1. THE SIMPLEX METHOD

2. STEP BY STEP GUIDE Formulate the problem
(i) Pick out important information
(ii) Formulate constraints
(iii) Formulate objective function
Introduce slack variables
Form initial tableau
Obtain new tableaux
(i) Identify pivotal column
(ii) Find θ-values
(iii) Identify pivotal row
(iv) Identify pivot
(v) Pivot
5. Get the solution

3. THE PROBLEM
A small factory produces two types of toys: cars and diggers. In the manufacturing process two machines are used: the moulder and the colouriser. A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser. The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day. Each digger gives a profit of £16 and each car gives a profit of £14. The profit needs to be maximised.
How do we formulate this problem?

4. STEP BY STEP GUIDE Formulate the problem
(i) pick out important information
(ii) formulate constraints
(iii) formulate objective function
Introduce slack variables
Form initial tableau
Obtain new tableaux
Get the solution

5. PICKING OUT IMPORTANT INFORMATION
A small factory produces two types of toys: cars and diggers. In the manufacturing process two machines are used: the moulder and the colouriser.
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.
Each digger gives a profit of £16 and each car gives a profit of £14.

6. PICKING OUT IMPORTANT INFORMATION
A small factory produces two types of toys: cars and diggers. In the manufacturing process two machines are used: the moulder and the colouriser.
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.
Each digger gives a profit of £16 and each car gives a profit of £14.

7. A digger needs 2 hours on the moulder and 1 hour on the colouriser
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.

8. PICKING OUT IMPORTANT INFORMATION
A small factory produces two types of toys: cars and diggers. In the manufacturing process two machines are used: the moulder and the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.
Each digger gives a profit of £16 and each car gives a profit of £14.

9. A digger needs 2 hours on the moulder and 1 hour on the colouriser
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.

10. STEP BY STEP GUIDE Formulate the problem
(i) pick out important information
(ii) formulate constraints
(iii) formulate objective function
Introduce slack variables
Form initial tableau
Obtain new tableaux
Get the solution

11. make two constraints from this information.
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.
Using the decision variables
d = number of diggers
c = number of cars
make two constraints from this information.

12. 2d + c ≤ 16 FORMING CONSTRAINT 1 THE MOULDER
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.
2d + c ≤ 16

13. d + c ≤ 9 FORMING CONSTRAINT 2 THE COLOURISER
A digger needs 2 hours on the moulder and 1 hour on the colouriser. A car needs 1 hour on the moulder and 1 hour on the colouriser.
The moulder can be operated for 16 hours a day and the colouriser for 9 hours a day.
d + c ≤ 9

14. STEP BY STEP GUIDE Formulate the problem
(i) pick out important information
(ii) formulate constraints
(iii) formulate objective function
Introduce slack variables
Form initial tableau
Obtain new tableaux
Get the solution

15. PICKING OUT IMPORTANT INFORMATION
A small factory produces two types of toys: cars and diggers. In the manufacturing process two machines are used: the moulder and the colouriser.
Each digger gives a profit of £16 and each car gives a profit of £14.

16. FORMING THE OBJECTIVE FUNCTION
Each digger gives a profit of £16 and each car gives a profit of £14.
Let Z be the total profit; formulate the objective function

17. FORMING THE OBJECTIVE FUNCTION
Each digger gives a profit of £16 and each car gives a profit of £14.
Z = 16d + 14c

18. THE LINEAR PROGRAMMING PROBLEM
MAXIMISE Z = 16d + 14c
subject to the constraints:
(i) 2d + c ≤ 16
(ii) d + c ≤ 9
(iii) c ≥ 0 , d ≥ 0
VERY IMPORTANT
DON’T FORGET YOUR NON – NEGATIVITY CONSTRAINTS !

19. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
Get the solution

20. INTRODUCING SLACK VARIABLES
To change inequalities (i) and (ii) into equations we add slack variables s and t
This gives:
(i) 2d + c + s = 16
(ii) d + c + t = 9

21. THE NEW LINEAR PROGRAMMING PROBLEM
MAXIMISE Z = 16d + 14c + 0s + 0t
subject to the constraints:
2d + c + s + 0t = 16
d + c + 0s + t = 9
c ≥ 0 , d ≥ 0 , s ≥ 0 , t ≥ 0

22. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
Get the solution

23. We want to put all the information in the
form of a table. This is called the
initial tableau.
To form the initial tableau we need to
change the objective function from
Z = 16d + 14c + 0s + 0t to
Z – 16d – 14c – 0s – 0t = 0

24. FORMING THE INITIAL TABLEAU
Label the table with your basic variables, s and t
and with your non – basic variables, d and c.
BASIC VARIABLES d c s t
VALUE Z

25. CLICK HERE BASIC VARIABLES The variables solved for are called
Non-basic variables are set to zero.
CLICK HERE

26. FORMING THE INITIAL TABLEAU
2d + 1c + 1s + 0t = 16
1d + 1c + 0s + 1t = 9
Z – 16d – 14c – 0s – 0t = 0
BASIC VARIABLES d c s t
VALUE Z 2 1 1 16 1 1 1 9
-16
-14

27. FORMING THE INITIAL TABLEAUZ t s
VALUE c d
BASIC VARIABLES 2 16 1 9
-16
-14
This is the objective row

28. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
(i) Identify pivotal column
(ii) Find θ-values
(iii) Identify pivotal row
(iv) Identify pivot
(v) Pivot
Get the solution

29. PIVOTAL COLUMN Z t s c d 2 16 1 9 -16 -14
We now need to find where to pivot and we start by
entering the basis by choosing the column with the most negative entry in the objective row. Z t s
VALUE c d
BASIC VARIABLES 2 16 1 9
-16
-14
This is the most negative coefficient with corresponding variable d and it’s column is called the pivotal column. d is now called the entering variable.

30. CLICK HERE ENTERING THE BASIS
Choose the non-basic variable that is to become a basic variable.
The most common rule for selecting this variable is to select the variable with the most negative entry in the objective row.
CLICK HERE

31. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
(i) Identify pivotal column
(ii) Find θ-values
(iii) Identify pivotal row
(iv) Identify pivot
(v) Pivot
Get the solution

32. FINDING θ-VALUES Z t s c d 2 1 1 16 1 1 1 9 -16 -14
You are now going to find the pivotal row and the leaving variable.
You need to find θ-values.
1. Identify positive entries in the pivotal column.
2. Divide each entry in value column by the corresponding positive entry in the pivotal column. Z t s
VALUE c d
BASIC VARIABLES 2 1 1 16 1 1 1 9
-16
-14

33. CLICK HERE LEAVING VARIABLE
The leaving variable is the variable in the
pivotal row chosen to leave the basis.
The entering variable is the new basic
variable.
CLICK HERE

34. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
(i) Identify pivotal column
(ii) Find θ-values
(iii) Identify pivotal row
(iv) Identify pivot
(v) Pivot
5. Get the solution

35. PIVOTAL ROW For row (i) For row (ii)
The row with the smallest θ-value is called the pivotal row.
Here the pivotal row is row (i)

36. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
(i) Identify pivotal column
(ii) Find θ-values
(iii) Identify pivotal row
(iv) Identify pivot
(v) Pivot
5. Get the solution

37. THE PIVOT The pivot! Z t s c d 2 1 1 16 1 1 1 9 -16 -14
VALUE c d
BASIC VARIABLES 2 1 1 16 1
The pivotal row 1 1 9
-16
-14
The pivotal column

38. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
(i) Identify pivotal column
(ii) Find θ-values
(iii) Identify pivotal row
(iv) Identify pivot
(v) Pivot
5. Get a solution

39. PIVOTING Replace the leaving variable with the entering variable.
2. Divide all entries in the pivotal row by the pivot. The pivot becomes 1.
3. Add suitable multiples of the pivotal row to all other rows until all entries, apart from the pivot, in the pivotal column are zero.

40. Step 1 - Replace the leaving variable with the entering variable.Z t s
VALUE c d
BASIC VARIABLES 2 16 1 9
-16
-14
Step 2 - Divide all entries in the pivotal row by the pivot. The pivot becomes 1.
BASIC VARIABLES d c s t
VALUE s d 1
1/2
1/2 8 t Z

41. PIVOTING Z t s c d 2 16 1 9 -16 -14 t 1/2 -1/2 1 1 VALUE
BASIC VARIABLES 2 16 1 9
-16
-14
Step 3 - Add suitable multiples of the pivotal row to all other rows until all
entries, apart from the pivot, in the pivotal column are zero.
row (ii) – ½ row (i)
gives t
1/2
-1/2 1 1

42. PIVOTING Z t 8
1/2 1 d
VALUE s c
BASIC VARIABLES t
1/2
-1/2 1

43. PIVOTING Z t s c d 2 16 1 9 -16 -14 Z -6 8 128 VALUE BASIC VARIABLES1 9
-16
-14
Step 3 - Add suitable multiples of the pivotal row to all other rows until all
entries, apart from the pivot, in the pivotal column are zero.
row (iii) + 8 row (i)
gives Z -6 8
128

44. This is our second tableauZ t 8
1/2 1 d
VALUE s c
BASIC VARIABLES t
1/2
-1/2 1 Z -6 8
128
This is our second tableau

45. PIVOTING Follow the rules for finding a pivot on your second tableau.
Pivot as before.
Continue this process until there are no negative entries in the objective row.
This will be your final tableau. This is called the optimal tableau.

46. OPTIMAL TABLEAU d c s t 1 -1 7 2 Z 12 140
BASIC VARIABLES d c s t
VALUE 1 -1 7 2 Z 12
140
Note there are no negative entries in the objective row.
Can you see the solution?

47. STEP BY STEP GUIDE Formulate the problem Introduce slack variables
Form initial tableau
Obtain new tableaux
5. Get the solution

48. OBTAINING THE SOLUTION
BASIC VARIABLES d c s t
VALUE 1 -1 2 12 d 7 c 2 Z
140
Remember that since s and t are now non–basic variables they are set to zero.
This corresponds to the solution:
s = 0, t = 0,
d = 7
c = 2
Z = 140

49. THE SOLUTION
Don’t forget to put your solution back into the context of the problem.
Z = 140
d = 7
c = 2
The maximum profit is £140
To make this profit the factory should produce 7 diggers and 2 cars.