DAM Core HO Graphica1 2014 With Answers

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GRAPHICAL SOLUTION-A graphical solution procedure is one method of solving two-variable LPP (Linear Programming

Problem). The primary purpose of this method is in helping to provide an understanding of1. What is involved in solving an LPP?

2. What information is available in the solution?

The step-by-step procedure is explained with the help of following example:

Ex: Two types of Television sets are produced with a profit of 6 units from eachtelevision of Type I and 4 units from each television of Type II. In addition, 2 and 3

units of raw materials are needed to produce one television of Type I and Type II,

respectively & 4 and 2 units of time are required to produce one television of Type Iand Type II, respectively. If 100 units of raw materials and 120 units of time are

available, how many units of each type of television should be produced to maximize

profit and still meet all constraints of the problem?

Step I: Write down the given information in a table format.

Given:

Type I Type II Available

Raw Material 2 3 100

Time 4 2 120

Profit/Unit 6 4

Here, the profit is to be maximized within the limited resources. This is the

maximization problem.

Step II: Initially the problem should be formulized i.e. objective function & constraintsshould be written.

In this example, the main objective is to produce more number of units of

Type I & II within limited resources.

Let X1, X2 be the number of units of Type I and Type II respectively.The objective function (the function to be optimized/maximized) is

Z = 6X1 + 4X2

Since each unit of Type I and Type II yields a profit of 6 and 4 units, this implies

thatMax Z = 6X1 + 4X2

Subject to constraints

2X1 + 3X2 ≤ 100 ………… Raw material 4X1+ 2X2 ≤ 120 …………. Time

X1, X2 ≥ 0 ...………. Non-negativity

Since only two variables are involved in this example, the problem can be solvedgraphically.

Step III: To solve it graphically i.e. to draw a graph, for all constraints draw the straight

line for where the constraints are satisfied exactly. Graph the constraints as if it

were equality.Therefore, the 1st constraint i.e. 2X1 + 3X2 ≤ 100 becomes

2X1 + 3X2 = 100

To graph the 1st constraint, determine the set of points that satisfy this constraint.

Check whether the origin (0,0) satisfies the constraint.

In this case, 2*0 + 3*0 = 0 ≤ 100

This implies that all points below the line 2X1 + 3X2 = 100 satisfy this

constraint. Now, to determine (X1, X2) co-ordinates, set X1=0 and find out X2 and vice-

versa.



Therefore, putting X1 = 0 in 2 X1 + 3X2= 100We get, 3X2 = 100

X2 = 33.33

And putting X2 = 0 in 2X1 + 3X2 = 100

We get, 2X1 = 100X1 = 50

Therefore, set of points is (50, 33.33).

Draw a line with (X1, X2) co-ordinates as (50, 33.33).Step IV: Now in the similar way determine the set of points by substituting X1 = 0 and

X2 = 0 simultaneously in the second constraint.

The 2nd

constraint is 4X1 + 2X2 = 120Putting X1 =0 in this constraint, we get

2X2 =120

X2=60

Putting X2 = 0, we get4X1 = 120

X1 = 30

Therefore, set of points is (30, 60).

Draw a straight line with (X1, X2) co-ordinates as (30, 60).Step V: Determine the region where X1≥0 & X2≥0.

X2

(0,60)

4X1 + 2X2=120

D (0,33,33) Region of “Feasible Solutions”

C (20,20)

2X1 + 3X2= 100

B (30,0) (30,0)

A (0,0) X1

Step VI: Take the region which is common to all the lines i.e. 2X1 + 3X2 ≤ 100;

4X1 + 2X2 ≤ 120 and X1, X2 ≥ 0. That region is called as “Feasible Region”.

Feasible region is a set of points that make all linear inequalities in the system truesimultaneously. That is, each point in this region satisfies all of the constraints and is a candidate for

providing the maximum profit. But the region contains infinity of feasible solutions. Therefore, there

is a need to find out the optimal solution. Find out the intersection points.

Step VII: Therefore, to find out the optimal solution, substitute the set of points of



intersection in the objective function i.e. Z = 6X1 + 4X2 The points of intersection are:

A (0, 0)

B (30, 0)

C (20, 20)D (0, 33.33)

Substitute these values in Z = 6X1 + 4X2

A (0, 0) Z = 6*0 + 4*0 = 0Z = 0

B (30, 0) Z = 6*30 + 4*0 = 180

Z = 180

C (20, 20) Z = 6* 20 + 4*20 = 200

Z = 200 D (0, 33.33) Z = 6*0 + 4*33.33 = 133.32

Z = 133.32

Z = 200 is the maximum value & is the optimal solution.Therefore, 20 units of Type I and 20 units of Type II should be

produced to yield a maximum profit of 200 units.

Minimization case Illustration:

Minimize Z= 10X1+4.5X2

Subject to constraints2X1+3X2≥3500

6X1+2X2≥7000

X1≥0 X2≥0 We first find the points to determine the line For constraint 1 2X1+3X2≥3500

When X1=0, X2=3500/3. The coordinates of the point are (0,3500/3)

When X2=0, X1=1750. The coordinates of the point are (1750,0)

We plot these points on the graph paper to show this constraint. Similarly for the second constraint

When X1=0, X2=7000/2. The coordinates of the point are (0,3500)When X2=0, X1=7000/6. The coordinates of the point are (7000/6,0)

We plot these points on the graph paper lines to show this constraint.

3500 A

3000

25006X1+2X2≥7000

2000

2X1+3X2≥3500

1500 feasible region



1000

500 B

C

500 1000 1500 2000

The shaded region is the feasible region. Infinitely many points satisfy the constraints. The objective is to

minimize the value of the objective function. The feasible region is bounded below. The points A, B and Care the points where the minimum value of the objective function may lie.

The feasible points are

Point Coordinates

A (0,3500)B (1000,500)

C (0,3500/3)

The value of the objective function at these points isPoint Coordinates Value of Z

A (0,3500) 15750B (1000,500) 12250* Minimum

C (1750,0) 17500The solution lies at point B. X1=1000 and X2=500 value of Z=12250

Graphical Method Problems

1.

Use the graphical method to solve the following LP problem.Maximize Z = 15x1 + 10x2

subject to the constraints

4x1 + 6x2 ≤ 360

3x1 + 0x2 ≤ 180

0x1 + 5x2 ≤ 200

and x1, x2 0.



2.

Use the graphical method to solve the following LP problem.Maximize Z = 2x1 + x2


x1 + 2x2 ≤ 10

x1 + x2 ≤ 6

x1 - x2 ≤ 2

x1 - 2x2 ≤ 1

and x1, x2 0.

3. Use the graphical method to solve the following LP problem.



Maximize Z = 10x1 + 15x2


2x1 + x2 ≤ 26

2x1 + 4x2 ≤ 56

- x1 + x2 ≤ 5

and x1, x2 0.

4. The ABC Company has been a producer of picture tubes for television sets and certain printed

circuits for radios. The company has just expanded into full scale production and marketing of AM

and AM-FM radios. It has built a new plant that can operate 48 hours per week. Production of an

AM radio in the new plant will require 2 hours and production of an AM-FM radio will require 3hours. Each AM radio will contribute Rs 40 to profits while an AM-FM radio will contribute Rs 80

to profits. The marketing department, after extensive research has determined that a maximum of 15AM radios and 10 AM-FM radios can be sold each week.

(a) Formulate a linear programming model to determine the optimum production mix of AM

and FM radios that will maximize profits.

(b) Solve the above problem using the graphic method.



5. Anita Electric Company produces two products P1 and P2. Produced and sold on a weekly basis. The

weekly production cannot exceed 25 for product P1 and 35 for product P2 because of limitedavailable facilities. The company employs total of 60 workers. Product P1 requires 2 man-weeks of

labour, while P2 requires one man-week of labour. Profit margin on P1 is Rs 60 and on P2 is Rs 40.

Formulate this problem as an LP problem and solve for maximum profit.

6. A local travel agent is planning a charter trip to a major sea resort. The eight day/seven-night

package includes the fare for round-trip travel, surface transportation, boarding and loading and



selected tour options. The charter trip is restricted to 200 persons and past experience indicates thatthere will not be any problem for getting 200 persons. The problem for the travel agent is to

determine the number of Deluxe, Standard, and Economy tour packages to offer for this charter.

These three plans differ according to seating and service for the fight, quality of accommodations,

meal plans and tour options. The following table summarizes the estimated prices for the three packages and the corresponding expenses for the travel agent. The travel agent has hired an aircraft

for the flat fee of Rs 2,00,000 for the entire trip.

Prices and Costs for Tour Packages per Person

Tour Plan Price(Rs)

Hotel Costs(Rs)

Meals & Other Expenses(Rs)

Delux 10,000 3,000 4,750Standard 7,000 2,200 2,500

Economy 6,500 1,900 2,200

In planning the trip, the following considerations must be taken into account:

(i) At least 10 per cent of the packages must be of the delux type.(ii) At least 35 per cent but no more than 70 per cent must be of the standard type.

(iii)

At least 30 per cent must be of the economy type.(iv) The maximum number of delux packages available in any aircraft is restricted to 60.

(v) The hotel desires that at least 120 of the tourists should be on the deluxe and standard packages together.

The travel agent wishes to determine the number of packages to offer in each type so as to maximize

the total profit.(a) Formulate this problem as a linear programming problem.

(b) Restate the above linear programming problem in terms of two decision variables, taking

advantage of the fact that 200 packages will be sold.

(c) Find the optimum solution using graphical method for the restated linear programming problem and interpret your results.



Ans : 20 – D , 100 – S, 80 – E

Profit = 280000 – 13000 = 267000.

7. Use the graphical method to solve the following LP Problem.Minimize Z = 3x1 + 2x2

Subject to the constraints

5x1 + x2 10

x1 + x2 6

x1 + 4x2 12

and x1, x2 0.



8. Use the graphical method to solve the following LP problem.Maximize Z = x1 + 2x2


- x1 + 3x2 ≤ 10

x1 + x2 ≤ 6

x1 - x2 ≤ 2

and x1, x2 0.



9. Cashewco has two grades of cashew nuts: Grade I – 750 kg and Grade II – 1,200 kg. These are to

be mixed in two types of packages of one kilogram each- economy and special. The economy packconsists of grade I and grade II cashews in the proportion of 1:3, while the special pack combines the

two in equal proportion. The profit margin on the economy and special packs is, respectively, Rs. 5

and Rs. 8 a pack.

(a) Formulate this as a linear programming problem.(b) Ascertain graphically the number of packages of economy and special types to be made that will

maximize the profits.



Would your answer be different if the profit margin on a special pack be Rs. 10?

Ans : NO Only total profit will change All other answers same.

10. Alpha Radio Manufacturing company must determine production quantities for this month for two

different models, A and B. Data per unit are given in the following table:

Model Revenue

(Rs)

Sub-assembly

time (hr)

Final assembly

time (hr)

Quality

inspection (hr)

A 250 1.0 0.8 0.5

B 300 1.2 2.0 0The maximum time available for these products is 1,200 hours for sub-assembly, 1,600 hours for final

assembly, and 500 hours for quality inspection. Orders outstanding require that at least 200 units of A and

100 units of B be produced. Determine the quantities of A and B that maximize the total revenue.



11. Attempt graphically the following problem:

Maximise 3x1 + 2x2

Subject to 2x1 + x2 12; x1 + x2 10; - x1 + 3x2 6; and x1, x2 0

12. Obtain graphically the solution to the following LPP:

Maximise Z = x1 + 3x2

Subject to



x1 + 2x2 9

x1 + 4x2 11

x1 – x2 2

x1, x2 0

13. A firm is engaged in producing two products : P1 and P2. The relevant data are given here:

Per Unit Product P1 Product P2

(i) Selling price Rs 200 Rs 240

(ii) Direct materials Rs. 45 Rs. 50(iii) Direct wages

Deptt A 8 hrs @ Rs. 2 / hr 10 hrs @ Rs 2 /hrDeptt B 10 hrs @ Rs 2.25 / hr 6 hrs @ Rs 2.25 /hr

Deptt C 4 hrs @ Rs 2.5 / hr 12 hrs @ Rs 2.5 / hr

(iv) Variable overheads Rs. 6.50 Rs 11.50Fixed overhead = Rs 2,85,000 per annum

No of employees in the three departments: Deptt A = 20

Deptt B = 15

Deptt C = 18 No. of hours / employee / week = 40 in each department

No. of weeks per annum = 50

(a) Formulate the given problem as a linear programming problem and solve graphically to determine(i) the product mix as will maximize the contribution margin of the firm

(ii) the amount of contribution margin and profit obtainable per year

(a) From the graph, do you observe any constraint that is redundant? Which one, if yes?



Yes the constraint 8X1 + 10 X2 <=40000 is redundant.

Optimal Profit = 437500 – 285000 =152500

14. A local business firm is planning to advertise a special sale on radio and television during a

particular week. A maximum budget of Rs. 16,000 is approved for this purpose. It is found thatradio commercials cost Rs. 800 per 30 second spot with a minimum contract of five spots.

Television commercials, on the other hand, costs Rs. 4,000 per spot. Because of heavy demand,

only four television spots are still available in the week. Also, it is believed that a TV spot is sixtimes as effective as a radio spot in reaching consumers. How should the firm allocate itsadvertising budget to attract the largest number of consumers? How will the optimal solution be

affected if the availability of TV spot is not constrained?



Extra Questions

1. M/s. P.M.S. Industries makes two kinds of leather purses for ladies. Purse type „A‟

is of high quality and purse type „B‟ is of lower quality. Contributions per unit were

Rs. 4 and Rs. 3 for purse type „A‟ and type „B‟ respectively. Each purse of type „A‟

requires two hours of machine time per unit & that of type „B‟ requires one hour perunit. The company has 1000 hours per week of maximum available machine time.

Supply of leather is sufficient for 800 purses of both types combined per week.Each type requires the same amount of leather. Purse type „A‟ requires a fancy zipand 400 such zips are available per week, there are 700 zips for purse type „B‟

available per week. Assuming no market or finance constraints recommend an

optimum product mix.

Solution:

Decision vari ables:

Let X1 be the number of units of purse type A

Let X2 be the number of units of purse type B

Th e L P model

Maximise Z = 4 x1 + 3 x2


2X1 + X2 ≤ 1000 (Machine time)

X1 + X2 ≤ 800 (Leather)

X1 ≤ 400 (Special zip for Purse A)



X2 ≤ 700 (Ordinary zip for Purse B)

Xi ≥ 0 (Non negativity restriction)

Scale

X-Axis 1 cm = 100 units Y-Axis 1 cm = 100 units

Draw the graph as indicated in Fig 4.1

The optimum solution will lie on one of the corner points of the shaded area OABCDE

Calculation of Z (Z = 4X1 + 3X2) at corner points

O (0,0); ZO = 4 x 0 + 3 x 0 = 0

A (400, 0); ZA = 4 x 400 = Rs. 1600

B (400, 200); ZB = 4 x 400 + 3 x 200 = Rs. 2200

C (200, 600); ZC = 4 x 200 + 3 x 600 = Rs. 2600

D (100, 700); ZD = 4 x 100 + 3 x 700 = Rs. 2500

0.0 100 200 300 400 500 600 700 800 900 1000

X1 Units

X2 Units

1000

800

600

400

200

O

X1 400

X2 700

Feasible Solution Area OABCDE

X1 + X2 800

2X1 + X2 1000

A

C

D

E



E (0, 700); ZE = 3 x 700 = Rs. 2100

Corner point C signifies the optimum solution because the corresponding value of Z ismaximum.

The optimum solution is 200 units of Purse Type „A‟ & 600 units of Purse Type B

The corresponding Total Contribution = Rs. 2600

2.

A catering manager is in the process of replacing the furniture in a canteen. Hewishes to determine how many tables of type „S‟ (seating 6) and how many o f type

„T‟ (seating 10) to buy.

He has to work under the following constraints:

(1) The canteen must be able to accommodate at least 60000 people.

(2) The available floor space of the canteen is at most 63000 sq. meters.

He estimates that each type „S‟ table needs 7 meters sq. of floor space while each type „T‟needs 9.

Advise the manager on how many tables of each type to buy if each type „S‟ costs Rs. 100

and each type „T‟ costs Rs. 190.

Solution:

(0, 0) 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

X1 Units

X2 Units

10000

8000

6000

4000

2000

O

Feasible Solution Area ABC

7X1 + 9X2 63000

6X1 + 10X2 60000

A

C



Decision vari ables:

Let X1 be the number of units of table type S

Let X2 be the number of units of table type T

Th e L P model

Minimise Z = 100 x1 + 190 x2


6X1 + 10X2 ≥60000 (People)

7X1 + 9X2 ≤ 63000 (Floor space)

Xi ≥ 0

Draw the Graph as indicated in Figure.

The optimum solution will lie on one of the corner points of the shaded area ABC

Calculation of Z (Z = 100X1 + 190X2) at corner points

A (5625, 2625); ZA = 100 x 5625 + 190 x 2625 = Rs. 1061,250

B (0, 7000); ZB = 190 x 7000 = Rs. 1330,000

C (0, 6000); ZC = 190 x 6000 = Rs. 1140,000

Corner point A signifies the optimum solution because the corresponding value of Z isminimum.

The optimum solution is 5625 units of Table type S & 2625 units of Table type T

The corresponding Total Cost = Rs. 1 061250

SOME SPECIAL CASES

Unbounded Solutions

When the value of a decision variable is permitted to increase infinitely without violatingany of the feasibility conditions, then the solution is said to be unbounded. Thus an

unbounded LPP occurs if it is possible to find arbitrarily large values of Z within the

feasible region. It is not possible to find a single optimal solution to an unbounded problemthough there are infinite feasible solutions for the same.



3. Use graphical method to solve the following LP problem

Maximise Z = 3x1 + 3x2


x1 - x2 ≤ 1

x1 + x2 ≥ 3

and x1, x2 ≥ 0

Solution:

The problem is depicted graphically in Fig. 4.3. The solution space is shaded and is bound by A and B from below.

It is noted here that the shaded convex region (solution space) is unbounded. The twocorners of the region are A = (0, 3) and B = (2, 1). The values of the objective function at

these corners are:

Z(A) = 6 and Z(B) = 8.

But there exist number of points in the shaded region for which the value of the objectivefunction is more than 8. For example the point (10,12) lies in the region and the function

value at this point is 70 which is more than 8. Thus both the variables x 1 and x2 can be

made arbitrarily large and the value of Z also increased. Hence, the problem has anunbounded solution.

Remark: An unbounded solution does not mean that there is no solution to the given LP problem, but impl ie s tha t there ex is t an in fi ni te number of so lu ti ons.

A (0,3)

1 2 3 4 X1

X2

3

2

1

0

X1 - X2 = 1

B (2,1)

X1 + X2 = 3

Unboun

ded



Infeasible Solution

Sometimes it is not possible to find a single solution that satisfies all the feasibilityconstraints. Such a problem does not have a feasible solution.

Please note that in the case of infeasiblity it is not possible to have a single feasiblesolution whereas in the case of unboundedness, it is possible to have infinite feasiblesolutions, but not a single optimal solution.

4. (Problem with inconsistence system of constraints) Use graphical method to solve

the following LP problem:

Maximise Z = 6x1 - 4x2


2x1 + 4x 2 ≤ 4

4x1 + 8x 2 ≥ 16

and x1, x2 ≥ 0

Solution:

The problem is shown graphically in Fig. 4.4. The two inequalities that form the constraint

set are inconsistent. Thus, there is no set of points that satisfies all the constraints. Hence,

there is no feasible solution to this problem.

1 2 3 4 X1

X2

3

2

1

0

4X1 + 8X2 = 16

2X1 + 4X2 = 4



Multiple Optimal Solutions

Can exist for a linear programming problem if

i. The given objective function is parallel to a constraint that forms the boundary of thefeasible solutions region

ii. The constraint should form a boundary on the feasible region in the direction of theoptimal movement of the objective function. In other words, the constraint must be a

binding co ns train t.

In case the slope of the objective function (represented by the iso-profit lines) is the sameas that of a constraint, then multiple optimal solutions might exist. For multiple optimal

solutions to exist, the following two conditions need to be satisfied:

(a) The objective function should be parallel to a constraint that forms an edge or boundar y on th e feas ible region; and

(b) The constraint should form a boundary on the feasible region in the direction of

optimal movement of the objective function. In other words, the constraint must be a binding co ns train t.

5. Solve graphically the following LPP:

Maximise Z = 8x1 + 16x2

Subject to

x1 + x2 ≤ 200

x2 ≤ 125

3x1 + 6x 2 ≤ 900

x1, x2 ≥ 0

Solution:

The constraints are shown plotted on the graph in Figure 4.5. Also, iso-profit lines have been graphed. We observe that iso -profi t lines are paral lel to the equat ion fo r th ird

constraint 3x1 + 6x 2 = 900. As we move the iso-profit line farther from the origin, it

coincides with the portion BC of the constraint line that forms the boundary of the feasible

region. It implies that there are an infinite number of optimal solutions represented by all

point s lying on the line segment BC, in cl ud ing the ex treme poin ts represented by B (50,

125) and C (100, 100). Since the extreme points are also included in the solutions, we may

disregard all other solutions and consider only these ones to establish that the solution to alinear programming problem shall always lie at an extreme point of the feasible region.

The extreme points of the feasible region are given and evaluated here.

Point x1 x2 Z = 8x1 + 16x2

0 0 0 0

A 0 125 2000

B 50 125 2400

C 100 100 2400 }



D 200 0 1600

The point B and C clearly represent the optima.

In this example, the constraint to which the objective function was parallel, was the one

which formed a side of the boundary of the space of the feasible region. As mentioned incondition (a), if such a constraint (to which the objective function is parallel) does not form

an edge or boundary of the feasible region, then multiple solutions would not exist.

6. Solve graphically:

Minimise Z = 6x1 + 14x2

Subject to

5x1 + 4x 2 ≥ 60

3x1 + 7x 2 ≤ 84

x1 + 2x2 ≥ 18

x1, x2 ≥ 0

Solution:

The restrictions in respect of the given problem are depicted graphically in Figure 4.6. Thefeasible area has been shown shaded. It may be observed here that although the iso -cost line

is parallel to the second constraint line represented by 3x 1 + 7x2 = 84, and this constraint

does provide a side of the area of feasible solutions, yet the problem has a unique optimal

solution, given by the point D. Here condition (b) mentioned earlier, is not satisfied. This is

100 200 300 X1

X2

200

150

100

50

0

X2 = 125 A

C

D E

Optimal Solutions

FEASIBLEIso-profit lines



because , being a mi nimisa ti on probl em, the opt imal movement of the ob ject ive functionwould be towards the origin and the constraint forms a boundary on the opposite side. Since

the constraint is not a binding one, the problem does not have multiple optima.

We can show the uniqueness of the solution by evaluating various extreme points as done

here.

Point x1 x2 Z = 6x1 + 14x2

A 8 5 118

B 84/23 240/23 168

C 28 0 148

D 18 0 108

The optimum solution is 108 units for x 1 = 18 and x2 = 0.

6 12 18 24 30 X1

X2

15

12

9

6

3

0

A

B

C D

FEASIBLE

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DAM Core HO Graphica1 2014 With Answers