Theorem 3.6 Every basic feasible solution to a LPP corresponds to a vertex of the set of feasible solution
2.10 Duality in LPP
preparing the simplex table and solving,
Cj → -5 1 0 -M 0 0
↓ Var→
Basis
↓
x1 x2 s1 A1 s2 s3 RHS R.R
-M A1 1 1 -1 1 0 0 2 2
0 s2 1 2 0 0 1 0 2 2
0 s3 2 1 0 0 0 1 2 1→
N.E.R
=Cj-Zj
5+M
↑
-1+M -M 0 0 0 Z=-2M
-M A1 0 1/2 -1 1 0 -1/2 1 2
0 s2 0 3/2 0 0 1 -1/2 1 2/3
→
5 x1 1 1/2 0 0 0 1/2 1 2
N.E.R
=Cj-Zj
0 (M-7)/2
↑
-M 0 0 -(M+5)/2 Z= -M+5
-M A1 0 0 -1 1 -1/3 -1/3 2/3
-1 x2 0 1 0 0 2/3 -1/3 2/3
5 x1 1 0 0 0 -1/3 2/3 2/3
N.E.R
=Cj-Zj
0 0 -M 0 -(-7+M)/3 -(M+11)/3 Z =
(8-2M)/3
As all NER entries are ≤ 0, the optimality criteria is satisfied, but the Z value in the final table contains the coefficient of the artificial variable. Hence, it is a case of no feasible solution.
program also gives the solution to its dual. Thus, whenever a linear program is solved by the simplex method, we are actually getting solutions for two linear programming problems. The original problem is called the primal problem.
Although the idea of duality is essentially mathematical, we shall see that duality has important interpretations which can help managers to answer questions about alternative courses of action and their relative values.
Let us understand the concepts and the managerial significance of duality with the help of the following example.
Example 2.24: ABC Company makes three products, T, C and B, which must be processed through the Assembly, Finishing and the Packaging departments. The three departments have maximum 60, 40 and 80 hours available. The profit on one unit of each of the products is Rs. 2 per T, Rs. 4 per C and Rs. 3 per B. The other data is given below.
Hours required for 1 unit of product
T C B Assembly 3 4 2 Finishing 2 1 2 Packaging 1 3 2
The problem can be formulated as:
Maximize: 2T + 4C + 3B subject to the constraints :
3T + 4C + 2B ≤ 60 2T + 1C + 2B ≤ 40 1T + 3C + 2B ≤ 80 and T, C and B ≥ 0
Let the slack variables SA, SF and SP be the unused hours in the three departments. So the above L.P becomes Maximize: 2T + 4C + 3B + 0SA + 0SF + 0SP
subject to the constraints :
3T + 4C +2B + SA = 60 2T + 1C + 2B + SF = 40
1T + 3C + 2B + SP = 80 and T, C and B ≥ 0
The following table gives the simplex solution of the above problem.
cj 2 4 3 0 0 0 RR
cB B xB T C B SA SF SP
0 SA 60 3 4 2 1 0 0 15 →
0 SF 40 2 1 2 0 1 0 40
0 Sp 80 1 3 2 0 0 1 80/3
z 0 cj - zj 2 4
↑
3 0 0 0
4 C 15 3/4 1 1/2 1/4 0 0 30
0 SF 25 5/4 0 3/2 -1/4 1 0 50/3 →
0 Sp 35 -5/4 0 1/2 -3/4 0 1 70
z 60 cj - zj -1 0 1
↑
-1 0 0
4 C 20/3 1/3 1 0 1/3 -1/3 0
3 B 50/3 5/6 0 1 -1/6 2/3 0
0 Sp 80/3 -5/3 0 0 -2/3 -1/3 1 z 228/3 cj - zj -11/6 0 0 -5/6 -2/3 0
This being a maximization problem, the optimality criteria is satisfied as all N.E.R entries, that is all cj - zj entries are ≤ 0.
Recall that
(a) Each positive number in the cj - zj row represents the net profit obtainable if 1 unit of the variable heading that column were added to the solution.
(b) Each negative number (a net loss) in the cj - zj row indicates the decrease in the total profit if 1 unit of the variable heading that column were added to the product mix. A negative number in the cj - zj row under one of the columns representing time has another interpretation also. A negative number here represents the amount of increase in total profit if the number of hours available in that department could be increased by 1.
We see from the table that the optimal solution is to produce 20/3 units of C, 50/3 units of B and no unit of T. The total contribution for this product mix is about Rs. 76.67. The values under the SA, SF and SP columns in the cj - zj row indicate that to remove 1 productive hour from each of the three departments would reduce the total contribution respectively, by Rs. 5/6, 2/3 and 0. This can be taken to mean also that if additional capital were available to expand productive time in these departments, the value of increased production to ABC Company, of 1 more hour in each of these departments would be Rs. 5/6, 2/3 and 0. i.e. adding another hour of Assembly time will increase profit by Rs. 5/6, adding another hour of Finishing will increase profit by Rs. 2/3 and adding another hour of packaging time will leave profit unchanged. These three values 5/6, 2/3 and 0 are called dual prices, shadow prices, or simply unit worth of a resource. To be more specific, if adding another hour to each department costs the same, we would add the time to the Assembly department, for there it is worth Rs. 5/6, which is more than 2/3 or 0.
This primal was concerned with maximizing the contribution from the three products; the dual will be concerned with evaluating the time used in the three departments to produce T, C and B.
The production manager of the ABC Company recognizes that the productive capacity of the three departments is a valuable resource to the firm; he wonders whether it would be possible to place a monetary value on its worth. He soon comes to think in terms of how much he would receive from another manufacturer, a renter who wants to rent all the capacity in ABC Company’s three departments. He reasons along the following lines.
Suppose the rental charges were Rs. A per hour of Assembly time, Rs. F per hour of Finishing and Rs. P per hour of Packaging time, then the cost to the renter of all the time would be
Total rent paid = 60A + 40F + 80P
and of course the renter would want to set the rental prices in such a way as to Minimize the total rent he would have to pay; so the objective of the dual is
Minimize : 60A + 40F + 80P
The production manager of ABC Company will not rent out his time unless the rent offered enables him to net as much as he would if he used the time to produce products T, C and B for ABC Company. This observation leads to the constraints of the dual.
To make one unit of T requires 3 Assembly hours, 2 Finishing hours and 1 packaging hour. The time that goes into making one unit of T would be rented out for Rs. 3A + 2F + 1P. If the manager used all that time to make T, he would earn Rs. 2 in contribution to profit., and so he will not rent out the time unless
3A + 2F + 1P ≥ 2
and this gives the first constraint in the dual. Similar reasoning with respect to C and B gives the other two dual constraints
4A + 1F + 3P ≥ 4 2A + 2F + 2P ≥ 3
So the entire dual problem which determines for the manager of the ABC Company, the value of the productive resources of the company (its plant hours) is:
Minimize 60A + 40F + 80P subject to the constraints :
3A + 2F + 1P ≥ 2 4A + 1F + 3P ≥ 4 2A + 2F + 2P ≥ 3 and A, F, and P ≥ 0
We add appropriate surplus and artificial variables as follows and then solve the dual problem. Only the initial and the final tables are shown below.
Minimize 60A + 40F + 80P subject to the constraints :
3A + 2F + 1P - S1 + A1 = 2 4A + 1F + 3P - S2 + A2 = 4 2A + 2F + 2P - S3 + A3 = 3 with A, F, and P ≥ 0
cj 60 40 80 0 0 0 M M M
cB B xB A F P S1 S2 S3 A1 A2 A3 RR
M A1 2 3 2 1 -1 0 0 1 0 0 2/3 →
M A2 4 4 1 3 0 -1 0 0 1 0 1
M A3 3 2 2 2 0 0 -1 0 0 1 3/2
z 9M cj-zj 60-9M
↑
40-5M 80-6M M M M 0 0 0
: : : : : : : : : : :
: : : : : : : : : : :
: : : : : : : : : : :
60 A 5/6 1 0 2/3 0 -1/3 1/6 0 1/3 -1/6
0 S1 11/6 0 0 5/3 1 -1/3 -5/6 -1 1/3 5/6
40 F 2/3 0 1 1/3 0 1/3 -2/3 0 -1/3 2/3
z 228/3 cj-zj 0 0 80/3 0 20/3 50/3 M M-20/3 M-50/3
This being a minimization problem, the optimality criteria is satisfied as all N.E.R entries, that is all cj-zj row entries are ≥ 0.
The optimum solution to the dual problem indicates that the worth to ABC Company, of 1 productive hour in the Assembly department is Rs. 5/6 (A = 5/6 in the final table), in the Finishing department is Rs. 2/3 and in the packaging department is Rs. 0 (T is not in the basis).
Of course, these are the same values we got by looking at the cj-zj row in the final table of the primal problem. Thus, when we solved the primal, we also got the solution to the dual. Does solving the dual also give us the solution to the primal? Yes, if we look at the values contained under the S1, S2 and S3 columns in the cj-zj row in the final table of solution of the dual, we find 0, 20/3 and 50/3, which are the optimal values for T, C and B in the primal.
Now look at the two problem formulations again.
Primal Problem
Maximize: 2T + 4C + 3B Subject to:
3T + 4C +2B ≤ 60 2T + 1C + 2B ≤ 40 1T + 3C + 2B ≤ 80 and T, C and B ≥ 0
Dual problem
Minimize: 60A + 40F + 80P Subject to :
3A + 2F + 1P ≥ 2 4A + 1F + 3P ≥ 4 2A + 2F + 2P ≥ 3 and A, F, and P ≥ 0
Some direct observations from the table above:
1. The objective function coefficients of the primal problem have become the right hand side constants of the dual. Similarly, the right hand side constants of the primal have become the objective function coefficients of the dual.
2. The inequalities have been reversed in the constraints.
3. The objective is changed from maximization in primal to minimization in dual.
4. Each column in primal corresponds to a constraint (row) in a dual. Thus, the number of dual constraints is equal to the number of primal variables.
5. Each constraint (row) in the primal corresponds to a column in the dual. Hence, there is one dual variable for every primal constraint.
6. The transpose of the technological (input-output) coefficient matrix of the primal becomes the technological (input-output) coefficient matrix of the dual.
Further economic interpretations of duality:
1. Suppose T, C, B is a feasible solution to the primal ( a level of output that can be achieved with the current resources) and A, F, P is a feasible solution of the dual
(a set of rents which would induce the manager to rent out the plant rather than to use it himself). Then, 2T + 4C + 3B ≤ 60A + 40F + 80P
2. The optimal values are the same in both problems. This is always the case. It means that the value to ABC Company, of all its productive resources is precisely equal to the profit the firm can make if it employs these resources in the best possible way. In this way, the profit made on the firm’s output is used to derive the imputed values of the inputs to produce that output.
3. In the above problem, the dual variable P was equal to 0 and not all the packaging time was used. This is entirely reasonable, since if the company already has excess packaging time, additional packaging time cannot be profitably used and so it worthless. This is half of what is called the principle of complimentary slackness.
4. A = 5/6, F = 2/3 and P = 0, 3A+2F+1P = 23/6. This shows us that the value of the time needed to produce one unit of T is Rs. 23/6. But one unit of T contributes a profit of Rs. 2. Since the time needed to produce T is worth more than the return on it, the optimal solution to the primal does not produce any T. This is the other half of the principle of complimentary slackness.
Now we look at the concept of duality by introducing mathematical rigour.
To every LPP, there is an associated LPP called dual of the given LPP. The given problem is called primal problem; i.e. every LPP when expressed in its standard form has associated unique LPP based on the same data.
These primal – dual pair are inter-related and variables of this pair have interesting implications in econometrics, production engineering etc. Let us define dual of
Primal : Maximize z = cTx, x є Rn
subject to the constraints : Ax ≤ b, x ≥ 0, b є Rm, A is m × n . Dual : Minimize z* = bTw, w є Rm
subject to the constraints : AT w ≥ c, w ≥ 0.
Note : In order to convert any problem into its dual, the primal LPP must be expressed in the maximization form with all constraints in ≤ type or = type. Thus, primal – dual pairs are as follows :
Primal problem Dual problem
1 Maximize z = cTx subject to the constraints : Ax ≤ b, x ≥ 0
Minimize z* = bTw subject to the constraints : AT w ≥ c, w ≥ 0
2 Maximize z = cTx subject to the constraints : Ax = b, x ≥ 0
Minimize z* = bTw subject to the constraints : AT w ≥ c, w is unrestricted.
3 Minimize z = cTx
subject to the constraints : Ax = b, x ≥ 0
Maximize z* = bTw subject to the constraints : AT w ≤ c, w is unrestricted.
4 Minimize (maximize) z = cTx subject to the constraints : Ax = b, x is unrestricted
Maximize (minimize) z* = bTw subject to the constraints : AT w = c, w is unrestricted.
Given an LPP, it can be directly converted into its dual using the following table :
Primal Problem Dual Problem
1 Maximization with constraints ≤ or = Minimization with constraints ≥ or = 2 No. of constraints No. of variables
3 Coefficients of the objective function RHS of constraints 4 Input – output matrix A Input – output matrix AT
5 j – th constraint of = type j – th variable unrestricted in sign 6 k – th variable unrestricted in sign k – th constraint of = type
Example 2.25 Write dual of LPP : Maximize z = 8x1 + 6x2
subject to the constraints : x1 – x2 ≤ 3/5, x1 – x2 ≥ 2, x1, x2 ≥ 0
Solution : To write given LPP in maximization form with all constraints ≤ - type; So primal is Maximize z = 8x1 + 6x2
subject to the constraints : x1 – x2 ≤ 3/5, - x1 + x2 ≤ -2, x1, x2 ≥ 0 Let w1 and w2 be dual variables. Then dual problem is
Minimize z* = 3/5w1 – 2w2
subject to the constraints : w1 – w2 ≥ 8, - w1 + w2 ≥ 6, w1, w2 ≥ 0 Example 2.26 Write dual of LPP :
Minimize z = 4x1 + 6x2 + 18x3
subject to the constraints : x1 + 3x2 ≥ 3, x2 + 2x3 ≥ 5, x1, x2 , x3 ≥ 0 Solution : Primal is
Minimize z = 4x1 + 6x2 + 18x3
subject to the constraints : x1 + 3x2 + 0x3 ≥ 3, 0x1 + x2 + 2x3 ≥ 5, x1, x2, x3 ≥ 0
Let w1 and w2 be dual variables corresponding to each of the primal constraint. Then dual problem is Maximize z* = 3w1 + 5w2
subject to the constraints : w1 + 0w2 ≤ 4, 3w1 + w2 ≤ 6, 0w1 + 2w2 ≤ 18. Rewriting the dual problem as Maximize z* = 3w1 + 5w2
subject to the constraints : w1 ≤ 4, 3w1 + w2 ≤ 6, 2w2 ≤ 18, w1, w2 ≥ 0,
Example 2.27 Write dual of LPP : Minimize z = 7x1 + 3x2 + 8x3
subject to the constraints : 8x1 + 2x2 + x3 ≥ 3, 3x1 + 6x2 + 4x3 ≥ 4, 4x1 + x2 + 5x3 ≥ 1, x1 + 5x2 + 2x3 ≥ 7, x1, x2 , x3 ≥ 0
Solution : Primal is
Minimize z = 7x1 + 3x2 + 8x3
subject to the constraints : 8x1 + 2x2 + x3 ≥ 3, 3x1 + 6x2 + 4x3 ≥ 4, 4x1 + x2 + 5x3 ≥ 1, x1 + 5x2 + 2x3 ≥ 7, x1, x2, x3 ≥ 0
Let w1, w2, w3 and w4 be dual variables corresponding to each of the primal constraint. Then dual problem is Maximize z = 3w1 + 4w2 + x3 + 7w4
subject to the constraints : 8w1 + 3w2 + 4w3 + w4 ≤ 7, 2w1 + 6w2 + w3 + 5w4 ≤ 3, w1 + 4w2 + 5w3 + 2w4 ≤ 8, w1, w2, w3, w4 ≥ 0 .
Example 2.28 Write dual of LPP : Maximize z = 3x1 + x2 + x3 – x4
subject to the constraints : x1 + 5x2 + 3x3 + 4x4 ≤ 4, x1 + x2 = - 1, x3 – x4 ≤ - 5, x1, x2 , x3, x4 ≥ 0 Solution : Primal as
Maximize z = 3x1 + x2 + x3 – x4
subject to the constraints : x1 + 5x2 + 3x3 + 4x4 ≤ 4, - x1 - x2 = 1, x3 – x4 ≤ -5, x1, x2, x3, x4 ≥ 0
Let w1, w2, and w3 be dual variables corresponding to each of the primal constraint. Then dual problem is Minimize z* = 4w1 + w2 – 5w3
subject to the constraints : w1 - w2≥ 3, 5w1 - w2≥ 1, 3w1 + w3 ≥ 1, 4w1 - w2 ≥ -1, and w1, w3 ≥ 0, and w2
unrestricted.
Example 2.29 Obtain dual of LPP : Minimize z = x1 - 3x2 - 2x3
subject to the constraints : 3x1 - x2 + 2x3 ≤ 7, 2x1 - 4x2 ≥ 12, - 4x1 + 3x2 + 8x3 = 10, x1, x2 ≥ 0 and x3 is unrestricted.
Solution : Putting x3 = x3′- x3″, we have primal as Minimize z = x1 - 3x2 – 2(x3′- x3″)
subject to the constraints : -3x1 + x2 - 2(x3′- x3″) ≥ - 7, 2x1 - 4x2 ≥ 12, - 4x1 + 3x2 + 8(x3′- x3″)= 10; x1, x2, x3′, x3″≥ 0
Also as the third constraint is an equality, we convert it into inequalities as follows : -4x1 + 3x2 + 8(x3′- x3″) ≤ 10 and -4x1 + 3x2 + 8(x3′- x3″) ≥ 10
Rewriting primal problem as minimization problem with all constraints ≥ - type : Minimize z = x1 - 3x2 – 2(x3′- x3″)
subject to the constraints : -3x1 + x2 - 2(x3′- x3″) ≥ - 7, 2x1 - 4x2 ≥ 12, 4x1 - 3x2 - 8(x3′- x3″)≥ - 10, -4x1 + 3x2 + 8(x3′- x3″) ≥ 10; x1, x2, x3′, x3″≥ 0
Let w1, w2, w3 and w4 be dual variables corresponding to each of the primal constraint. Then dual problem is Maximize z* = - 7w1 + 12w2 - 10w3 + 10w4
subject to the constraints : - 3w1 + 2w2 + 4w3 - 4w4 ≤ 1, w1 – 4w2 - 3w3 + 3w4 ≤ - 3, -2w1 - 8w3 + 8w4 ≤ - 2, 2w1 + 8w3 – 8w4 ≤ 2, w1, w2 , w3, w4 ≥ 0.
The third and the fourth constraints can be written as 2w1 + 8w3 – 8w4 = 2. From the objective function and the constraints, w3 and w4 can be put together by writing w = w3 – w4. So w becomes unrestricted in sign. Rewriting the dual problem as
Maximize z* = - 7w1 + 12w2 - 10w
subject to the constraints : - 3w1 + 2w2 + 4w ≤ 1, w1 – 4w2 - 3w ≤ - 3, 2w1 + 8w = 2, w1 ≥ 0, w2 ≥ 0 and w is unrestricted.
Instead of working out the dual in the above manner, the following way can also be applied by using the above table given.
Primal (minimize with ≥ ) Dual Minimize z = x1 - 3x2 - 2x3
subject to the constraints : -3x1 + x2 - 2x3 ≥ -7, 2x1 - 4x2 ≥ 12, 4x1 - 3x2 - 8x3 = -10,
x1, x2 ≥ 0 and x3 is unrestricted.
Maximize z* = - 7w1 + 12w2 - 10w subject to the constraints :
-3w1 + 2w2 + 4w ≤ 1, w1 – 4w2 - 3w ≤ - 3, -2w1 - 8w = -2,
w1 ≥ 0, w2 ≥ 0 and w is unrestricted
2.10.1 Duality Theorems :