The document discusses an optimization problem faced by a furniture company to maximize profit from chair and table production given machine time constraints. It presents the primal problem of maximizing profit and the dual problem of minimizing machine rental costs. The optimal solution from the primal problem is used to interpret the dual solution and find the shadow prices: M1 is worth Rs. 5 per hour, M2 is worth Rs. 0 per hour, and M3 is worth Rs. 2.5 per hour. The complementary slackness property and concept of binding vs. non-binding constraints is also explained.

1. EconomicInterpretationof Duality,
the concept of ShadowPrice and
the complementaryslacknesstheorem.
Compiled by
Preety Rateria (221097)
Ketan Bhasin (221064)
Nehal Khetan (220176)

2. Economic Interpretation of Dual Problem
Example:
A Company makes 2 types of furniture ̶ chairs and tables. The profit from the
products is Rs 20 per chair and Rs 30 per table. Both the products are processed
on three machines, M₁ and M₂ and M₃. The company’s objective is to maximize
profit assuming all the products produced will be sold.
Additional information regarding time required in hours for each product and
total time available in hours per week for machines is given below.
Machine Chair Table Total Available
time (in hours)
M₁ 3 3 36
M₂ 5 2 50
M₃ 2 6 60
The economic interpretation of dual is based directly upon the
interpretation of primal problem

3. Let X₁ and and X₂ be number of chairs and tables produced per week
Objective function: Max Profit = 20 X₁ + 30 X₂
Subject to constraints:
• 3 X₁ +3 X₂ ≤36
• 5 X₁ +2 X₂ ≤ 50
• 2 X₁ +6 X₂ ≤ 60
• X₁, X₂ ≥ 0
By introducing slack variables S₁, S₂ and S₃, the given problem can be restated
as:
Max Profit = 20 X₁ + 30 X₂ + 0S₁ + 0S₂ + 0S₃
Subject to
• 3 X₁ +3 X₂ + S₁ = 36
• 5 X₁ +2 X₂ + S₂ = 50
• 2 X₁ +6 X₂ + S₃ = 60
• X₁, X₂, S₁, S₂, S₃ ≥ 0

4. Initial Iteration
Basic
Variable
Cj
20 30 0 0 0 Solution
X₁ X₂ S₁ S₂ S₃ Xb
S₁ 0 3 3 1 0 0 36
S₂ 0 5 2 0 1 0 50
S₃ 0 2 6 0 0 1 60
Cj-Zj 20 30 0 0 0

5. Basic
Variable
Cj
20 30 0 0 0 Solution
X₁ X₂ S₁ S₂ S₃ Xb
X₁ 20 1 0 ½ 0 -1/4 3
S₂ 0 0 0 -13/6 1 ¾ 17
X₂ 30 0 1 -1/6 0 1/4 9
Cj-Zj 0 0 -5 0 -5/2
Final Iteration
Thus, to maximize weekly profit, 3 units of chairs and 9 units of tables have to
be produced to attain the profit of Rs 330

6. The Optimal Solution
X₁* = 3 (No. of chairs produced)
X₂* = 9 (No. of tables produced)
Profit = (3 X 20) + (9 X 30)
= Rs 330

7. Now, from the perspective of renting out all the machines, we need to find
the minimum weekly rental fee that should be charged per machine.
Assuming R to be the total weekly rental fee .
To determine R, the firm will assign values or worth to each hour of capacity
on machines M₁, M₂ and M₃.
Following assumptions are made:
Machines Worth for per
hour of capacity
(in Rs)
Hours
available per
week
Weekly worth
(in Rs)
M₁ Y₁ 36 36 Y₁
M₂ Y₂ 50 50 Y₂
M₃ Y₃ 60 60 Y₃
Dual Problem

8. Thus, the Dual optimizing equation is:
Min R = 36 Y₁ + 50 Y₂ + 60 Y₃
Constraints
Product Time worth Profit Constraint
Chair R = 3 Y₁ + 5 Y₂ + 3 Y₃ 20 3 Y₁ + 5 Y₂ + 3 Y₃ ≥ 20
Table R = 3 Y₁ + 2 Y₂ + 6 Y₃ 30 3 Y₁ + 2 Y₂ + 6 Y₃ ≥ 30
Non-negative Y₁, Y₂, Y₃ ≥ 0
Introducing slack and additional variables
3 Y₁ + 5 Y₂ + 3 Y₃ – S₁ + A₁ = 20
3 Y₁ + 2 Y₂ + 6 Y₃ – S₂ + A₂ = 30
Y₁, Y₂, Y₃, S₁, S₂, A₁, A₂ ≥ 0

9. Initial iteration
Basic
Variable
Cj
36 50 60 0 0 M M Solution
Y₁ Y₂ Y₃ S₁ S₂ A₁ A₂ Xb Ratio
A₁ M 3 2 2 -1 0 1 0 20
A₂ M 3 2 6 0 -1 0 1 30
Cj-Zj 36-6m 50-4m 60-8m m m 0 0 36-6m

10. Initial iteration
Basic
Variable
Cj
36 50 60 0 0 M M Solution
Y₁ Y₂ Y₃ S₁ S₂ A₁ A₂ Xb Ratio
A₁ M 3 2 2 -1 0 1 0 20 10
A₂ M 3 2 6 0 -1 0 1 30 5
Cj-Zj 36-6m 50-4m 60-8m m m 0 0 36-6m

11. First iteration
Basic
Variable
Cj
36 50 60 0 0 M Solution
Y₁ Y₂ Y₃ S₁ S₂ A₁ Xb Ratio
A₁ M 0 1
Y₃ 60 1/2 1/3 1 0 -1/6 0 5
Cj-Zj
New Value = Old Value - (
Corresponding key row element ∗ Corresponding key column element
Key element
)

12. First iteration
Basic
Variable
Cj
36 50 60 0 0 M Solution
Y₁ Y₂ Y₃ S₁ S₂ A₁ Xb Ratio
A₁ M 2 4/3 0 -1 1/3 1 10 5
Y₃ 60 1/2 1/3 1 0 -1/6 0 5 10
Cj-Zj 6-2M 30 -
4
3
𝑀 0 M 10-
𝑀
3
0

13. First iteration
Basic
Variable
Cj
36 50 60 0 0 M Solution
Y₁ Y₂ Y₃ S₁ S₂ A₁ Xb Ratio
A₁ M 2 4/3 0 -1 1/3 1 10 5
Y₃ 60 1/2 1/3 1 0 -1/6 0 5 10
Cj-Zj 6-2M 30 -
4
3
𝑀 0 M 10-
𝑀
3
0

14. Second iteration
Basic
Variable
Cj
36 50 60 0 0 Solution
Y₁ Y₂ Y₃ S₁ S₂ Xb Ratio
Y₁ 36 1 2/3 0 -1/2 1/6 5
Y₃ 60 0 1
Cj-Zj
New Value = Old Value - (
Corresponding key row element ∗ Corresponding key column element
Key element
)

15. Second and Final iteration
Basic Variable Cj
36 50 60 0 0 Solution
Y₁ Y₂ Y₃ S₁ S₂ Xb
Y₁ 36 1 2/3 0 -1/2 1/6 5
Y₃ 60 0 0 1 1/4 -1/4 5/2
Cj-Zj 0 26 0 3 9
Since all Cj-Zj values are zero or positive, therefore optimal solution has been
reached.

16. The optimal solution is that 1 hour of capacity on M₁ is worth Rs 5, 1 hour of
capacity on M₂ is worth NIL and 1 hour of capacity of M₃ is worth Rs 2.5
It can also be said that the shadow price of 1 hour of capacity of machines M₁,
M₂ and M₃ is Rs 5, NIL and 2.5 respectively.
Y₁* signifies an increase in the objective function by 5 if an additional hour
of M₁ is available
Y₂* signifies no change in the objective function if an additional hour of M₂
is available as it is fully utilised
Y₃* signifies an increase in the objective function by 2.5 if an additional
hour of M₃ is available

17. Therefore for the purpose of decision making, the interpretation is:
Machine Value added with
every additional
hour use in the
company
Interpretation
M₁ Rs 5 The machine will generate an additional profit of Rs 5
for every extra hour if used for manufacturing by the
company. Therefore, if the company intends to rent out
the machine, the minimum rent it should receive
should be more than Rs 5
M₂ Rs 0 The machine does not generate an additional profit for
every extra hour used for manufacturing by the
company. Therefore the company should rent out the
machine as the machine is not profitable for any
additional hour of company use.
M₃ Rs 2.5 The machine will generate an additional profit of Rs 2.5
for an extra hour of use by the company. Therefore, if
the company intends to rent out the machine, the
minimum rent it should receive should be more than
Rs 2.5

18. The existence of slack variable in the final simplex tableau indicates that the
resources have not been fully utilised.
Thus if the company plans to enlarge resources , it can look for only those resources
which are fully utilised. To access the value of additional resources, we can consider
the difference it would make if we provide an extra hour of each of the resources
fully utilised.
The increase in profit as a result of the additional unit of resource is seen to be the
marginal value of the resources and is referred to as the SHADOW PRICE for that
resource. Therefore, the maximum price that should be paid for additional hour
should not exceed the shadow price.
The shadow price for each resource is shown in the Cj-Zj row of the final simplex
tableau under the corresponding slack-variable. This value is always absolute in
nature.
Shadow Price

19. Complementary Slackness Property
Assume that the primal problem (Primal) has a solution X* and its dual
problem (Dual) has a solution Y*.
So, the property states :
1. If Xj* > 0, then the jth constraint in Dual is binding.
2. If the jth constraint in Dual is not binding, then Xj* = 0.
3. If Yi* > 0, then the ith constraint in Primal is binding.
4. If the ith constraint in Primal is not binding, then Yi* = 0.

20. With reference to the initial problem,
Since, Y₂* in Dual is 0, the 2nd constraint in Primal is not binding.
Similarly, Y₁* and Y₃* > 0 , 1st and 3rd constraints in Primal are binding.
 In the second constraint, shadow price, Y₂* is 0. So, if our optimum
solution has 5 X₁ +2 X₂ ≤ 50, we can’t get a better solution by permitting
5 X₁ +2 X₂ > 50.
 Such a constraint whose Shadow Price is 0 is called a non-binding
constraint

21. Thank You