# Answer in Operations Research for mekdi #249770

1.     A firm manufactures two products; the net profit on product 1 is Rupees 3 per unit and Rupees 5 per unit on product 2. The manufacturing process is such that each product has to be processed in two departments D1 and D2. Each unit of product1 requires processing for 1 minute at D1 and 3 minutes at D2; each unit of product 2 requires processing for 2 minutes at D1 and 2 minutes at D2. Machine time available per day is 860 minutes at D1 and 1200 minutes at D2. How much of product 1 and 2 should be produced every day so that total profit is maximum. Make the mathematical model for the given problem.

Let “x_1” and “x_2” be levels of production of two products, then

“z(x_1,x_2)=3cdot x_1+5cdot x_2”  is profit function and it should be maximized so “3cdot x_1+5cdot x_2rarr max”

For manufacturing  “x_1”unis of product 1 and “x_2” units of product 2 we need in

using “x_1+2cdot x_2” minutes on depatment D1 and “3cdot x_1+2cdot x_2” minutes on department D2 that have availiable machine time 800 and 1200 minutes per day correspondingly therefore we have two restrictions for the plan

“x_1+2cdot x_2le800;\n3cdot x_1+2cdot x_2le1200;\”

We also must take in account positiveness restrictions :

“x_1ge0,x_2ge0”

Thus we have math model of production:

“3cdot x_1+5cdot x_2rarr max”

“x_1+2cdot x_2le800;\n3cdot x_1+2cdot x_2le1200;\”

“x_1ge0,x_2ge0”

Standard form convenient for simplex method is

z->max;

“z-3cdot x_1-5cdot x_2=0;”

“x_1+2cdot x_2+1cdot t+0cdot s=800;\n3cdot x_1+2cdot x_2+0cdot t+1cdot s=1200;\nx_1,x_2,t,sge0”

To solve we use simplex method:

“defarraystretch{1.5}n begin{array}{c:c:c:c:c:c}n z & x_1 & x_2 &t&s&b\ hlinen 1 & -3 & -5 &0&0&0\n hdashlinen 0 & 1 & ! 2&1&0&800\n hdashlinen 0 & 3 & 2&0&1&1200\nend{array}”

“defarraystretch{1.5}n begin{array}{c:c:c:c:c:c}n z & x_1 & x_2 &t&s&b\ hlinen 1 & -1/2 & 0 &5/2&0&2000\n hdashlinen 0 & 1/2 & 1&1/2&0&400\n hdashlinen 0 & ! 2 & 0&-1&1&400\nend{array}”

“defarraystretch{1.5}n begin{array}{c:c:c:c:c:c}n z & x_1 & x_2 &t&s&b\ hlinen 1 & 0 & 0 &9/4&1&2100\n hdashlinen 0 & 0 & 1&3/4&-1&300\n hdashlinen 0 & 1 & 0&-1/2&1/2&200\nend{array}”

“x_1=200,x_2=300,z_{max}=2100”

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