CSC 445/545 Notes 3.5: Economic Interpretation of Dual Variables

Primal:	Dual:
Max cT x	Min bT y
subject to	subject to
A x <= b	AT y >= c
x >= 0	y >= 0

Manufacturing Problems

Suppose that:

• The problem is to maximize profit.
• Each x_j measures the level of output of the jth product.
• Each b_i specifies available amount of ith resource.
• Each a_ij is the units of resource i required for product j.
• Each c_j is profit in dollars per unit of product j.

How should the units of Yi be interpreted in order for the equations to make sense?

A1i* Y1  +  A2i*Y2  +  ...  + Ami*Ym   >= Ci

[Units res. 1 for product i] [ ? Y1 ] + 
[Units res. 2 for product i] [ ? Y2 ] + 
...
[Units res. m for product i] [ ? Ym ]  
>= [ Dollars per unit of product i]

To make this work, we need:

[Units res. 1/prod. i][Dollars/Unit res. 1]+ 
[Units res. 2/prod. i][Dollars/Unit res. 2]+ 
...
[Units res. m/prod. i][Dollars/Unit res. m]  
>=
[ Dollars / prod. i]

Let Y^* be the optimal solution to the dual.
With each extra unit of resource i, the profit increases by Yi^* dollars.
So, Yi^* represents the extra amount the firm should be willing to pay over and above the current trading price for each extra unit of resource i.

Definitions

Yi^* is called the marginal value of the ith resource. Marginal- difference between trading price and actual worth.
Yi^* is also called the shadow price of the ith resource.

Forestry Example: A forester has 100 acres of land and $4000 in capital and would like to plant X1 acres of hardwood and X2 acres of pine to maximize profit where

Wood type	Cost to harvest	Selling price	Profit
Hardwood	10	50	40
Pine	50	120	70

Product 1 : Hardwood.
Product 2 : Pine.
Resource 1: Acres of land.
Resource 2: Dollars. 

The linear programming problem is:

Maximize 40 X1  + 70 X2 

subject to

   X1    +    X2   <= 100 
10 X1    + 50 X2   <= 4000 

X1, X2 >= 0

Looking at the Units:

hard= acres of hardwood, pine= acres of pine.

Maximize 
(profit/hard)*(hard)+(profit/pine)(pine)
=profit

subject to

(acres/hard)  (hard) +   (acres/pine)(pine) 
    <= acres

(dollars/hard)(hard) + (dollars/pine)(pine) 
    <= dollars

The dual problem:

Minimize 100 Y1 + 4000 Y2

subject to
   
     y1 + 10 Y2  >=  40
     Y1 + 50 Y2  >=  70
     
Y1, Y2 >= 0

Looking at the units:

Minimize (acres)(?Y1)+ (dollars)(?Y2)

subject to
   
(acres/hard)(?Y1)+ (dollars/hard)(? Y2)
     >=(profit/hard)

(acres/pine)(?Y1)+ (dollars/pine)(? Y2)
     >=(profit/pine)

So the units are
Y1 : profit/acre
Y2 : profit/dollar

The final dictionary for this problem is:

X1 =    25.00 - 1.25 X3 + 0.02 X4 
X2 =    75.00 + 0.25 X3 - 0.02 X4 
----------------------------------
z  =  6250.00 -32.50 X3 - 0.75 X4 

The dual solution read from this is:
Y1= 32.5, Y2= 0.75

Conclusion:

• One extra acre would result in $32.50 extra profit.
• One extra dollar in capital results in 0.75 profit.
• Renting one acre at less than $32.50 results in more profit.
• Borrowing money at less than 0.75 interest rate results in more profit.

Notes 3.4 already covered the update formulas for this problem when there is a change to the bi's.