1. D1 Discrete Mathematics
Maximisation problems
The Simplex Algorithm

2. The basic idea
in maximisation problems we are trying to maximise some function (P, say – the objective function) of two or more variables, x and y
but there will be some constraints such as x + y ≤ 7
now, if x + y ≤ 7 then:
x + y + ‘something’ = 7
the ‘something’ (or slack variable) must be zero or greater
so, in the simplex algorithm, we change the inequalities to equations
then we use simultaneous equation techniques to get:
an equation which involves P, the slack variables and, at most, one of x or y
equations involving only one of x or y plus the slack variables
our solution can then be found
ideally the maximum value of P will occur when the slack variables are zero
the simplex algorithm is a way of achieving this, that can easily be programmed into a computer
let’s see how it works:

3. The simplex algorithm
formulate the maximising problem as a tableau using slack variables as necessary
ensure that all elements in the last column (except possibly the top one) are non-negative
select any column (except the last one) whose top element is negative (it may help to choose the lowest negative number e.g. -7 instead of -4; this may get the top row all non-negative more quickly)
for each row, divide the number in the last column by the number in the chosen column
in the selected column choose the positive number which gave you the smallest result; this is called the pivot
divide the pivot row by the pivot to give a new row
combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column
if all the top elements (except possibly the last one) are non-negative then the maximum has been reached; otherwise return to step 3
the last column contains the values of the objective function and the non-zero variables
note: this is essentially the method of elimination as applied to simultaneous equations

4. Example: Maximise 7x + 11y subject to 2x + 4y ≤ 60, 3x + 3y ≤ 60 and x, y ≥ 0
P = 7x + 11y
2x + 4y ≤ 60
3x + 3y ≤ 60
formulate the maximising problem (using slack variables as necessary)
P - 7x - 11y = 0
2x + 4y + s = 60
3x + 3y + t = 60 P x y s t l
ensure that all elements in the last column (except possibly the top one) are non-negative

5. Example: Maximise 7x + 11y subject to 2x + 4y ≤ 60, 3x + 3y ≤ 60 and x, y ≥ 0
P = 7x + 11y
2x + 4y ≤ 60
3x + 3y ≤ 60
formulate the maximising problem (using slack variables as necessary)
ensure that all elements in the last column (except possibly the top one) are non-negative
P - 7x - 11y = 0
2x + 4y + s = 60
3x + 3y + t = 60 P x y s t l 1 -7
-11 

6. Example: Maximise 7x + 11y subject to 2x+4y ≤ 60, 3x + 3y ≤ 60 and x, y ≥ 0
formulate the maximising problem (using slack variables as necessary)
ensure that all elements in the last column (except possibly the top one) are non-negative
P = 7x + 11y
2x + 4y ≤ 60
3x + 3y ≤ 60
P - 7x - 11y = 0
2x + 4y + s = 60
3x + 3y + t = 60 P x y s t l 1 -7
-11  2 4 60 

7. Example: Maximise 7x + 11y subject to 2x+4y ≤ 60, 3x + 3y ≤ 60 and x, y ≥ 0
formulate the maximising problem (using slack variables as necessary)
ensure that all elements in the last column (except possibly the top one) are non-negative
P = 7x + 11y
2x + 4y ≤ 60
3x + 3y ≤ 60
P - 7x - 11y = 0
2x + 4y + s = 60
3x + 3y + t = 60 P x y s t l 1 -7
-11  2 4 60  3 

8. select any column (except the last one) whose top element is negative
for each row, divide the number in the last column by the number in the chosen column
in the selected, column choose the positive number which gave you the smallest result; this is called the pivot P x y s t l 1 -7
-11  2 4 60  3 
0 ÷ -7 = 0
60 ÷ 2 = 30
60 ÷ 3 = 20

9. divide the pivot row by the pivotx y s t l 1 -7
-11  2 4 60  3 

10. divide the pivot row by the pivotx y s t l 1 -7
-11  2 4 60  3 
 =  ÷ 3

11. divide the pivot row by the pivotx y s t l 1 -7
-11  2 4 60  3 
1/3 20
 =  ÷ 3

12. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3 
1/3 20
 =  ÷ 3

13. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3 
 =  + 7 x 
1/3 20
 =  ÷ 3

14. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
1/3 20
 =  ÷ 3

15. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
 =  – 2 x 
1/3 20
 =  ÷ 3

16. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3

17. if all the top elements (except possibly the last one) are non-negative then the maximum has been reached; otherwise return to step 3 P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3

18. select any column (except the last one) whose top element is negative
for each row, divide the number in the last column by the number in the chosen column
in the selected, column choose the number which gave you the smallest result; this is called the pivot P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3

19. divide the pivot row by the pivotx y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3

20. divide the pivot row by the pivotx y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
 =  ÷ 2

21. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
-1/3 10
 =  ÷ 2

22. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
-1/3 10
 =  ÷ 2

23.     =  + 7 x   =  – 2 x   =  ÷ 3  =  + 4 x  ½  =  ÷ 2
combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
 =  + 4 x 
-1/3 10
 =  ÷ 2

24. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
180
 =  + 4 x 
-1/3 10
 =  ÷ 2

25. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
180
 =  + 4 x 
-1/3 10
 =  ÷ 2
 =  - 

26. combine appropriate multiples of the new row with all the other rows in order to reduce to zero all other elements in the pivot column P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
180
 =  + 4 x 
-1/3 10
 =  ÷ 2 -½
2/3
 =  - 

27. if all the top elements (except possibly the last one) are non-negative then the maximum has been reached; otherwise return to step 3
the last column contains the values of the objective function and the non-zero variables P x y s t l 1 -7
-11  2 4 60  3  -4
7/3
140
 =  + 7 x 
- 2/3 20
 =  – 2 x 
1/3
 =  ÷ 3
180
 =  + 4 x 
-1/3 10
 =  ÷ 2 -½
2/3
 =  - 

28. the last column contains the values of the objective function and the non-zero variables
now:
convert the tableau back to equation form P x y s t l 1 2
180
-1/3 10 -½
2/3

29. the last column contains the values of the objective function and the non-zero variables
now:
convert the tableau back to equation form P x y s t l 1 2
180
P + 2s + t = 180
-1/3 10 -½
2/3

30. the last column contains the values of the objective function and the non-zero variables
now:
convert the tableau back to equation form P x y s t l 1 2
180
P + 2s + t = 180
-1/3 10
y + ½s - 1/3t = 10 -½
2/3

31. the last column contains the values of the objective function and the non-zero variables
now:
convert the tableau back to equation form P x y s t l 1 2
180
P + 2s + t = 180
-1/3 10
y + ½s - 1/3t = 10 -½
2/3
x - ½s + 2/3t = 10

32. The last column contains the values of the objective function and the non-zero variables
The maximum will be reached when the slack variables are zero
since P = 180 – 2s – t
Which is why we needed those elements on the top row to be non-negative P x y s t l 1 2
180
P + 2s + t = 180
-1/3 10
y + ½s - 1/3t = 10 -½
2/3
x - ½s + 2/3t = 10

33. ½ y + ½s - 1/3t = 10 -½ x - ½s + 2/3t = 10 P x y s t l 1 2 180
the last column contains the values of the objective function and the non-zero variables P x y s t l 1 2
180
P + 2s + t = 180
P = 180
-1/3 10
y + ½s - 1/3t = 10 -½
2/3
x - ½s + 2/3t = 10

34. the last column contains the values of the objective function and the non-zero variablesP x y s t l 1 2
180
P + 2s + t = 180
P = 180
-1/3 10
y + ½s - 1/3t = 10
y = 10 -½
2/3
x - ½s + 2/3t = 10

35. the last column contains the values of the objective function and the non-zero variablesP x y s t l 1 2
180
P + 2s + t = 180
P = 180
-1/3 10
y + ½s - 1/3t = 10
y = 10 -½
2/3
x - ½s + 2/3t = 10
x = 10

36. There's a lot to do to apply the simplex algorithm, but
keep your head,
use the fraction button on your calculator,
take your time
and you should be ok!

37. That’s
all
folks!