LINEAR PROGRAMMING MEANING: it is a mathematical technique for finding the best use of an organization’s resources. It is designed to help managers in planning and decision making. It has demonstrated its value in different areas such as production, finance, marketing, research and personnel assignments. .

TERMINOLOGY AND REQUIREMENTS OF LPP OBJECTIVE FUNCTION PRESENCE OF CONSTRAINTS AND RESTRICTIONS ALTERNATIVE COURSE OF ACTION NON NEGATIVITY CONSTRAINTS LINEARITY FINITE NUMBER OF VARIABLES BASIC VARIABLES SLACK VARIABLES SURPLUS VARIABLES ARTIFICIAL VARIABLES OPTIMUM SOLUTION

APPLICATIONS OF LPP PRODUCT MIX DIET PROBLEM THE PORTFOLIO SELECTION PROBLEM MEDIA SELECTION BLENDING PROBLEM TRANSPORTATION PROBLEM TRAVELLING SALESMAN PROBLEM

PRODUCT MIX PROBLEM Q1. A firm is engaged in producing two products P1 and P2. Each unit of product P1requires 2kg of raw material and 4 labor hours for processing, where each unit of product P2 requires 5kg of raw materials and 3 labor hours of the same type. Every week the firm has the availability of 50kg of raw material and 60 labor hours. One unit of product P1sold earn profit rs20 and one unit of product P2 sold gives rs30 as profit.

SOLUTION : MAXIMISE Z = 20x1 + 30x2 SUBJECT TO 2X1 + 5X2 ≤ 50 4x1 + 3x2 ≤ 60 Where x1, x2 ≥ 0

BLENDING PROBLEM Q2. THE MANAGER OF AN OIL REFINERY MUST DECIDE ON THE OPTIMAL MIX OF 2 POSSIBLE BLENDING PROCESS OF WHICH THE INPUTS AND OUTPUTS PER PRODUCTION RUN ARE AS FOLLOWS: PROCESS INPUT CRUDE A CRUDE B OUTPUT GASOLINE A OUTPUT GASOLINE B 1 6 3 9 2 5 The maximum availability crude A & B are 250 units and 200 units. The market requirement shows that at least 150 units of gasoline X and 130 units of gasoline Y must be produce. the profit per production run from process 1&2 are Rs.40 & Rs.50.Formulate the problems for maximize profit.

SOLUTION: Max. Z= 40x1 +50x2 Subject to 6x1+5x2 ≤ 250 3x1+6x2 ≤ 200 6x1+5x2 ≥ 150 9x1+5x2 ≥ 130 Where x1,x2 ≥ 0

DIET PROBLEMS Q3. Vitamins A&B are found in two different food F1 & F2. one unit of food F1contains 2units vitamin A and 5 units vitamin B. One unit of F2 contains 4units of vitamin A and 2units of vitamin B. One unit of F1&F2 cost Rs.10 & Rs.12.5. the minimum daily requirements of vitamin A&B is 40&50units.Find out the optimal minimum of F1& F2 and minimum cost which meets the daily minimum requirements of vitamin A & B.

SOLUTION: FOOD A B COST PER UNIT F1 2 5 10 F2 4 12.50 MIN. REQ. 40 50

Min Z= 10X1+12.5X2 SUBJECT TO 2X1+4X2 ≥ 40 5X1+2X2 ≥ 50 WHERE X1, X2 ≥ 0

bond Type return Maturity time A industrial 12% 15 yrs B 15% 3yrs C Q4. PORTFOLIO INVESTMENT An investment group wants to invest up to Rs.10.00Lacs into various bonds. They are currently considering 4 bonds. bond Type return Maturity time A industrial 12% 15 yrs B 15% 3yrs C Govt. 18% 10yrs D 6% 4yrs The group has decided not to put less than half of its investment in the govt. bonds and that the average of the portfolio should not be more than 6 yrs. The investment should be such which maximizes the return .

MAX Z= .12X1+.15X2+.18X3+.6X4 SUBJECT TO X1+X2+X3+X4 ≤ 1000000 X3+X4 ≥ 500000 9X1-3X2+4X3-2X4 ≤ 0 WHERE X1,X2,X3,X4 ≥ 0

Q5. Media selection The owner of Spartan sports wishes to determine how many adv. to place in the selected 3 monthly magazines A, B and C. His objective is to adv. in such a way that total exposure to principal buyers of expensive sports goods is maximized. The quantum of readers for each magazine is known. The budgeted amount is at cost Rs. 1lakh for the adv. The owner has already decided that magazine A should have no more than 8 adv. And B&C each have at least 20 adv. Formulate L.P. problem. Magazine A B C READERS 100000 60 000 40000 PRINCIPAL BUYERS 12% 10% 7% COST PER ADV. 5000 4500 4250

MAXIMISE Z=12000X1+6000X2+2800X3 SUBJECT TO 5000X1+4500X2+4250X3 ≤ 100000 X1 ≤ 8 X2 ≥ 20 X3 ≥ 20 WHERE X1, X2, x3 ≥ 0

Q7. A dealer wishes to purchase a number of fans and sewing machine Q7. A dealer wishes to purchase a number of fans and sewing machine. He has only Rs.5760/- to invest and has space for almost 20 items. A fan cost him Rs.360/- and a sewing machine Rs.240/- his expectation is that he can sell a fan at profit of Rs.22/- and a sewing machine @Rs.18/-.how should he invest his money in order to maximize his profit.

SOLUTION MAX. Z=22x1+18x2 Subject to 360x1+240x2 ≤ 5760 or 3x1+2x2 ≤ 48 x1 +x2 ≤ 20 Where x1, x2 ≥ 0

X1 16 X2 24 X1 20 X2

COORDINATE VALUES PROFIT Z=22x1+18x2 (0,0) Z= 22x 0 + 18x 0 =0 (16,0) Z=22x 16+ 18x 0 =352 (8,12) Z=22x8+12x12= 392 (0,20) Z=22x0+ 18x20= 360

Q6. Solve graphically Min. Z=6X1+14x2 Subject to 5x1+4x2 ≥ 60 Where x1, x2 ≥ 0

SOLUTION: X1 12 X2 15 X1 28 X2 12 X1 18 X2 9

COORDINATES Z(MIN) Z=6x1+14x2 (8,5) Z=6x8+14x5= 118 (18,0) Z=6x18+14x0= 108 (28,0) Z=6x28+14x0= 168 (3,10) Z=6x3+14x10= 158

Ques. The standard weight of a special purpose brick is 5 kg and it contains two basic ingredients B1 and B2 costs Rs. 8 per kg. Strength considerations dictate that the brick contains not more than 4 kg of B1 minimum of 2 kg of B2. Since the demand for the product is likely to be related to the price of the brick, find out graphically minimum cost of the brick satisfying the above conditions. Solution The mathematical formulation of the given LPP is: Minimize Z = 5x₁ + 8x₂ (Cost function) Subject to x₁ ˂ 4 x₂ ˃ 2 x₁ + x₂ = 5 Where x₁ , x₂ ˃ 0 (Non negativity)

Making the inequalities of the constraints to equalities for the purpose to plot the above values on the graph, we get x₁ = 4 …. (a) x₂ = 2 …. (b) x₁ + x₂ = 5 …. (c) Now in equation (c) , if x₁ = 0 , x₂ = 5 if x₂ = 0 , x₁ = 5

x₂ Q(3,2) x₁ Since one of the constraints is equality constraints i.e. x₁ + x₂=5. There is no solution space rather a solution point which satisfies all the conditions i.e. point Q (3 , 2) . Ans. x₁ = 3 , x₂ = 2 , Z = 31

UNBOUNDED PROBLEM Ques. Solve the LPP by Graphic Method: Maximize Z = 5x₁ + 4x₂ Subject to 2x₁ - 4x₂ ˂ 1 2x₁ + 4x₂ ˃ 3 x₁ , x₂ > 0 Solution Making the inequalities into equalities: 2x₁ - 4x₂ = 1 x₁ = 0 x₂ = - 1/4 x₂ = 0 x₁ = 1/2

2x₁ + 4x₂ = 3 x₁ = 0 x₂ = 3/4 x₂ = 0 x₁ = 3/2

X₂ 2X₁+4X₂=3 2X₁-4X₂=1 X₁ It is not possible to find the farthest corner of the Unbounded Solution Region. The given LPP therefore has unbounded solution.

INFEASIBLE PROBLEM Ques. Solve graphically Find the maximum value of Z = -5x₂ Subject to x₁ + x₂ < 1 -0.5x₁ - 5x₂ < -10 x₁ , x₂ > 0 Solution Making inequalities of the constraints to equalities: x₁ + x₂ = 1 x₁ = 0 x₂ = 1 x₂ = 0 x₁ = 1

- 0.5x₁ - 5x₂ = -10 or .5x₁ + 5x₂ = 10 x₁ = o x₂ = 2 x₂ = 0 x₁ = 20

X₂ X₁+X₂=1 5X₁+5X₂=10 X₁ From the graph it is very much clear that there is no set of points that satisfy both the constraints. This means that there is infeasible solution.

MULTIPLE OPTIMAL SOLUTIONS Ques. Maximize Z = 4x₁ + 4x₂ Subject to. x₁ + 2x₂ < 10 6x₁ + 6x₂ < 36 x₁ < 4 x₁ , x₂ > 0 Solution Making inequalities to equalities x₁ + 2x₂ = 10 x₂ = 0 x₁ = 10 x₁ = 0 x₂ = 5

6x₁ + 6x₂ = 36 x₁ = 4 so (4,0) x₁ = 0 x₂ = 6 x₂ = 0 x₁ = 6

x₂ Feasible solution region B(2,4) C(4,2) x₁ The objective function touches point B and C simultaneously. So B and C represent optimal value. At B x₂ = 4 , x₁ = 2 , Z = 24 At x₁ = 4 , x₂ = 2 , Z = 24

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