STEP BY STEP GUIDE
1.
Formulate the problem
(i)
Pick out important information
(ii)
Formulate constraints
(iii)
Formulate objective function
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
(i)
Identify pivotal column
(ii)
Find θ-values
(iii)
Identify pivotal row
(iv)
Identify pivot
(v)
Pivot
5.
Get the solution
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?
STEP BY STEP GUIDE
1.
Formulate the problem
(i)
pick out important information
(ii)
formulate constraints
(iii)
formulate objective function
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
5.
Get the solution
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.
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.
• 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.
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.
• 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.
STEP BY STEP GUIDE
1.
Formulate the problem
(i)
pick out important information
(ii)
formulate constraints
(iii)
formulate objective function
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
5.
Get the solution
• 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.
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
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
STEP BY STEP GUIDE
1.
Formulate the problem
(i)
pick out important information
(ii)
formulate constraints
(iii)
formulate objective function
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
5.
Get the solution
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.
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
FORMING THE OBJECTIVE
FUNCTION
• Each digger gives a profit of £16 and each
car gives a profit of £14.
Z = 16d + 14c
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 !
STEP BY STEP GUIDE
1. Formulate the problem
2. Introduce slack variables
3. Form initial tableau
4. Obtain new tableaux
5. Get the solution
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
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
STEP BY STEP GUIDE
1. Formulate the problem
2. Introduce slack variables
3. Form initial tableau
4. Obtain new tableaux
5. Get the solution
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
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
s
t
Z
d
c
s
t
VALUE
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
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
FORMING THE INITIAL TABLEAU
BASIC
VARIABLES
d
c
s
t
VALUE
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
This is the objective row
STEP BY STEP GUIDE
1.
Formulate the problem
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
(i)
Identify pivotal column
(ii)
Find θ-values
(iii)
Identify pivotal row
(iv)
Identify pivot
(v)
Pivot
3.
Get the solution
PIVOTAL COLUMN
• 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.
BASIC
VARIABLES
d
c
s
t
VALUE
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
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.
STEP BY STEP GUIDE
1.
Formulate the problem
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
(i)
Identify pivotal column
(ii)
Find θ-values
(iii)
Identify pivotal row
(iv)
Identify pivot
(v)
Pivot
3.
Get the solution
FINDING θ-VALUES
• 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.
BASIC
VARIABLES
d
c
s
t
VALUE
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
STEP BY STEP GUIDE
1.
Formulate the problem
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
(i)
Identify pivotal column
(ii)
Find θ-values
(iii)
Identify pivotal row
(iv)
Identify pivot
(v)
Pivot
5.
Get the solution
PIVOTAL ROW
•
For row (i)


•
For row (ii)

=
16
2

8
9
1

9
• The row with the smallest θ-value is called the
pivotal row.
• Here the pivotal row is row (i)
STEP BY STEP GUIDE
1.
Formulate the problem
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
(i)
Identify pivotal column
(ii)
Find θ-values
(iii)
Identify pivotal row
(iv)
Identify pivot
(v)
Pivot
5.
Get the solution
THE PIVOT
The pivot!
The pivotal row
BASIC
VARIABLES
d
c
s
t
VALUE
s
2
1
0
16
t
1
-16
1
1
0
0
1
0
9
0
Z
-14
The pivotal column
STEP BY STEP GUIDE
1.
Formulate the problem
2.
Introduce slack variables
3.
Form initial tableau
4.
Obtain new tableaux
(i)
Identify pivotal column
(ii)
Find θ-values
(iii)
Identify pivotal row
(iv)
Identify pivot
(v)
Pivot
5.
Get a solution
PIVOTING
1. 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.
Step 1
- Replace the leaving variable with the entering variable.
BASIC
VARIABLES
d
c
s
t
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
VALUE
Step 2 - Divide all entries in the pivotal row by the pivot. The pivot
becomes 1.
BASIC
VARIABLES
d
c
s
t
VALUE
sd
1
1/2
1/2
0
8
t
Z
PIVOTING
BASIC
VARIABLES
d
c
s
t
VALUE
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
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
0
1/2
-1/2
1
1
PIVOTING
BASIC
VARIABLES
d
c
s
t
VALUE
d
1
1/2
1/2
0
8
t
0
1/2
-1/2
1
1
Z
PIVOTING
BASIC
VARIABLES
d
c
s
t
VALUE
s
2
1
1
0
16
t
1
-16
1
-14
0
0
1
0
9
0
Z
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
0
-6
8
0
128
BASIC
VARIABLES
d
c
s
t
VALUE
d
1
1/2
1/2
0
8
t
Z
0
0
1/2
-6
-1/2
8
1
0
1
128
This is our second tableau
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.
OPTIMAL TABLEAU
BASIC
VARIABLES
d
c
s
t
VALUE
d
1
0
1
-1
7
c
0
1
-1
2
2
Z
0
0
2
12
140
• Note there are no negative entries in the
objective row.
• Can you see the solution?
STEP BY STEP GUIDE
1. Formulate the problem
2. Introduce slack variables
3. Form initial tableau
4. Obtain new tableaux
5. Get the solution
OBTAINING THE SOLUTION
BASIC
VARIABLES
d
c
s
t
VALUE
d
c
1
0
1
-1
7
0
1
-1
2
Z
0
0
2
12
2
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
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.