Economic Interpretation of Duality, the concept of Shadow Price and the complementary slackness theorem.

Compiled byPreety Rateria (221097)Ketan Bhasin (221064)Nehal Khetan (220176)

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

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

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

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

The Optimal Solution

X₁* = 3 (No. of chairs produced)

X₂* = 9 (No. of tables produced)

Profit = (3 X 20) + (9 X 30) = Rs 330

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

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

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

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

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 - ()

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 - 0 M 10- 0

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 - 0 M 10- 0

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 - ()

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.

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

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

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

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.

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

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