The document provides step-by-step instructions for formulating and solving a linear programming problem to maximize profit for a toy factory. The problem is to determine the optimal number of dolls and cars to produce given constraints on machine hours and the profit earned from each item. The summary formulates the problem into a linear program with an objective function and constraints, introduces slack variables, and forms the initial simplex tableau to begin solving the problem.

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2. 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
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3. THE PROBLEM
• A small factory produces two types of toys: cars and
dolls. In the manufacturing process two machines
are used: the moulder and the assembler. A doll
needs 2 hours on the moulder and 1 hour on the
assembler. A car needs 1 hour on the moulder and 1
hour on the assembler. The moulder can be
operated for 16 hours a day and the assembler for 9
hours a day. Each doll gives a profit of Rs.16 and
each car gives a profit of Rs.14. The profit needs to
be maximised.
• How do we formulate this problem?
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4. 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
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5. PICKING OUT IMPORTANT
INFORMATION
• A small factory produces two types of toys: cars and
dolls. In the manufacturing process two machines
are used: the moulder and the assembler.
• A doll needs 2 hours on the moulder and 1 hour on
the assembler. A car needs 1 hour on the moulder
and 1 hour on the assembler.
• The moulder can be operated for 16 hours a day and
the assembler for 9 hours a day.
• Each doll gives a profit of Rs.16 and each car gives a
profit of Rs.14.
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6. PICKING OUT IMPORTANT
INFORMATION
• A small factory produces two types of toys: cars and
dolls. In the manufacturing process two machines
are used: the moulder and the assembler.
• A doll needs 2 hours on the moulder and 1 hour on
the assembler. A car needs 1 hour on the moulder
and 1 hour on the assembler.
• The moulder can be operated for 16 hours a day and
the assembler for 9 hours a day.
• Each doll gives a profit of Rs.16 and each car gives a
profit of Rs.14.
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7. • A doll needs 2 hours on the moulder
and 1 hour on the assembler. A car
needs 1 hour on the moulder and 1
hour on the assembler.
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8. PICKING OUT IMPORTANT
INFORMATION
• A small factory produces two types of toys: cars and
dolls. In the manufacturing process two machines
are used: the moulder and the assembler.
• The moulder can be operated for 16 hours a day and
the assembler for 9 hours a day.
• Each doll gives a profit of Rs.16 and each car gives a
profit of Rs.14.
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9. • A doll needs 2 hours on the moulder and 1 hour on
the assembler. A car needs 1 hour on the moulder
and 1 hour on the assembler.
• The moulder can be operated for 16 hours a day
and the assembler for 9 hours a day.
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10. 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
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11. • A doll needs 2 hours on the moulder and 1 hour on
the assembler. A car needs 1 hour on the moulder
and 1 hour on the assembler.
• The moulder can be operated for 16 hours a day
and the assembler for 9 hours a day.
• Using the decision variables
d = number of dolls
c = number of cars
make two constraints from this information.
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12. FORMING CONSTRAINT 1
THE MOULDER
• A doll needs 2 hours on the moulder and 1
hour on the assembler. A car needs 1 hour
on the moulder and 1 hour on the assembler.
• The moulder can be operated for 16 hours a
day and the assembler for 9 hours a day.
2d + c ≤ 16
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13. FORMING CONSTRAINT 2
The assembler
• A doll needs 2 hours on the moulder and 1
hour on the assembler. A car needs 1 hour
on the moulder and 1 hour on the assembler.
• The moulder can be operated for 16 hours a
day and the assembler for 9 hours a day.
d+c≤9
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14. 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
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15. PICKING OUT IMPORTANT
INFORMATION
• A small factory produces two types of toys: cars and
dolls. In the manufacturing process two machines
are used: the moulder and the assembler.
• Each doll gives a profit of Rs.16 and each car gives a
profit of Rs.14.
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16. FORMING THE OBJECTIVE
FUNCTION
• Each doll gives a profit of Rs.16 and each
car gives a profit of Rs.14.
• Let Z be the total profit; formulate the
objective function
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17. FORMING THE OBJECTIVE
FUNCTION
• Each doll gives a profit of Rs.16 and
each car gives a profit of Rs.14.
Z = 16d + 14c
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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 !
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19. STEP BY STEP GUIDE
1. Formulate the problem
2. Introduce slack variables
3. Form initial tableau
4. Obtain new tableaux
5. Get the solution
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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
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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
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22. STEP BY STEP GUIDE
1. Formulate the problem
2. Introduce slack variables
3. Form initial tableau
4. Obtain new tableaux
5. Get the solution
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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
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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
s
t
Z
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25. 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 1 0 1 9
Z -16 -14 0 0 0
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26. FORMING THE INITIAL TABLEAU
BASIC VALUE
VARIABLES
d c s t
s 2 1 1 0 16
t 1 1 0 1 9
Z -16 -14 0 0 0
This is the objective row
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27. 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
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28. 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 VALUE
VARIABLES
d c s t
s 2 1 1 0 16
t 1 1 0 1 9
Z -16 -14 0 0 0
This is the most negative coefficient with corresponding
variable d and it’s column is called the pivotal column. 29
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d
is now called the entering variable.

29. 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
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30. 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 VALUE
VARIABLES
d c s t
s 2 1 1 0 16
t 1 1 0 1 9
Z -16 -14 0 0 0
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31. 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
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32. PIVOTAL ROW
• For row (i) θ = 16
2 = 8
• For row (ii) θ = 9
1 = 9
• The row with the smallest θ-value is called
the pivotal row.
• Here the pivotal row is row (i)
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33. 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
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34. THE PIVOT
The pivot!
BASIC VALUE
VARIABLES d c s t
s 2 1 1 0 16
The pivotal row t 1 1 0 1 9
Z -16 -14 0 0 0
The pivotal column
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35. 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
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36. 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.
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37. Step 1 - Replace the leaving variable with the entering variable.
BASIC VALUE
VARIABLES
d c s t
s 2 1 1 0 16
t 1 1 0 1 9
Z -16 -14 0 0 0
Step 2 - Divide all entries in the pivotal row by the pivot. The pivot
becomes 1.
BASIC VALUE
VARIABLES
d c s t
s
d 1 1/2 1/2 0 8
t
Z
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38. PIVOTING
BASIC VALUE
VARIABLES
d c s t
s 2 1 1 0 16
t 1 1 0 1 9
Z -16 -14 0 0 0
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
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39. PIVOTING
BASIC VALUE
VARIABLES
d c s t
d 1 1/2 1/2 0 8
tt 0 1/2 -1/2 1 1
Z -16 -14 0 0 0
x16 16 8 8 0 128
0 -6 8 0 128
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40. PIVOTING
BASIC VALUE
VARIABLES
d c s t
d 1 1/2 1/2 0 8
tt 0 1/2 -1/2 1 1
Z
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41. BASIC VALUE
VARIABLES
d c s t
d 1 1/2 1/2 0 8
tt 0 1/2 -1/2 1 1
Z
Z 0 -6 8 0 128
This is our second tableau
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42. 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.
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43. BASIC
⇓ VALUE
VARIABLES
d c s t
d 1 1/2 1/2 0 8
⇒ tt 0 1/2 -1/2 1 1
Z
Z 0 -6 8 0 128
d 1 0 1 -1 7
c 0 1 -1 2 2
Z 0 0 2 12 140

44. 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?
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45. STEP BY STEP GUIDE
1. Formulate the problem
2. Introduce slack variables
3. Form initial tableau
4. Obtain new tableaux
5. Get the solution
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46. OBTAINING THE SOLUTION
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
• 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
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47. 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 Rs.140
• To make this profit the factory should
produce 7 dolls and 2 cars.
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48. PIVOTING
BASIC VALUE
VARIABLES
d c s t
s 2 1 1 0 16
t 1 1 0 1 9
Z -16 -14 0 0 0
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
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