + 10); 
    	prop(10, c) = c;
    end
    
    MaxProfit = RecursiveProfitRecursivo(1, 10)
    
    k1 = prop(1,10)
    
    kant=k1;
    
    y = 6 - Cost(kant,10);
    
    for q=2:10
        if(kant == 0)
        kant = kant + 1;
    end
    	kq=prop(q,y)
    	kant=kq;
    	y = y - Cost(kant,q);
    end %for i

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Function

    function y=RecursiveProfit(j,x)
    global M Kdep Cost Prof ValorF prop
    
    y=ValorF(j,x);
    
    if y~= -M
        return
    end %if
    
    auxMax=-M;
    decision=0;
    
    for k=1:Kdep(j)
    	if Prof(k,j) <= x-k
    		aux=Prof(k,j)+RecursiveProfit(j+1, (x - k));
    			if auxMax < aux 
    				auxMax=aux;
    				decision=k;
    			end %if aux
    		else break
    	end %if Cost   
    	
    end %for k
    
    ValorF(j,x)=auxMax;
    prop(j,x)=decision;
    y=auxMax;

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TommySzalapski

I'm guessing this is the offending line?
kq=prop(q,y)

So when you got y = 6 - Cost(kant,10); y must be 0?

PauloRP

ASKER

@TommySzalapski Sorry, i didn't understand. I'm way past my melting point with this one :(

TommySzalapski

Explain what this line is for:
y = 6 - Cost(kant,10);

Any time Cost(kant, 10) is 6 or more, you will see issues.

PauloRP

ASKER

@TommySzalapski y is supposed to be the money available minus the cost associated with a "decision" for the problem. I based my code on some code for a similar problem (resource alocation). I forgot to edit that line.
In the original problem that expression is:

Money to invest (6) minus where to invest, and compute that 3 times (3 decisions to be made). In my version there are 10 decisions. k is supposed to be "decision", k1 "decision 1" and kant "previous decision".

Thank you very much for your help. I don't have words to express my gratitude. I'm really stuck and i'm on a very tight deadline :(

SOLUTION

hypercube

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hypercube

I have the code running but am still hoping to confirm the objective statement re "is x a single number or 10 numbers?"

PauloRP

ASKER

Hey fmarshal. Sorry for not responding. The solution itself is 1 number, MaxProfit. But part os the solution is deciding how many fishes to catch each year, prop.

Thanks.

ASKER CERTIFIED SOLUTION

hypercube

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