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Situation 2 AC, a medical devices supplier, must deliver a component to three ma

ID: 3059928 • Letter: S

Question

Situation 2 AC, a medical devices supplier, must deliver a component to three manufacturing sites. AC can purchase as much as 20 tons of the component from a supplier in town A and 18 tons in town B. AC requires 14, 8, and 12 tons, at sites 1, 2, and 3, respectively. The purchase price per ton at each town and the delivering cost per ton are given in the table below. Due to recent county regulations, the total amount of component delivered from town A must be at least 60% of the total component distributed. Distribution Cost per Ton Site Town180 260 Price per ton $180 $250 $210 B $240 $260 $220 $400 5540 a) Formulate a standard form model (not the spreadsheet model) that will help AC determine how much to distribute from each town to each site to minimize the total cost of distributing the component. b) Identify and describe the decisions variables. c) Solve and explain the output [results] of the standard model using Solver.

Explanation / Answer

Let xij tons of the component be delivered from town i to site j where i= A,B and j=1,2,3.

AC can purchase as much as 20 tons of the component from a supplier in town A i.e. xA1+xA2+xA3<=20

Similarly, xB1+xB2+xB3<=18.

AC requires 14, 8, 12 tons at site 1,2,3 reapectively.

Thus, xA1+xB1=14, xA2+xB2=8, xA3+xB3=12 (Since, the total number of components delivered to site 1 is xA1+xA2 ans so on)

The total number of components distributed=xA1+xA2+xA3+xB1+xB2+xB3

The total number of components distributed from town A=xA1+xA2+xA3

Hence, we have

xA1+xA2+xA3>=(60/100)(xA1+xA2+xA3+xA1+xA2+xA3

=> 0.4(xA1+xA2+xA3)>=0.6(xA1+xA2+xA3)

Cost involved in supplying one component form town i to site j=price per ton in town i+Distribution cost per ton from town i to site j, i=A,B, j=1,2,3.

Therefore, the total cost of distribution=xA1+xA2+xA3+xB1+xB2+xB3

Thus the probem can be formulated as

Minimize c=xA1+xA2+xA3+xB1+xB2+xB3

subject to

xA1+xA2+xA3<=20

xB1+xB2+xB3<=18.

xA1+xB1=14,

xA2+xB2=8,

xA3+xB3=12

0.4(xA1+xA2+xA3)-0.6(xB1+xB2+xB3)>=0

b) The decision variables are the number of tons to be supplied from town i to site j, i=A,B j=1,2,3

c) The results are computed using the solver

xA1, xA2,....., xB3 shows the requires number of components. with total 22740 the total cost

xA1 xA2 xA3 xB1 xB2 xB3 Total Availability Decision Variables 14 6 0 0 2 12 Contribution 580 650 610 780 800 760 22740 Constraint 1 1 1 1 0 0 0 20 20 Constraint 2 0 0 0 1 1 1 14 18 Constraint 3 1 0 0 1 0 0 14 14 Constraint 4 0 1 0 0 1 0 8 8 Constraint 5 0 0 1 0 0 1 12 12 Constraint 6 0.4 0.4 0.4 -0.6 -0.6 -0.6 0 0