Mada za sehemu hiiLinear ProgrammingMada 3
- Linear Programming Problems Formulation
- Graphical solution
- Transportation problems
Linear Programming is a branch of mathematics which enables one to solve problems in which the greatest or minimum/least value of a certain quantity is required under some given limitations or constraints.
Steps in formulating a linear programming problem.
- Read the problem several times and assess what is known and what is to be determined.
- Identify the unknown quantities and assign variables to them, be careful about the units.
- Determine the objective function; it involves the quantity to be maximized or minimized.
- Translate the constraints into linear inequalities. Constraints are limitations or restrictions to the problem; for each constraint the units must be the same.
- Graph the constraints and find the feasible solution.
- Find the corner points of the feasible solution. These are points of intersection of the graph.
- Evaluate the objective function. The highest value of the objective function has to be maximized or smallest value to be minimized.
Example
In a big organization, decisions about distribution in order to realize maximum profit or reduce costs of production are done by use of linear programming.
Limitations/constraints are translated by linear inequalities.
Greatest value or least value will be expressed as a function (called the objective function).
Drawing of linear inequalities
Example 01
Draw and show the half plane represented by 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
8(0)+2(0)≥16
0≥16 (False)
Example 02
Determine the solution set of the simultaneous inequalities
x+y≥3
x−2y≤9
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
At x-intercept y=0 at y-intercept, x=0
x=9 −2y=9
y=−4.5
Using (0, 0) as a test point
x+y≥3: 0+0≥3, 0≥3 (F)
x−2y≤9: 0−(2)(0)≤9, 0≤9 (T)
- The clear part is the solution set
- The solution set is called the feasible region
Example
Find the maximum and minimum value of c=4x+3y+38 subjected to
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
y=0 and y=6
Line y=6
For 0≤x≤5
x=0 and x=5
Line x=5
Test points: x≥0, shade left of x=0; y≥0, shade above the x-axis
| Corner points | C = 4x + 3y + 38 |
|---|---|
| A (5, 0) | 4(5) + 3(0) + 38 = 58 |
| 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 |
Therefore, the maximum value of c=76 and occurs at (5, 6).
The minimum value of c=53 and occurs at (0, 5).
Example
Find the maximum and minimum values of the given functions and the value of 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
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