Assignment Type : Optional

Topic : DYNAMIC PROGRAMMING

1) A freight company with one ship wishes to increase its fleet to five after four years. At

the beginning of each year a decision is made whether to buy any ship. These ships

arrive immediately and are in service for that year. The earning in each year for

different fleet size is shown in the data below (all in some monetary units):

Year Fleet Size

1 2 3 4 5

1 20 35 40 40 40

2 25 45 60 60 60

3 20 35 50 60 60

4 25 35 50 40 40

The capital cost of each ship is 35 monetary units and no more than two ships can be

purchased in any year. The cash balance at the end of each year is given by:

Cash carried forward + Earnings – Capital cost of ships

If this is surplus, 10% is added to its value to represent other investment earnings during

the same year. If a deficit, 20% is charged against (in the same year) . The company

starts with no cash or deficit. The objective is to have maximum cash balance at the end

of the fourth year. Use Dynamic Programming to suggest the optimal buying policy for

the purchase of ships. Please state explicitly, stages, states, decision variables and the

recursive relationship.

3) The sentry company manufactures a home fire protection device, which consists of

three major electronic components. The reliability of this home fire protection device can

be improved by installing several parallel units of one or more of the three major

components. The following table summarizes the probability that the respective

components will function properly if they consist of one, two, or three parallel units.

Number of parallel units COMPONENTS

Probability of functioning properly

1 2 3

1 0.80 0.80 0.85

2 0.85 0.90 0.90

3 0.90 0.95 0.95

The probability that the fire protection device will function properly is the product of the

probabilities associated with the individual components functioning properly. If the total

number of components that can be put in the system is restricted to five, use Dynamic

programming to determine how many parallel units should be installed for each of the

three components in order to maximize the probability that the fire protection device will

function properly.

4) The altitude of an airplane flying between two cities A & G, separated by a distance

of 1500 Km, can be changed at points B, C, D, E & F (see fig.). The fuel cost involved in

changing from one altitude to another between any two consecutive points is given in the

Table below. (The table also gives the cost of staying at the same altitude level between

two consecutive points). Determine the ideal altitude of the airplane at the intermediate

points for minimum total fuel cost using Dynamic programming. Define your states,

stages, and the recursive relationship.

Table: Fuel cost between any two consecutive points

From altitude (Meters) To altitude (Meters)

0 1000 2000 3000 4000 5000

0 --- 4000 4800 5520 6160 6720

1000 800 1600 2680 4000 4720 6080

2000 320 480 800 2240 3120 4640

3000 0 160 320 560 1600 3040

4000 0 0 80 240 480 1600

5000 0 0 0 0 160 240

Figure: Altitude of Airplane

AL

TIT

UD

E I

N M

ET

ER

S

5000 * * * * *

4000 * * * * *

3000 * * * * *

2000 * * * * *

1000 * * * * *

* * * * * *

A B C D E F G

(A – Starting point; G – Stopping point; B, C, D, E, & F- Intermediary points)

Distance between any two consecutive points is 250 Km. (For e.g. distance between A &

B is 250 Km, between B&C is 250 Km, and so on).

5) (Stage coach problem). A mythical salesman had to travel from East to West of the

United states some 100 years ago. Although his starting point and destinations were

fixed, he had considerable choice as to which territories (or cities) to travel through on

route. The possible routes are shown in the figure where each city is represented by a

number block. Thus four ‘stages’ were required to travel from his embarkation in city 1

to his destination city 10. The cost of traveling from one city to the other in any feasible

route is given in the diagram (cost expressed in some monetary units). The salesman’s

decision problem was to choose the best possible route from city 1 to 10 so that the

overall cost of traveling was the minimum.

Figure:

6) A fruit seller must allocate four boxes of perishable fruit between three stores. His

expected profit from leaving boxes at each store is shown below:

No. of boxes Store 1 Store 2 Store 3

0 0 0 0

1 4 3 6

2 6 5 8

3 7 7 9

4 7 8 9

How should he allocate the boxes among the three stores so as to get the maximum

profit?

7) A firm has eight commercial “spots” on the local TV, all at peak viewing hours. The

estimated increase in profit for using them to advertise their four products is shown below

(in some monetary units). Use Dynamic programming to suggest the optimal number of

exposures of the various products.

Product No. of Exposures

1

2

3

4

5

6

7

8

10

9

2

5

1

102

12

10

5

7

15

13

7

5

3

4

7 1

4

1

1 2 3 4 5 6 7 8

A 1 2 4 7 11 16 17 18

B 2 3 5 8 11 14 16 17

C 1 3 5 9 11 13 16 19

D 3 5 7 11 14 15 16 16

8) An Airline wishes to operate five planes over four routes. Average cash revenue per

year, comprising total receipt for passengers and freight, less variables running cost per

plane are given below (all fig. in some monetary units).

Route Number of planes operating

1 2 3 4 5

A 10 17 24 29 32

B 12 20 26 32 34

C 8 14 19 23 22

D 14 22 30 36 38

Route D being a status route (or socially justifiable route similar to the North- Eastern

area of India for Indian Airlines), the management would want to have at least one plane

operating in that route. List all possible optimal allocations of five planes over the four

routes.

9.) A government space project is conducting research on a certain engineering problem

that must be solved before man can fly safely to Mars. Three research teams are currently

trying three different approaches for solving this problem. The estimate has been made

that, under present circumstances, the probability that the respective teams- call them 1,

2, and 3 will not succeed is 0.40, 0.60, and 0.80 respectively.

Thus, the current probability that all three teams will fail is (0.40) (0.60) (0.80) = 0.192.

Since the objective is to minimize this probability, the decision has been made to assign

two more top scientists among the three teams to lower it as much as possible.

The table below gives the estimated probability that the respective teams will fall

when 0, 1, or 2 additional scientists are added to that team. The problem is to determine

how to allocate the two additional scientists to minimize the probability that all the three

reams will fail.

Table: Data on the Government Space Project Problem

Probability of Failure

No. of new Scientists Team

1 2 3

0 0.40 0.60 0.80

1 0.20 0.40 0.50

2 0.15 0.20 0.30

10) There is a log of 20’ length having a large end diameter of 32” and a small end

diameter of 22”. For simplicity assume that taper is linear, i.e.; there is uniform reduction

in the diameter of the log from one end to the other; in this example, for every 2’ length

of the log there is a difference of 1” in diameter. The value of small logs cut from this log

depends upon its diameter. In particular, from plywood and saw milling industry’s point

of view it depends upon the small end diameter. Assume that the log is to be cut into

small logs of 8’, 6’, 4’, and 2’ lengths and its values are as follows:

Small end Diameter

Length 21” – 24”

(Rs.)

25” – 28”

(Rs.)

29” – 32”

(Rs.)

2’ 150 170 200

4’ 400 450 520

6’ 650 720 800

8’ 950 1070 1200

Observe that there are several ways in which we can cut this 20’ long log into small logs.

However, each such cutting strategy generates a value for the entire log. (It also restricts

the possible uses, once a cut is made). The objective is to devise a cutting strategy which

maximizes this value. Use Dynamic programming to decide the optimal cutting strategy.

11) A steel Rolling Mill has two stands S1 & S2. Red hot billets of steel of square cross

section of 3” x 3” enter the first of these stands. Maintaining the square cross section, it is

then rolled progressively smaller by the two stands till the required thickness is achieved.

Each of these stands can reduce the thickness by a maximum of ¼” square in steps of

(1/16”, 2/16”, 3/16”, and 4/16”). The speed required by these stands, in feet per minute, to reduce various thicknesses are

as follows:

Reduction in square cross-section (inches)

Stand 1/16 2/16 3/16 4/16

S1 12 10 8 6

S2 11 9 7 5

Suppose the required output thickness is a billet of 2 ¾” x 2 ¾” (i.e. 11/4” x 11/4”), what is the fastest aped possible and what should be the speed setting for the stands?

Dia 22” Dia 32”

12) A steel rolling Mill has 10 stands. A red hot billet of steel of square cross section of

3’’ x 3” enters the first of these stands. Maintaining the square cross-section it is then

rolled progressively smaller by other stands, till the required thickness is achieved. The

ultimate required thickness might be anywhere between ¼” to 1 ½” in steps of ¼”. Each

‘stand’ can reduce the thickness by 1/16” to ½” in steps of 1/16”. However, the bigger the

reduction in thickness, the slower the stands will rotate because of the torque stresses.

The Engineers of the fir, can supply sets of tables for each stand which would give the

fastest speed of output of the material given the input and output thickness. There is no

stock piling possible in between any two stands. Find the fastest speed at which the whole

mill can operate for each required thickness and show a way to determine the “settings”

of the stands. Find the fastest speed at which the whole mill can operate for each required

thickness and show a way to determine the “settings” of the stands.