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Operations Research (MTH601)
259
Segment IX: Dynamic
Programming
Lectures 42- 43
INTRODUCTION
Dynamic programming is basically a mathematical technique developed by Richard Bellman and his
associates at the Rand Corporation. This technique is a powerful tool for making a sequence of interrelated
decisions. There is no standard mathematical formulation of the dynamic programming problem, which is in
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contrast to linear programming. It is a general type of approach to problem solving and each problem has to
be analyzed depending on the conditions pertaining to the problem and the particular equations used must be
developed to suit the problem. In this way one should take care to formulate a dynamic programming problem,
using the method of recursion.
Dynamic programming provides a solution with much less effort than exhaustive enumeration. In
dynamic programming we start with a small portion of the problem and find the optimal solution for this smaller
problem. We then gradually enlarge the problem finding the current optimal solution from the previous one,
until we solve the original problem in its entirety. In this connection we refer to Bellman's principle of
optimality, which states:
"An optimal policy has the property that, whatever the initial state and initial decision are, the
remaining decisions must constitute an optimal policy with respect to the state resulting from the first decision".
Dynamic programming technique can be applied to problems of inventory control, production
planning, chemical reactor design, heat exchanger designs, business situation to take an optimal decision for
investments etc.
A number of illustrative examples are presented for developing dynamic programming procedure.
We consider the following problem called "Stage coach problem" to illustrate the concepts of
Example
dynamic programming. A salesman has to travel between points A to B indicated by the network shown in
figure
2
5
8
1
3
6
9
A
B
7
4
The distance to be traveled by the salesman from various states to other states, is given below.
2
3
4
5
6
7
8
9
10
1 20 40 30
2 70 40 60
5 10  40
8
30
3 30 20 40
6 60 30
9
40
4 40 10 50
7 30 30
Which route minimizes the total distance traveled from A to B.
Solution:
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First point we note in this problem is that the decision, which is best for each successive stage, need
not yield the over-all optimal decision.
If we follow the policy of minimum distance at each state, we land in a path A269B, with a
total distance travelled as 130 km. However it should be evident that sacrificing a little on one is less distant that
A26 = (2 + 4).
We can solve the problem by trial and error. The number of possible routes is 18 in this example and
having to calculate the total distance for all routes is a tiresome task. Instead of exhaustive enumeration of all
the routes, it is better to start with a small problem and find the optimal solution for this smaller problem. Thus
we extend this for the entire problem. This is what we try to do in dynamic programming.
The problem is divided into a number of stages and proceeds backwards from final destination. In this
example if we were in 8 or 9 and the final destination is B we have to travel only one stage to complete the
journey. This we call as one stage problem indicating that there is one more stage to go to complete the journey.
If we were in 5 or 6 or 7 we have to travel 2 stages to reach the final destination. Like this we have 4 stages from
A to B.
Let xn(n = 1, 2, 3, 4) be the immediate destination when there are n more stages to travel. Thus, the
route selected would be 1x4x3x2x1, where x1 = 10. Let fn(x, xn) be the total cost of the best over-all
policy for the last n stages, given that the salesman is in the state s and selects xn as the immediate destination
called the state. Given s and n, let xn* denote the value of xn which minimizes fn(s1, xn), and let fn*(s) be the
corresponding minimum value of fn(x, xn). Thus, fn*(s) = fn(s, xn*). The objective is to find f4*(1) and the
corresponding policy. In dynamic programming we successively find f1*(s), f2*(s), f3*(s) and f4*(1).
The problem is divided into four stages. In one stage problem, we have one more stage to go, we can
write the solution to one-stage problem as follows.
s
f1*(s)
x1*
8
30
10
9
40
10
When the salesman has two more stages to go, the solution requires a little analysis. When two stages are ahead
to reach final destination he may occupy one of the states 5 or 6 or 7. If he is in state 5, he must go to either state
8 or 9 at a cost of 10 or 40 respectively. If he selects state 8, the minimum additional distanced to travel after
reaching there, is given in the above table as 30 so that the total distance for this decision would be 10 + 30 =
40. If he selects state 9, the total distance is 40 + 40 = 80. Comparing the two cases, it is better if he would
choose 8, x2* = 8, since it gives the minimum total distance, f2*(5) = 40. In the same way for s = 6 and s = 7, we
get the following results for the two stage problem shown below.
x2
f2(s, x2)  =
Cs x2
+
f1*(x2)
f2*(s) x2*
s
8
9
5
10 + 30 = 40
40 + 40 = 80
40
8
6
60 + 30 = 90
30 + 40 = 70
70
9
7
30 + 30 = 60
30 + 40 = 70
60
8
The solution for the three-stage problem is obtained in a similar fashion. In the three-stage problem, we
have f3(s, x3) = Csx3 + f2*(x3). To illustrate, if the salesman is in state 2, and selects to go to state 5 next, the
minimum total distance, f3(2, 5) would be the cost of the first stage C25 = 70 plus the minimum distance from
state 5 onward f2*(5) = 4 so that f3*(2, 5) = 70 + 40 = 110, similarly f5*(2, 6) = 40 + 70 = 110 and f3*(2, 7) = 60
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+ 60 = 120, so that the minimum total distance from state 2 onward is f3*(2) = 110 and the immediate
destination should be x3* = 5 or 6. The results are tabulated below.
x3
f3(s, x3)  =
Csx3  +  f2*(x3)
f3*(s)  x3*
s
5
6
7
2
70 + 40 = 110  40 + 70 = 110  60 + 60 = 120
110  5 or 6
3
30 + 40 = 70  20 + 70 = 90  40 + 60 = 120
70  5
4
40 + 40 = 80  10 + 70 = 80  50 + 60 = 110
80  5 or 6
Continuing in this way, we move to the four-stage problem. The optimum distance travelled given the
immediate destination, is again the sum of the distances of the first stage plus the minimum distance thereafter.
The results are tabulated in table.
x4
f4(s, x4)
=
C sx4 + f3*(x4)
s
2
3
4
f4* (x) x4*
1
20 + 110 = 130
40 + 70 = 110  30 + 80 = 110
110  3 or 4
We can summarize the optimal solution. From the four stage problem, we infer that the salesman
should go initially to either state 3 or state 4. If he chooses x4* = 3, the three-stage problem result for s = 3 is x4*
= 5. From this we go to the two stage problem which gives x*2 = 8 for s = 5 and the single stage problem yields
x1* = 10 for s = 8. Hence the optimal route is 135810. If the salesman selects x4*=4, this leads to the
other two optimal routes, 145810 and 146910. They all yield a total of f4*(1) = 110.
FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
The physical interpretation to the abstract structure of dynamic programming problems can be provided
by the example of stagecoach problem discussed in the previous section. Any problem in dynamic programming
can be formulated with its basic structure similar to that of the stagecoach problem. The basic features, which
characterize dynamic programming problems, are given in the following.
1.
The problem can be divided up into stages, with a policy decision required at each stage.
2.
Each stage has a number of states associated with it.
3.
The effect of the policy decision at each stage is to change the current state into a state associated with
the next stage.
4.
Given the current state, an optimal policy for the remaining stage, is independent of the policy adopted
in previous stages.
5.
The procedure of solving the problem begins by finding the optimal policy for each state of the last
stage.
6.
A recursive formula can be framed to identify the optimal policy for each state with (n - 1) stages
remaining.
7.
Using the recursive relationship the procedure is to move backward stage by stage, until it finds the
optimal policy when starting at the initial stage.
Six units of capital is available to invest in four business ventures. The returns from each unit
Example
of investment in all the four ventures are given in the table below. Find how should the capital be allocated to
business proposals in order to maximize profit.
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Expected returns from
Business proposals
Unit
A
B
C
D
0
0
0
0
0
1
5
2
6
2
2
6
4
7
3
3
7
6
8
4
4
8
8
8
5
5
8
9
8
6
6
8
10
8
6
The problem has 4 stages, each business proposal representing a stage and we have six states
Solution:
with each stage.
One stage problem
Here we consider only one business proposal namely D (working backward) we can allot any unit of
capital and the expected returns are as shown in table. For business D alone,
S
f1*(s)
x1*
0
0
0
1
2
1
2
3
2
3
4
3
4
5
4
5
6
5
6
6
5,6
Two-stage problem:
Here we consider two business proposals D and C. The capital is to be allocated to D and C in many
possible ways. If a certain unit of capital is allotted to C (second new business) then the remaining capital only
can be allocated to D and we have several combinations. The convenient way to understand about the returns is
to analyse all possible combinations. Let x1, be the amount allotted to D and x2 to C.
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Then profit function;
f2 (s, x2) = p(x2) + f1*(s, x2).
Capital, s,
Allotment
Allotment
Net Return from
x2*
for C
for D
two ventures
s = x2 + x1
x2
x1
f2(s)
s=0
0
0
0
s=1
0
1
0+2=2
1
0
6 + 0 = 6*
1
s=2
0
2
0+3=3
1
1
6 + 2= 8*
1
2
0
7+0=7
s=3
0
3
0+4=7
1
2
6 + 3 = 9*
1
2
1
7 + 2 = 9*
1, 2
3
0
8+0=8
s=4
0
4
0+5=5
1
3
6 + 4 = 10 *
1
2
2
7 + 3 = 10 *
1, 2
3
1
8 + 2 = 10 *
1, 2, 3
4
0
8+0=8
s=5
0
5
0+6=6
1
4
6 + 5 = 11*
1
2
3
7 + 4 = 11*
1,2
3
2
8 + 3 = 11*
1, 2,3
4
1
8 + 2 = 10
5
0
8+0=8
s=6
0
6
0+6=6
1
5
6 + 6 = 12*
1
2
4
7 + 5 = 12*
1, 2
3
3
8 + 4 = 12*
1, 2, 3
4
2
8 + 3 = 11
5
1
8 + 2 = 10
6
0
8+2=8
Now we turn to a three-stage problem. Here we consider three business ventures. The capital can be
allotted to business venture B (= x3) and the remaining amount to C and D combined.
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The profit function is given by
f3(s, x3) = p(x3) + f2*(s-x3)
Capital, s,
Allotment
Allotment
Net Return from
x3*
for B
for C and
three ventures
D
s = x3 + (s-x1)
(x3)
(s-x3)
f3(s)
s=0
0
0
0
s=1
0
1
0 + 6 = 6*
0
1
0
2+0=2
s=2
0
2
2 + 6 = 8*
0
1
1
4+0=4
0, 1
s=3
0
3
0+9=9
1
2
2 + 8 = 10*
1
2
1
4 + 6 = 10*
1, 2
3
0
6+0=6
s=4
0
4
0 + 10 = 10
1
3
2 + 9 = 11
2
2
4 + 8 = 12 *
2
3
1
6 + 6 = 12 *
2, 3
4
0
8+0=8
s=5
0
5
0 + 11 = 11
1
4
2 + 10 = 12
2
3
4 + 9 = 13
3
2
6 + 8 = 14*
3
4
1
8 + 6 = 14*
3,4
5
0
9+0=9
s=6
0
6
0 + 12 = 12
1
5
2 + 11 = 13
2
4
4 + 10 = 12*
3
3
6 + 9 = 15
4
2
8 + 8 = 16*
4
5
1
9 + 6 = 15
6
0
10 + 0 = 10
Now finally, we come to the four-stage problem, in which we consider all the business proposals A, B, C and D.
A certain capital is allotted to A and the remaining for B, C and D together for which the optimum result can be
taken from the three-stage problem.
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Table of Contents:
  1. Introduction:OR APPROACH TO PROBLEM SOLVING, Observation
  2. Introduction:Model Solution, Implementation of Results
  3. Introduction:USES OF OPERATIONS RESEARCH, Marketing, Personnel
  4. PERT / CPM:CONCEPT OF NETWORK, RULES FOR CONSTRUCTION OF NETWORK
  5. PERT / CPM:DUMMY ACTIVITIES, TO FIND THE CRITICAL PATH
  6. PERT / CPM:ALGORITHM FOR CRITICAL PATH, Free Slack
  7. PERT / CPM:Expected length of a critical path, Expected time and Critical path
  8. PERT / CPM:Expected time and Critical path
  9. PERT / CPM:RESOURCE SCHEDULING IN NETWORK
  10. PERT / CPM:Exercises
  11. Inventory Control:INVENTORY COSTS, INVENTORY MODELS (E.O.Q. MODELS)
  12. Inventory Control:Purchasing model with shortages
  13. Inventory Control:Manufacturing model with no shortages
  14. Inventory Control:Manufacturing model with shortages
  15. Inventory Control:ORDER QUANTITY WITH PRICE-BREAK
  16. Inventory Control:SOME DEFINITIONS, Computation of Safety Stock
  17. Linear Programming:Formulation of the Linear Programming Problem
  18. Linear Programming:Formulation of the Linear Programming Problem, Decision Variables
  19. Linear Programming:Model Constraints, Ingredients Mixing
  20. Linear Programming:VITAMIN CONTRIBUTION, Decision Variables
  21. Linear Programming:LINEAR PROGRAMMING PROBLEM
  22. Linear Programming:LIMITATIONS OF LINEAR PROGRAMMING
  23. Linear Programming:SOLUTION TO LINEAR PROGRAMMING PROBLEMS
  24. Linear Programming:SIMPLEX METHOD, Simplex Procedure
  25. Linear Programming:PRESENTATION IN TABULAR FORM - (SIMPLEX TABLE)
  26. Linear Programming:ARTIFICIAL VARIABLE TECHNIQUE
  27. Linear Programming:The Two Phase Method, First Iteration
  28. Linear Programming:VARIANTS OF THE SIMPLEX METHOD
  29. Linear Programming:Tie for the Leaving Basic Variable (Degeneracy)
  30. Linear Programming:Multiple or Alternative optimal Solutions
  31. Transportation Problems:TRANSPORTATION MODEL, Distribution centers
  32. Transportation Problems:FINDING AN INITIAL BASIC FEASIBLE SOLUTION
  33. Transportation Problems:MOVING TOWARDS OPTIMALITY
  34. Transportation Problems:DEGENERACY, Destination
  35. Transportation Problems:REVIEW QUESTIONS
  36. Assignment Problems:MATHEMATICAL FORMULATION OF THE PROBLEM
  37. Assignment Problems:SOLUTION OF AN ASSIGNMENT PROBLEM
  38. Queuing Theory:DEFINITION OF TERMS IN QUEUEING MODEL
  39. Queuing Theory:SINGLE-CHANNEL INFINITE-POPULATION MODEL
  40. Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
  41. Replacement Models:ITEMS DETERIORATING WITH TIME VALUE OF MONEY
  42. Dynamic Programming:FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
  43. Dynamic Programming:Analysis of the Result, One Stage Problem
  44. Miscellaneous:SEQUENCING, PROCESSING n JOBS THROUGH TWO MACHINES
  45. Miscellaneous:METHODS OF INTEGER PROGRAMMING SOLUTION