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Linear Programming. Introduction. Mathematical programming is used to find the best or optimal solution to a problem that requires a decision or set of decisions about how best to use a set of limited resources to achieve a state goal of objectives. Steps involved in mathematical programming - PowerPoint PPT Presentation
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Linear Programming
Wednesday, April 19, 2023
1Dr.G.Suresh Kumar@KL University
Introduction• Mathematical programming is used to find the best or
optimal solution to a problem that requires a decision or set of decisions about how best to use a set of limited resources to achieve a state goal of objectives.
• Steps involved in mathematical programming
– Conversion of stated problem into a mathematical model that abstracts all the essential elements of the problem.
– Exploration of different solutions of the problem.
– Finding out the most suitable or optimum solution.
• Linear programming requires that all the mathematical functions in the model be linear functions.
Wednesday, April 19, 2023 2Dr.G.Suresh Kumar@KL University
General LPP
Max/min Z = c1x1 + c2x2 + ... + cnxn
subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
: am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
and xj ≥ 0, j = 1, 2, 3, …, n.xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients
Wednesday, April 19, 2023
3Dr.G.Suresh Kumar@KL University
Formation of LPP • A manufacturer produces two types of models M1
and M2. Each M1 model requires 4 hours of grinding and 2 hours of polishing; whereas M2 model requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and 3 polishers. Each grinder works for 40 hours a week and each polisher works for 60 hours a week. Profit on an M1 model is Rs. 3 and on M2 model is Rs. 4. Whatever is produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the two types of models so that he make the maximum profit in a week.
Wednesday, April 19, 2023
4Dr.G.Suresh Kumar@KL University
Solution• Let x be the number of M1 models and y be the
number of M2 models produced in a week.
• Then the weekly profit is
Z = 3x + 4y.• The total number of grinding hours needed in a
week to produce x and y is
4x + 2y.• The total number of polishing hours needed in a
week to produce x and y is
2x + 5y.
Wednesday, April 19, 2023
5Dr.G.Suresh Kumar@KL University
Solution cont…
• Since the number of grinding hours is not more than 80 and the number of polishing hours not more than 180, therefore
4x + 2y ≤ 80
2x + 5y ≤ 180.
• Also, the negative number of models are not produced, we must have
x ≥ 0 and y ≥ 0.
Wednesday, April 19, 2023
6Dr.G.Suresh Kumar@KL University
Solution cont…
• Now the problem is to find x, y which
Maximize Z = 3x + 4y
Subject to 4x + 2y ≤ 80
2x + 5y ≤ 180
and x, y ≥ 0.
Wednesday, April 19, 2023
7Dr.G.Suresh Kumar@KL University
Problem#2• An aeroplane can carry a maximum of 200
passengers. A profit of Rs.400 is made on each first class ticket and a profit of Rs. 300 is made on each economy class ticket. The airline reserves at least 20 seats for first class. However, at least 4 times as many passengers prefer to travel by economy class than by the first class. How many tickets of each class must be sold in order to maximize profit for the airline ? Formulate the problem as L.P. model and then solve.
Wednesday, April 19, 2023
8Dr.G.Suresh Kumar@KL University
Solution
• Maximize Z = 400x + 300y
subject to x + y ≤ 200
x ≥ 20
y ≥ 4x
and x, y ≥ 0.
Wednesday, April 19, 2023
9Dr.G.Suresh Kumar@KL University
Problem#3• Consider the following problem faced by a
production planner in a soft-drink plant. He has two bottling machines A and B. A is designed for 8-ounce bottles and B for 16-ounce bottles. However, each can be used on both types with some loss of efficiency. The following data is available :
Machine 8-ounce bottle 16-ounce bottle
A 100/ minute 40/ minute
B 60/ minute 75/ minute
Wednesday, April 19, 2023
10Dr.G.Suresh Kumar@KL University
Problem#3The machines can be run 8 hours per day, 5 days
per week. Profit on 8-ounce bottle is 15 paise and on a 16-ounce bottle is 25 paise. Weekly production of the drink cannot exceed 300,000 ounces and the market can absorb 25,000 8-ounce bottles and 7,000 16-ounce bottles per week. The planner wishes to maximize his profit subject to all production and marketing restrictions. Formulate the problem as L.P model.
Wednesday, April 19, 2023
11Dr.G.Suresh Kumar@KL University
Solution
• Maximize Z = 0.15x+0.25y
subject to 2x + 5y ≤ 480,000
5x + 4y ≤ 720,000
8x + 16y ≤ 300,000
x ≤ 25, 000
y ≤ 7,000
and x, y ≥ 0.
Wednesday, April 19, 2023
12Dr.G.Suresh Kumar@KL University
General LPP
Max/min Z = c1x1 + c2x2 + ... + cnxn
subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
: am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients
Wednesday, April 19, 2023
13Dr.G.Suresh Kumar@KL University
Solutions of LPP
• A set of values x1, x2, …, xn which satisfies the constraints of LPP is called its solution.
• Any solution to a LPP which satisfies the non-negativity restrictions is called its feasible solution.
• Any feasible solution which maximizes or minimizes the objective function is called its optimal solution.
Wednesday, April 19, 2023
14Dr.G.Suresh Kumar@KL University
Solutions of LPP• A solution is obtained by setting n-m variables to zero
(if m equality constraints and n variables of LPP) is called a basic solution.
• Basic solution of LPP is also satisfies the non-negativity restrictions is called basic feasible solution.
• Basic feasible solution which maximizes or minimizes the objective function is called its optimal solution.
• A basic feasible solution in which all basic variable are non-zero is called a non-degenerate basic feasible solution otherwise it is a degenerate basic feasible solution.Wednesday, April 19, 20
2315Dr.G.Suresh Kumar@KL
University
Problem#1
• Find all the basic solutions of the following system of equations identifying in each case the basic and non-basic variables.
2x1 + x2 + 4x3 = 11
3x1 + x2 + 5x3 = 14
Ans : (3, 5, 0) basic x1, x2 non-basic x3.
(0, 3, -1) basic x2, x3 non-basic x1.
(1/2, 0, 5/2) basic x1, x3 non-basic x2. Wednesday, April 19, 2023
16Dr.G.Suresh Kumar@KL University
Problem#2
• Find an optimal solution to the following LPP by computing all basic solutions and then finding one that maximizes objective function.
Max. Z = 2x1 + 3x2 + 4x3 + 7x4
Subject to 2x1 + 3x2 - x3 + 4x4 = 8
x1 - 2x2 + 6x3 -7x4 = -3
and x1, x2, x3, x4 ≥ 0.
Ans : (0, 0, 44/17, 45/17) and Z=28.9Wednesday, April 19, 2023
17Dr.G.Suresh Kumar@KL University
Slack variables
• If the constraints of a general LPP be
ai1x1 + ai2x2 + ... + ainxn ≤ bi (i = 1, 2,…), then the non-negative variables si which satisfy
ai1x1 + ai2x2 + ... + ainxn + si = bi (i = 1, 2,…)
are called slack variables.
Wednesday, April 19, 2023
18Dr.G.Suresh Kumar@KL University
Surplus variables
• If the constraints of a general LPP be
ai1x1 + ai2x2 + ... + ainxn ≥ bi (i = 1, 2,…), then the non-negative variables si which satisfy
ai1x1 + ai2x2 + ... + ainxn - si = bi (i = 1, 2,…)
are called surplus variables.
Wednesday, April 19, 2023
19Dr.G.Suresh Kumar@KL University
Canonical form of LPP
The canonical form of LPP has
• Objective function is of maximization type.
• All constraints are of ≤ type.
• All variables are non-negative.
Wednesday, April 19, 2023
20Dr.G.Suresh Kumar@KL University
Standard form of LPP
The standard form of LPP has
• Objective function is of maximization type
• All constraints are expressed as equations
• Right hand side of each constraint is non-negative
• All variables are non-negative.
Wednesday, April 19, 2023
21Dr.G.Suresh Kumar@KL University
Simplex Method
• Step#1: (i) Check whether the objective function is to be maximized or minimized. If minimized, then convert it into maximization using
Min. Z = - Max. (-Z).
(ii) Check whether all b’s are positive. If negative, then multiply by ‘-1’ both sides of the corresponding constraint.
Wednesday, April 19, 2023
22Dr.G.Suresh Kumar@KL University
Simplex Method
Step#2: Express the problem in the standard form.
Convert all inequalities into equations by introducing slack or surplus variables in the constraints of the LPP.
Step#3 : Find an initial basic feasible solution and form the initial simplex table.
Wednesday, April 19, 2023
23Dr.G.Suresh Kumar@KL University
Initial Simplex Table
cj c1 c2 … cn 0 0 … 0 Mini
CB xB values a1 a2 … an an+1 an+2 … an+m ratio
0 s1 b1 a11 a12 … a1n 1 0 … 0
0 s2 b2 a21 a22 … a2n 0 1 … 0
… … …. … … … … … … … …
0 sm bm am1 am2 … amn 0 0 … 1
Zj-cj -c1 -c2 … -cn 0 0 … 0
Wednesday, April 19, 2023
24Dr.G.Suresh Kumar@KL University
Simplex Method
Step#4 : Optimality test
Compute Zj-cj, where Zj = ∑ cBiaij.
If all Zj-cj ≥ 0, then the current basic feasible solutions is optimal.
If even one Zj-cj < 0, then the current basic feasible solution is not optimal and proceed to the next step.
Wednesday, April 19, 2023
25Dr.G.Suresh Kumar@KL University
Simplex MethodStep#5: (i) identifying incoming vector
Compute Zk-ck = Min. {Zj-cj/ Zj-cj < 0}, then ak is the entering(incoming) vector. If there is a tie select any one.
(ii) identifying outgoing vector
Compute Min.{xBi/aik, aik > 0} = xBr/ark, then ar is the out going vector. The element ark is called key element.
If all aik≤0, then the problem has an unbounded solution and we stop the process.
Wednesday, April 19, 2023
26Dr.G.Suresh Kumar@KL University
Simplex Method
(iii) iterate towards an optimal solution
Covert the key element to unity by dividing the key row by the key element. Then make all other elements of the key column zero by subtracting proper multiples of key row from other rows.
Step#6 : Go to step#4 and repeat the procedure until either an optimal solution or an unbounded solution is obtained.
Wednesday, April 19, 2023
27Dr.G.Suresh Kumar@KL University
Problem#1
Maximize Z = 5x1 + 3x2
Subject to x1 + x2 ≤ 2
5x1 + 2x2 ≤ 10
3x1 + 8x2 ≤ 12
and x1, x2 ≥ 0.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
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Solution
• In standard form
Maximize Z = 5x1 + 3x2+0s1+0s2+0s3
Subject to x1 + x2 +s1 +0s2 +0s3= 2
5x1 + 2x2 +0s1+s2 +0s3=10
3x1 + 8x2 +0s1 +0s2+s3= 12
and x1, x2 , s1, s2, s3≥ 0.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
29
Initial Simplex Table
cj 5 3 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 ratio
0 s1 2 1 1 1 0 0 2
0 s2 10 5 2 0 1 0 2
0 s3 12 3 8 0 0 1 4
Zj-cj -5 -3 0 0 0
Wednesday, April 19, 2023
30Dr.G.Suresh Kumar@KL University
1
First Simplex Table
cj 5 3 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 ratio
5 x1 2 1 1 1 0 0
0 s2 0 0 -3 -5 1 0
0 s3 6 0 5 -3 0 1
Zj-cj 0 2 5 0 0
Wednesday, April 19, 2023
31Dr.G.Suresh Kumar@KL University
Solution
• Since all Zj-cj ≥ 0, the current basic feasible solution is optimal.
• Therefore the optimal solution to the given problem is
x1= 2, x2 = 0 and
maximum Z= 5(2)+3(0)=10.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
32
Problem#2
Minimize Z = x1 - 3x2 + 3x3
Subject to 3x1 - x2+ 2x3 ≤ 7
2x1 + 4x2 ≥ -12
-4x1 + 3x2 + 8x3≤ 10
and x1, x2, x3 ≥ 0.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
33
Solution
• In standard form
Maximize Zʹ = -x1+3x2 -3x3+0s1+0s2+0s3
Subject to 3x1 - x2 +2x3+s1 +0s2 +0s3= 7
-2x1 - 4x2+0x3+0s1+s2 +0s3=12
-4x1 + 3x2+8x3+0s1 +0s2+s3= 10
and x1, x2, x3, s1, s2, s3≥ 0.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
34
Initial Simplex Table
cj -1 3 -3 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 a6 ratio
0 s1 7 3 -1 2 1 0 0 --
0 s2 12 -2 -4 0 0 1 0 --
0 s3 10 -4 3 8 0 0 1 10/3
Zj-cj 1 -3 3 0 0 0
Wednesday, April 19, 2023
35Dr.G.Suresh Kumar@KL University
3
Improved Simplex Table-I
cj -1 3 -3 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 a6 ratio
0 s1 31/3 5/3 0 14/3 1 0 1/3 31/5
0 s2 76/3 -22/3 0 32/3 0 1 4/3 --
3 x2 10/3 -4/3 1 8/3 0 0 1/3 --
Zj-cj -3 0 11 0 0 1
Wednesday, April 19, 2023
36Dr.G.Suresh Kumar@KL University
5/3
Improved Simplex Table-II
cj -1 3 -3 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 a6 ratio
-1 x1 31/5 1 0 14/5 3/5 0 1/5
0 s2 352/5 0 0 156/5 22/5 1 14/5
3 x2 58/5 0 1 32/5 4/5 0 3/5
Zj-cj 0 0 82/5 9/5 0 8/5
Wednesday, April 19, 2023
37Dr.G.Suresh Kumar@KL University
Solution
• Since all Zj-cj ≥ 0, the current basic feasible solution is optimal.
• Therefore the optimal solution to the given problem is
x1= 31/5, x2 = 58/5, x3=0 and
maximum Zʹ= -1(31/5)+3(58/5)-3(0)=143/5.
Minimum Z = -(Max.Zʹ) = -143/5
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
38
Problem#3• A farmer has 1,000 acres of land on which he can
grow corn, wheat or soya-bean. Each acre of corn costs Rs. 100 for preparation, requires 7 man-days of work and yields a profit of Rs. 30. An acre of wheat costs Rs. 120 for preparation, requires 10 man-days of work and yields a profit of Rs. 40. An acre of soya-beans costs Rs. 70 for preparation, requires 8 man-days of work and yields a profit of Rs. 20. If the farmer has Rs. 100,000 for preparation and can count on 8,000 man-days of work, how many acres should be allocated to each crop to maximize profit?
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
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SolutionLet x, y, z be the number of acres of corn,
wheat, soy-bean respectively. Then the given problem in LPP model:
Maximize Z = 30x + 40y + 20z
Subject to 100x +120y+70z ≤ 1,00,000
7x + 10y + 8z ≤ 8,000
x + y + z ≤ 1000
and x, y, z ≥ 0.
Ans: x = 250, y = 625, z = 0.Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
40
Initial Simplex Table
cj 30 40 20 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 a6 ratio
0 s1 10000 100 12 7 1 0 0 833.3
0 s2 8000 7 10 8 0 1 0 800
0 s3 1000 1 1 1 0 0 1 1000
Zj-cj -30 -40 -20 0 0 0
Wednesday, April 19, 2023
41Dr.G.Suresh Kumar@KL University
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
42
Improved Initial Simplex Table
cj 30 40 20 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 a6 ratio
0 s1 4000 16 0 -26 1 -12 0 250
40 x2 800 7/10 1 8/10 0 1/10 0 1142
0 s3 200 3/10 0 1/5 0 -1/10 1 666.6
Zj-cj -2 0 12 0 4 0
Wednesday, April 19, 2023
43Dr.G.Suresh Kumar@KL University
Improved Initial Simplex Table
cj 30 40 20 0 0 0 Mini
CB xB values a1 a2 a3 a4 a5 a6 ratio
30 x1 250 1 0 -13/8 1/16 -3/4 0
40 x2 625 0 1 31/16 -7/16 5/8 0
0 s3 125 0 0 11/16 -3/16 1/8 1
Zj-cj 0 0 5/2 1/8 5/2 0
Wednesday, April 19, 2023
44Dr.G.Suresh Kumar@KL University
Solution
• Since all Zj-cj ≥ 0, the current basic feasible solution is optimal.
• Therefore the optimal solution to the given problem is
x1= 250, x2 = 625, x3=0 and
Maximum Z= 30(250)+40(625)+20(0)
= 32500Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
45
Problem#4
• A firm produces three products which are processed on three machines. The relevant data is given below
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
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Machine Time per unit (Minutes) Machine capacity
(minute/day)Product A Product B Product C
M1 2 3 2 440
M2 4 -- 3 470
M3 2 5 -- 430
Problem#4
The profit per unit for product A, B and C is Rs. 4, Rs. 3 and Rs.6 respectively. Determine the daily number of units to be manufactured for each product. Assume that all the units produced are consumed in the market.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
47
SolutionLet x1, x2, x3 be the number of units of Products A, B, C to be
produced, then the problem in LP model
Maximize Z = 4x1 + 3x2 + 6x3
Subject to 2x1 + 3x2+ 2x3 ≤ 440
4x1 + 3x3 ≤ 470
2x1 + 5x2 ≤ 430
and x1, x2, x3 ≥ 0.
The optimal solution x1=0, x2=380/9, x3=470/3
and Zmax= 1066.67 rupees.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
48
Artificial variable Techniques
• Big M-Method or Method of Penalties :
Step#1: Express the problem in standard form
Step#2: Add non-negative variable to the left hand side of all constraints which are of (≥) or (=) type. Such new variables are called artificial variables. The penalty or cost for these variables in the objective function is –M, M>0 and large.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
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Big M-Method (Charne’s Penalty)
Step#3 : solve the modified LPP by simplex method.
At any iteration of the simplex method, one of the following three cases may arise:
Case(i)
There remains no artificial variable in the basis and the optimality condition satisfied. Then the solution is optimal solution.
Wednesday, April 19, 2023
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Big M-MethodCase(ii): There is at least one artificial variable in
the basis at zero level and optimality is satisfied. Then the solution is degenerate optimal basic feasible solution.
Case(iii) : There is at least one artificial variable in the basis at non-zero level and optimality satisfied. Then the problem has no feasible solution.
Step#4: Continue simplex method until either optimal solution or unbounded solution is obtained
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
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Problem#1Maximize Z = 3x1 + 2x2
Subject to 2x1 + x2 ≤ 2
3x1 + 4x2 ≥ 12
And x1, x2 ≥ 0.
Sol: In standard form
Maximize Z = 3x1 + 2x2+ 0s1+ 0s2
Subject to 2x1 + x2 + s1 + 0s2 = 2
3x1 + 4x2 + 0s1 – s2 = 12
And x1, x2 , s1, s2 ≥ 0.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
52
Solution After introducing the artificial variable, Modified
LPP
Maximize Z = 3x1 + 2x2+ 0s1+ 0s2+ 0A
Subject to 2x1 + x2 + s1 + 0s2 + 0A= 2
3x1 + 4x2 + 0s1 – s2 + A = 12
And x1, x2 , s1, s2 , A ≥ 0.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
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Initial Simplex Table
cj 3 2 0 0 -M Mini
CB xB values a1 a2 a3 a4 a5 ratio
0 s1 2 2 1 1 0 0 2
-M A 12 3 4 0 -1 1 3
Zj-cj -3-3M -2-4M 0 M 0
Wednesday, April 19, 2023
54Dr.G.Suresh Kumar@KL University
1
Improved Simplex Table
cj 3 2 0 0 -M Mini
CB xB values a1 a2 a3 a4 a5 ratio
2 x2 2 2 1 0 0 0
-M A 4 -5 0 -4 -1 1
Zj-cj -1+5M 0 4M M 0
Wednesday, April 19, 2023
55Dr.G.Suresh Kumar@KL University
Solution
• Since all Zj – cj ≥ 0 and artificial variable A is presented in the basis at non zero level. Therefore, the problem has no feasible solution.
Wednesday, April 19, 2023
Dr.G.Suresh Kumar@KL University
56
Problem• The following table gives the various
vitamin contents of three types of food and daily requirements of vitamins along with cost per unit. Find the combination of food for minimum cost.
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Dr.G.Suresh Kumar@KL University
57