# Setup a network Flow for this Problem using nodes and links but do not solve for optimal solution

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A manufacturer must produce a certain product in sufficient quantity to meet contracted sales in the next four months. The production facilities available for this product are limited, but by different amounts in the respective months. The unit cost of production also varies according to the facilities and personnel available. The product can be produced in one month and then held for sale in a later month, but at an estimated storage cost of $1 per unit per month. No storage cost is incurred for goods sold in the same month in which they are produced. There is presently no inventory of this product, and none is desired at the end of the four months. Pertinent data are given in the table below.

Month.....Contracted Sales.....Max Production.....Unit Cost of Production.....Unit Storage Cost Per Month

1.................20................................40.............................14............................................1

2.................30................................50.............................16............................................1

3.................50................................30.............................15............................................1

4.................40................................50.............................17............................................1

The approach is to create a node for each month, plus a supply node from which the production comes, and a demand node which receives the contracted sales. Lower, upper bounds and cost should be given to each link. An artificial back link from the demand node to the supply node is needed for a fully circulating network. Links connecting the nodes for each month represent carryover inventory.

Setup a network Flow for this Problem using nodes and links but do not solve for optimal solution.

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The solution assists with setting up a network flow for the problem using nodes and links but not solving for the optimal solution.

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