Operations Research

Operations Research Assignment

Group No. 4 Batch 2010 - 13Group Member Name Roll No.BIJOY SAHAHIRAK GHOSHSANKARSHAN SENGUPTASIDHARTH JAISWAL

Assignment: Interpret third table (starting from replacement ratio of 2nd

table) in the Simplex method solution of below problem:

• A firm is engaged in producing two products A & B.• Each unit of product A requires 2 kg of raw materials & 4 labour hours for

processing , whereas each unit of product B requires 3 kg of raw materials & 3 hours of labour, of the same type.

• Every week , the firm has an availability of 60 kg of raw materials & 96 labour hours .

• One unit of product A sold yield Rs 40 & one unit of product B sold gives Rs 35 as profit.

• Formulate this problem as a LPP to determine as to how many units of each of the products should be produced per week so that the firm can earn the maximum profit.

• Assume that there is no marketing constraint so that all that is produced can be sold

Answer:

Below is the 2nd table for the above LPP Problem:

Simplex Table 2

Operations Research

Step 3:

Simplex Table 2 (Non-optimal solution)

BasisCj 40 35 0 0 Replacement Ratio

XBi/air, air > 0

Xj X1 X2 X3 X4CB XB

C3=0 X3=12 0 3/2  1 -1/2 12/(3/2) = 8

C1=40 X1=24 1 3/4 0 1/4 24/(3/4) = 32

Zj 40 30 0 10

Δj =Cj-ZJ 0 5 0 -10  

Justification for Replacement Ratio:

Here from Table2, it can be concluded that x2 will be the entering vector as this is a maximization problem and Δj>0 only for this column. Now it needs to be determined

Which will be the leaving vector (out of x1 and x3) How much of the entering variable x2 will be used for the next step

For x2 (Product B) in row1, 1 unit of Product B requires net 3/2 kg of raw material to be used from the unused pool of raw material (x3).Hence no. of units of Product B that can be produced= No. of units of raw material available/Net Raw Material required per unit of B = 12/(3/2) = 8

For x2 (Product B) in row 2, labor being fully exhausted (as here x4=0), ¾ units of x1 is required to be given up to release sufficient labor to produce 1 unit of x2.So, no. of units of Product B that can be produced=No of units of x1 being produced/Units of x1 required to produce 1 unit of x2= 24/(3/4) = 32Out of the 2 rows, row 1 results in lesser production of x2 and is hence the limiting condition. To produce 32 units of x2 would require 32*3/2=48 kg of raw material which is not available in this case (only 12 kg are available).Hence x3 is the leaving variable and quantity of Product B in the next table would be 8 units.

Simplex Table 3

Step 3:

Operations Research

Simplex Table 2 (Optimal solution)

BasisCj 40 35 0 0

Xj X1 X2 X3 X4CB XB

C2=35 X2=8 0 1 2/3 -1/3

C1=40 X1=18 1 0 -1/2 1/2

Zj 40 35 10/3 25/3

Δj =Cj-ZJ 0 0 -10/3 -25/3

Justification for Table 3

Quantities for x1 and x2

X2 =8 has been deduced from the replacement ratio of the previous tableAs both raw material and labor hours are fully used up, so 18 units of x1 is required. This follows from below:2*18(x1) + 3*8(x2) = 604*18(x1) + 3*8(x2) = 96

Column Headed x1:• A ‘1’ corresponding to x1 in this column implies that each unit of product A

added to the current product mix would replace one unit of same product in current solution. This happens since both raw material and labor are fully exhausted (x3=x4=0) and hence there is no spare input left to produce A

• Since adding a unit of A would necessarily involve reducing a unit of same product, this would imply no change in production of B which is indicated by element 0 against x2.

Column Headed x2:• A ‘1’ corresponding to x2 in this column implies that each unit of product B

added to the current product mix would replace one unit of same product in current solution. This happens since both raw material and labor are fully exhausted (x3=x4=0) and hence there is no spare input left to produce B

• Since adding a unit of B would necessarily involve reducing a unit of same product, this would imply no change in production of A which is indicated by element 0 against x1.

Column Headed x3: Keeping in mind that both raw material and labor are completely

exhausted, to free up 1 kg of raw material for other purposes, certain quantity of x2 needs to be given up.

Operations Research

The quantity of x2 given up will result in some extra labor hours freed, which along with excess raw material may be used in some extra production of Product A.

To release 1 kg of raw material, 2/3 unit of x2 is freed which results in 2 kg raw material and 2 labor hours to be released. Out of this, 1 kg of raw material and 2 kg of labor hours is available for production of x1.

As 2 kg raw material and 4 units of labor hour are required to produce 1 unit of Product A, therefore 1 kg of raw material and 2 kg of labor hours can be used to produce an extra amount of ½ unit of Product A.

Values 2/3 and -1/2 under the x3 column reflect the above facts.

Column Headed x4: Keeping in mind that both raw material and labor are completely

exhausted, to free up 1 labor hour for other purposes, certain quantity of x1 needs to be given up

The quantity of x1 given up will result in some extra labor hours freed, which along with excess raw material may be used in some extra production of Product B(x2).

To release 1 labor hour, ½ unit of x1 is freed which results in 1 kg of raw material and 2 labor hour to be released. Out of this, 1 kg raw material and 1 labor hour is available for production of Product B (x2)

As 3 kg raw material and 3 labor hours are required to produce 1 unit of Product B, hence 1 unit of raw material and 1 labor hour can be used to produce 1/3 unit of Product B in excess.

Values of -1/3 and ½ under x4 indicate the above facts

Δj Elements:

Δj=Cj-Zj

Zj represents loss of profits that results from addition of one unit of variable heading a particular column

For Column headed x1 Addition of one unit of product A causes a reduction of one unit of same

product and a zero change in x2 Reduction in Product A results in a loss of profit of Rs 40 per unit. Hence, for x1, the value equals 40

.For Column headed x2

Addition of one unit of product B causes a reduction of one unit of same product and a zero change in x1

Reduction in Product B results in a loss of profit of Rs 35 per unit. Hence, for x2, the value equals 35

For Column headed x3 Addition of one unit of x3 results in reduction in production of B by 2/3 unit

and increase in production of A by ½ unit. Reduction by 2/3 unit of B results in loss of 2/3*35

Operations Research

Addition of ½ unit of A results in profit of ½*40 So net loss of profit for x3 equals (-1/2*40 + 2/3*35) = Rs 10/3

For Column headed x4 Addition of one unit of x4 results in reduction in production of A by 1/2 unit

and increase in production of B by 1/3 unit. Reduction by 1/2 unit of A results in loss of 1/2*40 Addition of 1/3 unit of B results in profit of 1/3*35 So net loss of profit for x4 equals (1/2*40 - 1/3*35) = Rs 25/3

• Cj value for various columns indicate profit associated with in it values of variables involved.

• The values are 40, 35, 0 and 0 respectively for x1, x2, x3 and x4.• Thus, if one unit of x1 is added to the solution, it would add Rs. 40 to the

profit whereas addition of the unit also involves a loss of profit of Rs. 40 as indicated by Zj values for this variable.

• Therefore, net change to profit on adding one unit of x1 is given by Δj=Cj-Zj = 40 – 40 = 0.

• Similarly for x2, net change to profit on adding one unit of x2 is given by Δj=Cj-Zj = 35 – 35 = 0.

• For x3, net change is Δj = 0-10/3= -10/3• For x4, net change is Δj = 0-25/3= -25/3• Thus all Δj values are either zero or negative thus giving us the optimum

feasible solution of the problem, viz. x1=18 kg and x2=8 labor hours.

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