Download Optimization Techniques

Dynamic Programming
Applications
Water Allocation –
Numerical Example
1
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Objectives
To demonstrate the water allocation problem
through a numerical example using
Backward approach
Forward approach
2
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Numerical Problem


Consider a canal supplying water for three different crops
Maximum capacity of the canal is 4 units of water.
Field 3
x3
x1
Field 1
x2
Field 2

3
Optimization Problem: Determine the optimal allocations xi to each
crop that maximizes the total net benefits from all the three crops
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Numerical Problem …contd.
Net benefits from producing the crops can2 be expressed as a function of
the water allotted. NB1 ( x1 )  5 x1  0.5 x1
NB2 ( x2 )  8 x2  1.5 x2
NB3 ( x3 )  7 x3  x3
2
2
The net benefit values
are calculated for each
crop and are as shown
4
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Numerical Problem …contd.

Representation of the problem as a set of nodes and links
x1
x2
4
0
x3
4
0
1
0
2
3
4
1
2
3
3
1
4
2
2
2
2
1
1
1
3
4
1
0
Crop 1
5
0
Crop 2
D Nagesh Kumar, IISc
1
0
0
0
Crop 3
Optimization Methods: M6L4
Numerical Problem …contd.
6

The values inside the node show the value of state variable at
each stage

Number of nodes for any stage corresponds to the number of
discrete states possible for each stage.

The values over the links show the different values taken by
decision variables corresponding to the value taken by state
variables
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Numerical Problem: Solution by Backward
recursion
Sub-optimization function for the 3rd crop:
f 3 ( S3 )  max NB3 ( x3 ) with the range of
x3
0 x3  S 3
7
D Nagesh Kumar, IISc
S 3 from 0 to 4.
Optimization Methods: M6L4
Solution by Backward recursion … contd.
Next, by considering last two stages together, the sub-optimization
function is f ( S )  max NB ( x )  f ( S  x )
2
2
x2
x2  S 2
2
2
1
2
2
The calculations are shown below
Table 3
8
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Backward recursion … contd.
9
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Backward recursion … contd.
Considering all the three stages together
f1 (Q)  max NB1 ( x1 )  f 2 (Q  x1 ) with S1  Q  4
x1
0 x1 Q
10
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Backward recursion … contd.
Backtrack through each table,

Optimal allocation for crop 1, x1 = 1 and S1 = 4
Thus, S2  S1  x1  3
From 2nd stage, the optimal allocation for crop 2, x2 = 1.
Now, S3  S2  x2  2
From 3rd stage calculations, x3 * = 2
Maximum total net benefit from all the crops = 21
11
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Numerical Problem: Solution by Forward
recursion
Start from the first stage and proceed towards the final stage
Suboptimization function for the first stage
f1 ( S1 )  max NB1 ( x1 )
x1
x1  S1
Range of values for S1 is from 0 to 4
The calculations are shown in the table
12
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Forward recursion … contd.
13
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Forward recursion …
contd.
f 2 (S2 ) 
 NB2 ( x2 )  
max 

x2
f
(
S

x
)
2
2 
x2  S 2  1
14
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Forward recursion …
contd.
f 3 ( S3 )  max NB3 ( x3 )  f 2 ( S3  x3 ) with S3  4
x3
x3  S 3 Q
15
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Solution by Forward recursion …
contd.
Backtrack through each table,
Optimal allocation for crop 3, x3 * = 2 and S 3 = 4
Then, S 2  S3  x3  2

x
The optimal allocation for crop 2, 2  1
Now, S1  S2  x2  1

From 1st stage calculations, x1  1
Maximum total net benefit from all the crops = 21
These solutions are the same as those we got from backward recursion
method.
16
D Nagesh Kumar, IISc
Optimization Methods: M6L4
Thank You
17
D Nagesh Kumar, IISc
Optimization Methods: M6L4