Dynamic Programming and
Applications
Water Allocation - I
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Objectives
 To discuss the Water Allocation Problem
 To explain and develop recursive equations for backward
approach
 To explain and develop recursive equations for forward
approach
2
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Introduction
 Dynamic Programming : Sequential or multistage decision making
process
 Water Allocation problem is solved as a sequential process using dynamic
programming
Water Allocation Problem
 Consider a canal supplying water for three different crops
 Maximum capacity of the canal is Q units of water.
 Amount of water allocated to each field as xi
Field 3
x3
x1
Field 1
x2
Field 2
3
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Water Allocation Problem…
 Net benefits from producing the crops can be expressed as a function of
the water allotted.
NB1 ( x1 )  5 x1  0.5 x1
2
NB2 ( x2 )  8 x2  1.5 x2
NB3 ( x3 )  7 x3  x3
2
2
 Optimization Problem: Determine the optimal allocations xi to each
crop that maximizes the total net benefits from all the three crops
4
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Solution using Dynamic Programming

Structure the problem as a sequential allocation process or a multistage decision
making procedure.

Allocation to each crop is considered as a decision stage in a sequence of
decisions.

Amount of water allocated to crop i is xi

Net benefit from this allocation is NBi(xi)

Introduce one state variable Si :- Amount of water available to the remaining (3-i)
crops

State transformation equation can be written as
Si 1  Si  xi
defines the state in the next stage
5
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Sequential Allocation Process
The allocation problem is shown as a sequential process
6
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Backward Recursive Equations
Objective function: To maximize the net benefits
3
max  NBi ( xi )
i 1
Subjected to the constraints
x1  x2  x3  Q
0  xi  Q
for i  1,2,3
Let f1 (Q) be the maximum net benefits that can be obtained from
allocating water to crops 1,2 and 3
 3

f1 (Q )  max  NBi ( xi )
x1  x2  x3  Q

x1 , x2 , x3  0  i 1
7
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Backward Recursive Equations…
Transforming this into three problems each having only one decision



f1 (Q)  max  NB1 ( x1 )  max  NB2 ( x2 )  max NB3 ( x3 )
x1 
x2
x3

0 x1 Q 
0 x2 Q  x1  S 2 
0

x

S
3
2  x2  S 3

variable
Now starting from the last stage, let f 3 ( S3 ) be the maximum net benefits
from crop 3.
State variable S 3 for this stage can vary from 0 to Q
Thus,
f 3 ( S 3 )  max NB3 ( x3 )
x3
0 x3  S 3
8
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Backward Recursive Equations…
But S3  S 2  x2 . Therefore f 3 ( S3 )  f 3 ( S 2  x2 )
Hence,


f1 (Q)  max  NB1 ( x1 )  max NB2 ( x2 )  f 3 ( S 2  x2 )
x1
x2

0 x1 Q 
0

x

Q
 x1  S 2
2

Now, let f 2 (S2 ) be the maximum benefits derived from crops 2 and 3 for a
given quantity S 2 which can vary between 0 and Q
Therefore,
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f 2 (S2 ) 
max
x2
0 x2 Q  x1  S 2
Water Resources Planning and Management: M4L2
NB2 ( x2 )  f 3 ( S 2  x2 )
D Nagesh Kumar, IISc
Backward Recursive Equations…
Now since S2  Q  x1 , f1 (Q) can be rewritten as
f1 (Q)  max NB1 ( x1 )  f 2 (Q  x1 )
x1
0 x1 Q
Once the value of f 3 ( S3 ) is calculated, the value of f 2 ( S 2 ) can be
determined, from which f1 (Q) can be determined.
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Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Forward Recursive Equations
Let the function f i ( Si ) be the total net benefit from crops 1 to i for a
given input of Si which is allocated to those crops.
Considering the first stage,
f1 ( S1 )  max NB1 ( x1 )
x1
x1  S1
Solve this equation for a range of S1 values from 0 to Q
Considering the first two crops, for an available quantity of S 2 ,
f 2 (S2 ) can be written as
f 2 ( S 2 )  max NB2 ( x2 )  f1 ( S 2  x2 )
x2
x2  S 2
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Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Forward Recursive Equations…
S2
ranges from 0 to Q
Considering the whole system, f 3 ( S3 ) can be expressed as,
f 3 ( S3 )  max NB3 ( x3 )  f 2 ( S3  x3 )
x3
x3  S 3 Q
If the whole Q units of water should be allocated then the value of S 3
can be taken as equal to Q
Otherwise, S 3 will take a range of values from 0 to Q
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Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc
Thank You
Water Resources Planning and Management: M4L2
D Nagesh Kumar, IISc