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LINEAR PROGRAMING

- 1. LINEAR PROGRAMING Definition Is a branch of mathematics which enables one to solve problems which either the greatest or minimum/least value of a certain quantity is required under some given limitations or constraints. Example In a big – organization, decision about distribution in order to realize maximum profit or reduce costs of productionare done by use of linear programming. Limitation/ constraints are translatedbylinear inequalities. Greatest value or least value will be expressedas a function (calledthe objective function) Introduction Drawing of linear inequalities Example 01 Draw and show the half plane representedby 8x+2y ≥16 Solution For 8x +2y≥16; draw 8x+ 2y = 16 For x – intercept, y = 0 8x = 16 X = 2 For y – intercept, x = 0 2y = 16 Y = 8 Using (0, 0) as a test point
- 2. 8(0) +2(0)≥ 16 0 ≥ 16 (False) Example 02 Determine the solutionset of the simultaneous inequalities Solution x + y ≥ 3 draw x + y = 3 (full line) x + y = 3 At x – intercept y= 0 at y – intercept x= 0 x = 3 y = 3 For x – 2y ≤ 9, x – 2y = 9
- 3. At x – intercept y= 0 at y – intercept, x= 0 X =9 -2y= 9 Y = Using (0, 0) as a test point Using (0,0) as a test point x + y ≥ 3 x – 2y≤ 9 0 + 0 ≥ 3 0 – (2)(0) ≤ 9 0 ≥ 3 (F) 0 ≤ 9 (T) The clear part is the solutionset
- 4. The solutionset is calledthe feasible region Questions Draw by shading unwanted regions of the half planes represented by following simultaneous inequalities (i) y≥ 2x – 1, y ≤ -1 (ii) y≤ 2x – 1, y ≥ x – 3, y ≥ -1 (iii) y < 2x – 1, y ≤ -1 (iv) 6x + 9y ≥ 12 0.4x+ 0.1y≥ 0.2 32x+ 10y≥ 20 Evaluationofa functionsatisfiedbythe givenset ofinequalities →Example Find the maximum and minimum value of c = 4x + 3y + 38 subjectedto x + y ≥ 5 0 ≤ y ≤ 6 0 ≤ x ≤ 5 x ≥ 0, y ≥ 0 Solution For x + y ≥ 5 x + y = 5 When x = 0, y = 5 y = 0, x = 5 For 0 ≤ y ≤ 6 0 = y = 6 Line y = 6 For 0 ≤ x ≤ 5 O = x = 5 Line x = 5 Test points x ≥ 0, shade left of x = 0 (0, 0) y ≥ 0, shade below of x = axis
- 5. Corner points C = 4x + 3y + 38 A (5, 0) 4 (5) + 3 (0) +38 = 58
- 6. B (5, 6) 4 (5) + 3 (6) + 38 = 76 C (0, 6) 4 (0) + 3 (6) +38 = 56 D (0, 5) 4 (0) + 3 (5) + 38 = 53 : . The maximum value of c = 76 and occurs at (5, 6) The minimum value of c = 53 and occurs at (0, 5) Questions 1. Find the maximum and minimum values of the given functions and the value x and y where they occur (i) Z = 4x+ 3y Subject to x + 2y ≤ 10 3x + y ≤ 5 x ≥ 0, y ≥ 0 (ii) P = 134x+ 20y Subject to x + y ≤ 160 10 ≤ x ≤ 60 0 ≤ y ≤ 120 (iii) T = 4x +7y Subject to. x + y ≤ 18 5 ≤ x ≤ 10 3 ≤ y ≤ 10 x ≥ 0, y ≥ 0 (iv) P = 2x + 4y Subject to 2x + 3y ≥ 3 -5x+ 4y ≤ 0 3x + 4y ≤ 18 X ≥ 0, y ≥ 0. FORMULATION OF A LINEAR PROGRAMMING PROBLEM. Steps in formulatinga linear programming problem. 1. * Read the problem several times and assess what is known and what is to be determined.
- 7. 2. * Identify the unknown quantities and assign variables to them, be careful about the units. 3. * Determine the objective function;it involves the quantity to be maximized or minimized. 4. * Translate the constraints into linear inequalities, * Constraints are limitation or restrictions to the problem for each constraints the units must be same. 5. * Graph the constraints andfind the feasible solution 6. * Find the corner points of the feasible solution. These are points of intersectionof the graph 7. * Evaluate the objective function. Highest value of the objective function has to be maximized or smallest value to be minimized. Example A student has 120 shillings to spend on exercise books. At a school shop an exercise book costs 8 shillings, at stationery store an exercise book costs 12 shillings. The school has only 6 exercise books and the student wants to obtain the greatest number of exercise books possible usingthe money. Find the greatest number of exercise books he canbuy. Solution Let x be number of exercise books to be bought at school shop Let y be number of exercise books to be bought at stationeryshop →Objective function Let f (x, y) = objective function Then f (x, y) = x + y →Constrains or linear inequalities 8x + 12y ≤ 120 x ≤ 6 Non – Negative constraints x ≥ 0 y ≥ 0 →Equations 8x + 12y = 120 4x + 6y = 60 2x + 3y = 30 When x = 0, y = 10 y = 0, x = 15 x = 6, y = 0
- 8. x = 6 Corner points F (x, y) = x + y A (0, 0) 0 + 0 = 0 B (6, 0) 6 + 0 = 6 C ( 6, 6) 6 + 6 = 12 D (0, 10) 0+10 = 10
- 9. : . The greatest numbers of exercise books he can buy are 12 books 6 from the school shop and 6 from stationery. Example Student in a certain class are about to take a certain test of BAM which has two sections A and B; where in section A each question worth 10 marks while in section B; each worth 25 marks. The student must do at least 3 questions of section A; but not more than 12. A student must also do 4 questions from section B but not more than 15. In addition students cannot do more than 20 questions. How many questions of each type should the student do to obtain the maximum scores? Solution Let x be number of questions to be done in sectionA Let y be number of questions to be done in sectionB →Objective function f (x, y) = 10x+ 25y →Constrains 3 ≤ x ≤ 12 4 ≤ y ≤ 15 x + y ≤ 20 x ≥ 0, y ≥ 0 Maximize f (x, y) = 10x+ 25ysubject to; 3 ≤ x ≤ 12 4 ≤ y ≤ 15 x + y ≤ 20 x ≥ 0, y ≥ 0 Equations 3 = x = 12 4 = y = 15 x + y = 20 x = 20, y = 0 x = 0, y = 20 x = 0
- 10. y = 0 Corner Points f(x, y) = 10x+ 25y A (3, 4) 10 (3) + 25 (4) = 130 B (12, 4) 10 (12) + 25 (4) = 220 C (12, 8 ) 10 (12) + 25 (8) = 320 D (5, 15) 10 (5) + 25 (15) = 425 E (3, 15) 10 (3) + 25 (15) = 405 The student should do 5 questions from section A and 15 questions from section B to obtain maximum score of 425. Diet problems onlinear programming problem Example 01 A doctor prescribes a special diet for patients containing the following number of units of Vitamin A and B per kg of two types of foodF1 and F2 Type of Food Vitamin A Vitamin B F1 20 units/kg 7 units/kg F2 15 units /kg 14 units /kg
- 11. If the minimum daily intake required is 120 units of A and 70 units of B, what is the least total mass of fooda patient must have so as to have enough of these vitamins? EXAMPLE 02 Rice and beans provide maximum levels of protein, calories and vitamin B2. If used as a staple diet. The food values per kg of uncooked rice and beans are as shown in the table below. Protein/kg Calories/kg Vitamin B2/kg Price kg Rice 60g 3200 cal 0.4 400 Beans 90g 1000 cal 0.1 500 Min daily req. 120 2000 cal 0.2 What is the lowest cost of diet meeting, these specifications? Solution. Let x be number of kg of rice to be bought Let y be number of kg of beans to be bought →Objective function 400x+ 500y Constrains 60x+ 90y≤ 120 3200x+ 1000y≤ 2000 0.4x + 0.1y≤ 0.2 Minimize f (x, y) = 400x+ 500ySubject to; 60x+ 90y≤ 120 3200x+ 1000y≤ 2000
- 12. 0.4x+ 0.1y≤ 0.2 x ≥ 0, y ≥ 0 For 60x+ 90y≥ 120 60x+ 90y= 120 2x + 3y = 4 When x = 0, y = 1.3 Y = 0, x = 2 For 3200x+ 1000y≥ 2000 32x+ 10y= 20 16x+ 5y = 10 When x = 0, y = 2 Y = 0, x = 0.63 For 0.4x + 0.1y ≤ 0.2 0.4x+ 0.1y= 0.2 When x = 0, y = 2 When y = 0, x = 0.5
- 13. Corner points F (x, y) = x + y A (0, 8) 0 + 8 = 8 B (3.6, 3.2) 3.6 + 3.2 = 6.8 C (10,0) 10 + 0 = 10 : . The least total mass a patient should have is 6.8kg i.e. 3.6kg of food 1 and 3.2 kg of food2. Question 1. A doctor prescribes that in order to obtain adequate supply of vitamin A and C his patient should have portions of food 1 and food 2. The number of units of vitamin A and C are given in the followingtable A C
- 14. Food1 3 2 Food2 1 7 The doctor prescribes a minimum of 14 units of vitamin A and 21 units of vitamin C. What are the least portions of food1 and food2 that will fit the doctor’s prescriptions? LINEAR PROGRAMMING PROBLEMS 1. Two printers N and T produce three types of books. N produces 80 types I books per day, 10 type II books per day and 20 types III books per day, while T produces 20 types I books per day 10 type II books per day and 70 types III books per day. The orders placed are 1600 type I, 500 type II and 2100 type III books. The daily operating costs for N shs. 10,000/=, for T shs, 20,000/= how many days should each printer operate to meet the orders at aminimum cost. 2. A small textile company manufactures three different size of shirts, Large (L), medium (M) and small (S) at two different plants A and B. The number of shirts of each size producedand the cost of productionper day are as follows! A B Monthly demand Large size per day 50 60 2500 Medium size per day 100 70 3500 Small size per day 100 200 7000 Production Cost per day T. shs. 2500 3500 _ (i) How many days per month should each factory operate in order to minimize total cost. (ii) What is the minimum cost of production Solution01 Let x be number of days printer N shouldoperate
- 15. Let Y be number of days printer T should operate →Objective function(f(x, y)) 10000x+ 20000y →Constrains 80x + 20y≥ 1600 10x + 10y≥ 500 20x + 70y≥ 2100 X ≥ 0, y ≥ 0 →Equations 80x + 20y≥ 1600 80x+ 20y= 1600 8x + 2y = 160 When x = 0, y = 80 y = 0, x = 20 10x + 10y≥ 500 10x+ 10y= 500 x + y = 50 When x = 0, y = 50 y = 0, x = 50 20x + 70y≥ 2100 20x+ 70y= 2100 2x + 7y = 210 When x = 0, y = 30
- 16. y = 0, x = 105 Corner points F (x, y) = 10000x+ 20000y A (0, 80) 10000 (0)+ 20000 (80) = 1,600,000 B (28, 22) 10000 (28) + 20000(22) = 720,000 C (105, 0) 10,000 (105)+ 20000 (0)= 1,050,000 Printer N should be operated for 28 days and printer T should work for 22 days to meet the orders at minimum cost. Solution02 Let x be number of days per month factoryA shouldoperate Let y be number of days per month factoryB shouldoperate →Objective function
- 17. F (x, y) = 2500x+ 3500y →Constrains 50x + 60y≥ 2500 100x+ 70y ≥ 3500 100x+ 200y≥ 7000 x ≥ 0, y ≥ 0 Minimize f (x, y) = 2500x+ 3500y Subject to 50x+ 60y≥ 2500 100x+ 70y≥ 3500 100x+ 200y≥ 7000 X ≥ 0, y ≥ 0 →Equations 50x + 60y= 2500 When y = 0, x = 50 X = 0, y = 41.7 100x+ 70y = 3500 When y = 0, x = 35 x = 0, y = 50 100x+ 200y= 7000 When y = 0, x = 70 X = 0, y = 35
- 18. Corner points F (x, y) = 2500x+ 3500y A (0,50) 2500 (0) + 3500(50)= 175,000 B (15, 30) 2500 (15) + 3500 (30) = 142,500 C (20,25) 2500 (20) + 3500 (25) = 137,500 D (70, 0) 2500 (70) + 3500 (0)= 175,000 Factory A should operate for 20 days and factory B should operate for 25 days in order to minimize total cost. →Minimum cost of productionis 137,500 3. In a certain garage the manager had the following facts floor space required for a saloon is 2m2 and for a lorry is 3m2. Four technicians are required to service a saloon car and three technicians for a lorry per day. He has a maximum of 24m2 of
- 19. a floor space and a maximum of 36 technicians available; in addition he is not allowed to service more Lorries than saloon cars. The profit for serving a saloon car is 40,000/= and a lorry is 60,000/=. How many motor vehicles of each type should be serviceddaily in order to maximize the profit? Solution Let x be number of salooncars to be serviceddaily Let y be number of Lorries to be serviceddaily →Objective function F (x, y) = 40000x+ 60000y →Constrains 2x + 3y ≤ 24 4x + 3y ≤ 36 x ≥ y x ≥ 0 and y ≥ 0 Maximize f (x, y) = 40000x+ 60000y Subject to 2x + 3y ≤ 24 4x + 3y ≤ 36 x ≥ y x ≥ 0 and y ≥ 0 →Equations 2x + 3y ≤ 24 When x = 0, y = 8
- 20. Y = 0, x = 12 4x + 3y ≤ 36 When x = 0, y = 12 y = 0, x = 9 x = y Corner points F (x, y) = 40000x+ 60000y A (0, 0) 40000 (0) + 60000(0) = 0 B (4.8, 4.8) 40000 (4.8) + 60000 (4.8) = 480,000 C (6,4) 40000 (6)+ 60000 (4) = 480,000 D (9, 0) 40000 (9) + 60000 (0) = 360,000 6 salooncars and 4 Lorries shouldbe serviceddaily to maximize profit to 480,000/=
- 21. More example A builder has two stores, one at S1 and the other at S2. He is building houses at P1, P2, and P3. He needs 5 tons of bricks at P1, 6 tons of bricks at P2 and 4 tons of bricks at P3. The stores contain 9 tons of bricks at S1 and 6 tons of bricks at S2. The transport cost per ton are shown in the diagram To From P1 P2 P3 S1 6/= 3/= 4/= S2 4/= 2/= 6/= How does the builder send his bricks at a minimum cost?What is the minimum overall cost? Solution Let the builder send x tons of bricks from S1 to P1 and y tons of bricks from S1 to P2 Then the transportationof bricks to P1, P2 and P3 will be as follow: - To From P1 P2 P3 S1 X Y 9 – (x + y) S2 5 – x 6 – y 4 – [9 – (x + y)] The constrains are obtainedas follows x ≥ 0, y ≥ 0 9 – (x + y) ≥ 0 i.e. x + y ≤ 9 5 – x ≥ 0 i.e. x ≤ 5 6 – y ≥ 0 i.e. y ≤ 6 4 – [9 - (x + y)} ≥ 0 i.e. x + y ≥ 5
- 22. Objective function F (x, y) = 6x + 3y + 4(9 - (x + y)) + 4 (5 – x) + 2(6 – y) +6 [4 – (9 - (x + y)] = 6x + 3y +36 – 4x+ 4y + 20 – 4x+ 12 – 2y + 24 – 54 + 6x+ 6y F (x, y) = 4x – 3y+ 38 Minimize f (x, y) = 4x – 3y + 38 Subject to x + y ≤ 9 x + y ≥ 5 x ≤ 5, y ≥ 0 y ≤ 6, y ≥ 0 →Equation x + y = 9 When x = 0, y = 9 y = 0, x = 9 x + y = 5 x = 0, y = 5 y = 0, x = 9 x = 5 y = 6
- 23. Corner points F (x, y) = 4x + 3y + 38 A (0, 5) 4 (0) + 3 (5) + 38 = 53 B (0, 6) 4 (0) + 3 (6) + 38 = 56 C (3, 6) 4 (3) + 3 (6) + 38 = 68 D (5, 4) 4 (5) + 3 (4) + 38 = 70 E (5, 0) 4 (5) + 3 (0) + 38 = 58 The builder should send the bricks of tons as follows To From P1 P2 P3 S1 0 5 4 S2 5 1 0
- 24. The overall minimum cost is 53/= EXERCISE There is a factory located at each of the places P and Q. From these location a certain commodity is delivered to each of the three deports situated at A, B and C. The weekly requirements of the deports are respectively 5,5 and 4 unit of the commodity while the production capacity of the factories P and Q are 8 and 6 units respectively, just sufficient for requirement of deports. The cost of transportationper unit is given. To From A B C P 16 10 15 Q 10 12 10 Formulate this linear programming problem and how the commodities can be transported at minimum cost. What is the overall minimum cost?