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Example of Model Construction (1 of 4)

• Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

• Information and Data:
• Business firm makes and sells a steel product
• Product costs $5 to produce
• Product sells for $20
• Product requires 4 pounds of steel to make
• Firm has 100 pounds of steel
• Business Problem:
• Determine the number of units to produce to make the most profit, given the limited amount of steel available.

• Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

• A variable is a symbol used to represent an item that can take on any value.
• Parameters are known, constant values that are often coefficients of variables in equations.
• Data are pieces of information from the problem environment.
• A model is a functional relationship that includes variables, parameters, and equations.

• Example of Model Construction (2 of 4)

• Variables: X = # units to produce (decision variable)
• Z = total profit (in $)
• Model: Z = $20X - $5X (objective function)
• 4X = 100 lb of steel (resource constraint)
• Parameters: $20, $5, 4 lbs, 100 lbs (known values)
• Formal Specification of Model:
• maximize Z = $20X - $5X
• subject to 4X = 100

• Example of Model Construction (3 of 4)

• Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

• Example of Model Construction (4 of 4)

• Solve the constraint equation:
• 4x = 100
• (4x)/4 = (100)/4
• x = 25 units
• Substitute this value into the profit function:
• Z = $20x - $5x
• = (20)(25) – (5)(25)
• = $375
• (Produce 25 units, to yield a profit of $375)

• Model Solution:

• Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

• Model Building:
• Break-Even Analysis (1 of 9)

• Used to determine the number of units of a product to sell or produce that will equate total revenue with total cost.
• The volume at which total revenue equals total cost is called the break-even point.
• Profit at break-even point is zero.