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# Dynamic Programming - Recursive Implementation

Posted on 2013-11-04
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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;
``````

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)}
``````

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);
``````
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
``````

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
0
Question by:PauloRP
• 6
• 4
• 4

LVL 25

Expert Comment

The "book" appears to be in Portuguese which is more than difficult for me to comprehend....
0

Author Comment

What do you mean? The book is in english. Plus, i already explained my problem in plain english.
0

LVL 37

Expert Comment

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.
0

Author Comment

@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 :(
0

LVL 37

Expert Comment

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
0

Author Comment

@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
``````

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;
``````
0

LVL 37

Expert Comment

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?
0

Author Comment

@TommySzalapski Sorry, i didn't understand. I'm way past my melting point with this one :(
0

LVL 37

Expert Comment

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

Any time Cost(kant, 10) is 6 or more, you will see issues.
0

Author Comment

@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 :(
0

LVL 25

Assisted Solution

Fred Marshall earned 500 total points
Yes, the English here is fine. I was only referring to the link you provided, as a reference no doubt, and the result.  The change from google.pt to google.com seems to work.

It seems you've made good progress on this.  I'm still trying to wrap my head around your code.  If the objective function is this:

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

Then I would read it this way:  "Maximum profit over 10 years given b fishes at the beginning of the ith year ....."  I'm unclear on this.
It seems to me that you want:
Maximum profit over 10 years given a catch size of x_i for the ith year in each of the 10 years.
I'm assuming that the catch each year is different because you said:
decide the number of fish to catch each year
instead of "decide the equal number of fish to catch per year each and every year".  But, perhaps I made a bad interpretation of the objective.  Which is it?
0

LVL 25

Expert Comment

I have the code running but am still hoping to confirm the objective statement re "is x a single number or 10 numbers?"
0

Author Comment

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.
0

LVL 25

Accepted Solution

Fred Marshall earned 500 total points
Yes.  I understand the overall objective.  What I still don't understand are the specifications.  If x represents the number of fish caught in one year, is the specification to maximize total profit based on:

x_i+0,  x_i+1, x_i+2, ....... x_i+9 based on the x_i being *all equal* or 10 separate values?
0

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