Verify Unit Of Measure In A Multivariate Equation © Dale R. Geiger 20111

You can’t…. += ?? © Dale R. Geiger 20112

Terminal Learning Objective Task: Verify Unit Of Measure In A Multivariate Equation Condition: You are a cost advisor technician with access to all regulations/course handouts, and awareness of Operational Environment (OE)/Contemporary Operational Environment (COE) variables and actors. Standard: with at least 80% accuracy: Solve unit of measure equations Describe key cost equations © Dale R. Geiger 20113

Importance of Units of Measure You can’t add apples and oranges but you can add fruit Define the Unit of Measure for a cost expression Use algebraic rules to apply mathematical operations to various Units of Measure © Dale R. Geiger 20114

Adding If two components of the cost expression have the same unit of measure, they may be added together Example: Smoky Mountain Inn Depreciation on building$60,000 per year Maintenance person’s salary $30,000 per year Cleaning person’s salary $24,000 per year Real estate taxes $10,000 per year Depreciation, maintenance, cleaning, and taxes are all stated in $ per year, so they may be added to equal $124,000 per year © Dale R. Geiger 20115

Adding If two components of the cost expression have the same unit of measure, they may be added together Example: Smoky Mountain Inn Laundry service $4.00 per person-night Food $6.00 per person-night Laundry and food are both stated in $ per person- night, so they may be added to equal $10 per person-night © Dale R. Geiger 20116

Subtracting If two components of the cost expression have the same unit of measure, they may be subtracted Example: Selling price is $10 per widget Unit cost is $3.75 per widget Since both Selling price and Unit cost are stated in $ per widget, they may be subtracted to yield Gross Profit of $6.25 per widget © Dale R. Geiger 20117

Dividing “Per” represents a division relationship and should be expressed as such Example: Cost per unit = Total $ Cost / # Units Total Cost = $10,000 # Units = 500 $10,000/500 units = $20/unit © Dale R. Geiger 20118

Cancelling If the same Unit of Measure appears in both the numerator and denominator of a division relationship, it will cancel Example: $25 thousand 10 thousand units = $2.50/unit © Dale R. Geiger 20119

Multiplication When multiplying different units of measure, they become a new unit of measure that is the product of the two factors Example: 10 employees * 40 hrs = 400 employee-hrs 2x * 3y = 6xy © Dale R. Geiger

Cross-Cancelling When multiplying two division expressions, common Units of Measure on the diagonal will cancel Example: Variable Cost Variable cost $4/unit * 100 units = $4 * 100 units Unit 1 = $400 © Dale R. Geiger

Factoring If the same unit of measure appears as a factor in all elements in a sum, it can be factored out Example: $4/hour + $6/hour = $/hr *(4 + 6) © Dale R. Geiger

Check on Learning If two components of a cost expression have the same unit of measure, they may be either or. Which mathematical operation using two different units of measure results in a new unit of measure? © Dale R. Geiger

Proving a Unit of Measure What is the cost expression for a driving trip? The cost will be the sum of the following components: $ Gasoline + $ Insurance + $ Driver’s time © Dale R. Geiger

Variables Affecting Cost of Trip All of the following items will affect our trip’s cost: Distance in miles (represented by x) Gas usage in miles per gallon (represented by a) Cost per gallon of gas in dollars (represented by b) Insurance cost in dollars per mile (represented by c) Average speed in miles per hour (represented by d) Driver’s cost in dollars per hour (represented by e) © Dale R. Geiger

Cost of Gasoline Cost of gasoline = # gallons * $/gallon # gallons = x miles ÷ a miles/gallon When dividing fractions, invert the second fraction and multiply Cost of gasoline = x miles * gallon /a miles * b $/gallon © Dale R. Geiger

Cost of Gasoline Example Your car gets 20 miles/gallon, gas costs $4/gallon and you drive 100 miles: 100 miles * gallon/20 miles * $4/gallon (100 miles * gallon/20 miles) * $4/gallon (100 * gallon/20) * $4/gallon (100 5* gallon)/20 * $4/gallon (5* gallon)* $4/gallon 5 gallons * $4/gallon 5 * $4 = $20 © Dale R. Geiger

Cost of Insurance Cost of Insurance = Insurance $/mile * miles on trip Insurance $/mile = Total Insurance $ /year Total miles /year So: c $/mile * x miles © Dale R. Geiger

Cost of Driver’s Time Cost of Driver’s Time = # hours * $/hour # hours = x miles ÷ d miles/hour Or: x miles * hour/d miles Hours/mile * $/hour * miles on trip So: hours/d mile * e $/hour * x miles © Dale R. Geiger

Cost Expression gallon/ a miles * x miles * b $/gal + c $/mile * x miles + hours/d mile * e $/hour * x miles © Dale R. Geiger

Proving the Unit of Measure Cost of Gasoline+Cost of Insurance+Cost of Driver’s Time x miles*gallon a miles *b $ gal +c $ mile *x miles+hour d miles *e $ hour *x miles © Dale R. Geiger

Proving the Unit of Measure Cost of Gasoline+Cost of Insurance+Cost of Driver’s Time x miles*gallon a miles *b $ gal +c $ mile *x miles+hour d miles *e $ hour *x miles * ( Cost of Gasoline /mile + Cost of Insurance /mile + Cost of Driver’s Time /mile ) x miles* ( gallon a miles * b $ gal +c $ mile + hour d miles * e $ hour ) © Dale R. Geiger

Proving the Unit of Measure x miles* ( Cost of Gasoline /mile + Cost of Insurance /mile + Cost of Driver’s Time /mile ) x miles * ( gallon a miles * b $ gal +c $ mile + hour d miles * e $ hour ) x miles* ( gallon a miles * b $ gal + c $ mile + hour d miles * e $ hour ) x miles* ( b $ a miles + c $ mile + e $ d miles ) x miles* ( baba * $ mile + c * $ mile + eded * $ mile ) © Dale R. Geiger

Proving the Unit of Measure x miles* $ mile * ( Gasoline+Insurance+ Driver’s Time ) x miles* $ mile * ( baba + c + eded ) x miles 1 * $ mile * ( baba + c + eded ) $ * x * ( baba + c + eded ) x miles* ( Cost of Gasoline /mile + Cost of Insurance /mile + Cost of Driver’s Time /mile ) x miles* ( baba * $ mile + c * $ mile + eded * $ mile ) © Dale R. Geiger

Plugging Values into the Equation $ * x * ( baba + c + eded ) What is the cost of the trip if: The distance (x) is 300 miles The car gets 25 miles per gallon (b) The cost of a gallon of gas is $4 The insurance cost per mile (c) is $.05 The driver’s cost per hour is $20 (e) The average speed is 80 miles per hour (d) © Dale R. Geiger

Plugging Values into the Equation $ * x * ( baba + c + eded ) $ * 300 * ( ) What is the cost of the trip if: The distance (x) is 300 miles The car gets 25 miles per gallon (a) The cost of a gallon of gas is $4 (b) The insurance cost per mile (c) is $.05 The driver’s cost per hour is $20 (e) The average speed is 80 miles per hour (d) © Dale R. Geiger

Check on Learning What is the procedure when dividing by a fractional unit of measure? © Dale R. Geiger

The Value of Equations Equations represent cost relationships that are common to many organizations Examples: Revenue – Cost = Profit Total Cost = Fixed Cost + Variable Cost Beginning + Input – Output = Ending © Dale R. Geiger

Input-Output Equation Input-Output Equation Beginning + Input – Output = End If you take more water out of the bucket than you put in, what happens to the level in the bucket? © Dale R. Geiger

Applications of Input-Output Account Balances What are the inputs to the account in question? Raw materials? Work In process? Finished goods? Your checking account? What are the outputs from the account? © Dale R. Geiger

Applications of Input-Output Gas Mileage: Miles/Gallon = Miles Driven/Gallons Used Calculate Miles Driven using the odometer How do you know Gallons Used? If you always fill the tank completely, then: Beginning + Input – Output = Ending Or, chronologically: Beginning – Output + Input = Ending Full Tank – Gallons Used + Gallons Added = Full Tank – Gallons Used + Gallons Added = 0 Gallons Used = Gallons Added © Dale R. Geiger

Using the Input-Output Equation If any three of the four variables is known, it is possible to solve for the unknown The beginning balance on your credit card is $950. During the month you charge $300 and make a payment of $325. At the end of the month your balance is $940. What was the finance charge? What are the inputs? Charges and finance charge What are the outputs? Payments © Dale R. Geiger

Using the Input-Output Equation If any three of the four variables is known, it is possible to solve for the unknown The beginning balance on your credit card is $950. During the month you charge $300 and make a payment of $325. At the end of the month your balance is $940. What was the finance charge? What are the inputs? Charges and finance charge What are the outputs? Payments © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Using the Input-Output Equation Set up the equation: Beginning + Inputs – Outputs = Ending Beg + Charges + Finance Charges – Payments = End $950 + $300 + Finance Charge – $325 = $940 $ Finance Charge – $325 = $940 $925 + Finance Charge = $940 Finance Charge = $940 – $925 Finance Charge = $15 © Dale R. Geiger

Check on Learning What are three useful equations that represent common cost relationships? © Dale R. Geiger

Practical Exercises © Dale R. Geiger