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 USX Steel expects to faces the following demand during the next 4 months: 300 tons; 150 tons: 250 tons; and 100 tons, respectively

Accounting Nov 16, 2020

 USX Steel expects to faces the following demand during the next 4 months: 300 tons; 150 tons: 250 tons; and 100 tons, respectively. At the beginning of month 1, USX has 12 workers. During any month, a worker at USX can produce up to 20 tons of steel. Each worker is paid $6500 per month. At the start of each month, a worker can be hired at the cost of $4300 or fired at the cost of $5400. Demand may be backlogged at a cost of $110 per month per ton of steel; and the cost of holding a ton of steel in inventory for one month is $120. The raw material used to produce a ton of steel costs $510. All demand must be met at the end of month 4. Formulate an LP to help USX Steel minimize the total costs during the next 4 months. A. Define the decision Variables B. Write out the objective function C. Write out all the constraints

Expert Solution

ANSWER

A.

Decision variables:

Hi be the number of workers to be hired at the start of month i

Fi be the number of workers to be fired at the start of month i

Wi be the number of workers available at the start of month i

Pi be the tons of steel produced in month i

Vi be the inventory at the end of month i

Bi be the backlog at the end of month i

-------------------------------------------

B.

Objective function:

Min 6500W1+6500W2+6500W3+6500W4

+4300H1+4300H2+4300H3+4300H4

+5400F1+5400F2+5400F3+5400F4

+510P1+510P2+510P3+510P4

+120V1+120V2+120V3+120V4

+110B1+110B2+110B3+110B4 (Total cost over the four months)

-------------------------------------------

C.

Constraints:

W1-H1+F1 = 12   (number of workers in month 1)

W2-H2+F2-W1 = 0   (number of workers in month 2)

W3-H3+F3-W2 = 0   (number of workers in month 3)

W4-H4+F4-W3 = 0   (number of workers in month 4)

P1-20W1 <= 0 (production capacity in month 1)

P2-20W2 <= 0 (production capacity in month 2)

P3-20W3 <= 0 (production capacity in month 3)

P4-20W4 <= 0 (production capacity in month 4)

P1-V1+B1 = 300 (demand of month 1)

P2-V2+B2+V1-B1 = 150 (demand of month 2)

P3-V3+B3+V2-B2 = 250 (demand of month 3)

P4-V4+B4+V3-B3 = 100 (demand of month 4)

B4 = 0   (all demand must be met at the end of month 4, which means backlog = 0)

All variables >= 0

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