Linear Programming Definition
Added on 2022-08-16
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Contents
Question:1 : ....................................................................................................................................... 3
Question:2: ........................................................................................................................................ 6
Question: 3: ....................................................................................................................................... 7
Question: 4: ....................................................................................................................................... 8
Question:1 : ....................................................................................................................................... 3
Question:2: ........................................................................................................................................ 6
Question: 3: ....................................................................................................................................... 7
Question: 4: ....................................................................................................................................... 8
Question:1 :-
(Solution)
a. The problem, which was formulated in the question in shown below,
Amount (litres) per 1000 kg of A and B
Sheep Cow Goat Cost ($/kg)
A 30 60 40 5
B 80 40 70 8
Before going forward with Linear Programming problems, we need to focus on its
mother distribution, which is Operations Research or OR. Operations Research is a type of
technique, which is used in organizations to solve complex business-oriented problems. World
war two was the time of inception of such a versatile study process which now a days, is been
utilized to solve numerous types of problem in different domains.
Linear programming falls under OR methods. It works with a single objective function,
e.g. maximization of profit or minimization of cost of the production, with the help of some
decision variables. Why the name linear? The answer is that the objective function of the
problem , the constraints are linear function of the decision variables.
A general and perhaps the simplest form of a Mathematical Programming Problem is
shown below,
min or max f(x1,.........,xn)
s.t. gi(x1; : : : ; xn) {= or >= or <=} bi
Where ,
x belongs to X
And Linear programming assumes the linearity of both the functions f and gi.
Looking at the problem we can defiantly conclude that the problem satisfies all the
requirement, to be called and solved as a Linear Programming Problem.
b. Now we need to formulate the Linear Programming Problem,
The objective is to minimize the total cost of production of cheese. The problem is
formulated below,
Minimize,
Z = 5𝑥1 + 8𝑥2
Subject to,
30𝑥1 + 80𝑥2 ≥ 45
60𝑥1 + 40𝑥2 ≥ 50
40𝑥1 + 70𝑥2 ≤ 60
(Solution)
a. The problem, which was formulated in the question in shown below,
Amount (litres) per 1000 kg of A and B
Sheep Cow Goat Cost ($/kg)
A 30 60 40 5
B 80 40 70 8
Before going forward with Linear Programming problems, we need to focus on its
mother distribution, which is Operations Research or OR. Operations Research is a type of
technique, which is used in organizations to solve complex business-oriented problems. World
war two was the time of inception of such a versatile study process which now a days, is been
utilized to solve numerous types of problem in different domains.
Linear programming falls under OR methods. It works with a single objective function,
e.g. maximization of profit or minimization of cost of the production, with the help of some
decision variables. Why the name linear? The answer is that the objective function of the
problem , the constraints are linear function of the decision variables.
A general and perhaps the simplest form of a Mathematical Programming Problem is
shown below,
min or max f(x1,.........,xn)
s.t. gi(x1; : : : ; xn) {= or >= or <=} bi
Where ,
x belongs to X
And Linear programming assumes the linearity of both the functions f and gi.
Looking at the problem we can defiantly conclude that the problem satisfies all the
requirement, to be called and solved as a Linear Programming Problem.
b. Now we need to formulate the Linear Programming Problem,
The objective is to minimize the total cost of production of cheese. The problem is
formulated below,
Minimize,
Z = 5𝑥1 + 8𝑥2
Subject to,
30𝑥1 + 80𝑥2 ≥ 45
60𝑥1 + 40𝑥2 ≥ 50
40𝑥1 + 70𝑥2 ≤ 60
Where, 𝑥1 , 𝑥2 > 0
c. For the solution of the Linear Programming Problem, we need to use
Graphical Method here,
We are taking a choice of Z, let it be 40.
Thus, we get the equation ,
5𝑥1 + 8𝑥2 = 40
Or, 𝑥1 + 𝑥2 = 1
8 5
Which gives the following two co-ordinates, wiz. (8,0) and (0,5) . Now, with the
given three lines, one of them defines the feasible region.
It can be observed, that as we move away from origin (i.e. (0,0)) the value
gradually decreases.
We have to solve the following equations to get the feasible region,
−30𝑥1 − 80𝑥2 + 45 = 0
−60𝑥1 − 40𝑥2 +50 =0
And 40 𝑥1 + 70𝑥2 − 60 = 0
From 1 and 2 we get,
𝑥1 −4000−1800 = 𝑥2 −1500−2700 = 1
−1200−4800
Solving this we get , (x1,x2) = (58/60 , 42/60)
From 1 and 3 we get,
𝑥1
Solving this we get,
4800+3150 = 𝑥2 1800+1800 = 1
−2100−3200
(x1,x2) =(-7950/5300, -3600/5300)
From 2 and 3 we get,
𝑥1
Solving this we get,
2400+3500 = 𝑥2 3600+2000 = 1
−4200−1600
(x1,x2) = (-5900/5800, -5600/5800)
Now from the objective function we are replacing the values of the points
respectively to get the optimal solution,
c. For the solution of the Linear Programming Problem, we need to use
Graphical Method here,
We are taking a choice of Z, let it be 40.
Thus, we get the equation ,
5𝑥1 + 8𝑥2 = 40
Or, 𝑥1 + 𝑥2 = 1
8 5
Which gives the following two co-ordinates, wiz. (8,0) and (0,5) . Now, with the
given three lines, one of them defines the feasible region.
It can be observed, that as we move away from origin (i.e. (0,0)) the value
gradually decreases.
We have to solve the following equations to get the feasible region,
−30𝑥1 − 80𝑥2 + 45 = 0
−60𝑥1 − 40𝑥2 +50 =0
And 40 𝑥1 + 70𝑥2 − 60 = 0
From 1 and 2 we get,
𝑥1 −4000−1800 = 𝑥2 −1500−2700 = 1
−1200−4800
Solving this we get , (x1,x2) = (58/60 , 42/60)
From 1 and 3 we get,
𝑥1
Solving this we get,
4800+3150 = 𝑥2 1800+1800 = 1
−2100−3200
(x1,x2) =(-7950/5300, -3600/5300)
From 2 and 3 we get,
𝑥1
Solving this we get,
2400+3500 = 𝑥2 3600+2000 = 1
−4200−1600
(x1,x2) = (-5900/5800, -5600/5800)
Now from the objective function we are replacing the values of the points
respectively to get the optimal solution,
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