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Learn how to recognize and solve breakeven analysis problems to maximize profit, minimize costs, and determine breakeven values in linear and nonlinear scenarios. Gain insights into fixed and variable costs, revenue optimization, and the time value of money.
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CTC 475 Review B/C ratios Use incremental; don’t rank
CTC 475 Breakeven Analyses
Objectives Know how to recognize and solve breakeven analysis problems: • Maximize profit • Minimize costs • Maximize revenues • Determine breakeven values • Determine average costs
Fixed and Variable Costs • Fixed costs do not vary in proportion to the quantity of output: • Insurance • Building depreciation • Some utilities • Variable costs vary in proportion to quantity of output • Direct Labor • Direct Material
Fixed & Variable Costs • Fixed costs are expressed as one number • $200 • Variable costs are expressed as an amount per unit • $10 per unit
Total Costs (TC) Total Costs (TC) at a unit of production = Fixed Costs (FC) + Variable Costs (VC) * # of Units Produced
Total Costs As currently defined total costs are linear with respect to units produced
Can Decrease Total Costs by Lowering Variable Cost ($10 to $8)
Total Revenue (Linear) • Total Revenues = price (p) times number of units sold (D) • If I sell 100 units at $20 per unit then total revenue = $2000
Breakeven • Breakeven occurs at the point where TR=TC • If a company can sell more than the breakeven point then the company makes a net profit (NP) • If a company sells less than the breakeven point then the company loses money • NP=TR-TC
Breakeven Point Ways to lower the breakeven point: • Reduce fixed cost • Reduce variable cost • Increase revenue per unit
Linear Breakeven • Let D = # of Units that can be sold • TR = $5D • TC = $300 + $3.50D • Set TR=TC and solve for D to find the breakeven • D=200 units
Linear Breakeven ExampleDetermine net profit (D=1000) NP = TR-TC TR=$5*1000 = $5000 TC=$300+$3.5*1000 = $3800 NP=$1200 ($5000-$3800)
Nonlinear Breakeven Usually there is a relationship between price (p) and number of units that can be sold (D-for demand) • If price is high demand is low • If price is low demand is high
Price – Demand Relationshipp=a-b*D a-price at which demand=0 b-slope
Price-Demand Equation • Price (p) = a – b *D • Now let’s take a look at the TR equation: • TR=pD • But p=a-bD (price and demand are related) • Therefore TR=(a-bD)(D) or • TR=aD-bD2
Max. Revenue D high; p low High Sells Low revenue D low; p high Low Sells Low Revenue
Maximizing Nonlinear Revenue • TR=aD-bD2 • Take derivative of TR w/ respect to D ; set derivative to zero and solve for D • Derivative=a-2bD=0 (will give zero slope) • D=a/2b • 2nd derivative will tell you whether you have a max. (deriv. is neg) or min. (deriv. is pos)
Breakeven Example - Nonlinear • Given: • t is the number of tons sold per season • Selling Price = $800-0.8t • TC=$10,000+$400t • Maximize revenue and profit; find breakeven pts. • Calculations: • TR=Selling Price *t = $800t-0.8t2 • NP=TR-TC=-0.8t2+400t-10,000
Maximize Revenue (Calculus) • TR = $800t-0.8t2 • Set deriv = 0 and solve for t • Deriv of TR w/ respect to t =800-1.6t • t=500 tons • Substitute t into TR equation to get TR=$200,000 • Substitute t into NP equation to get NP=$-10,000 • Lost money even though revenue was maximized • Better to maximize net profit
Maximize Profit (Calculus) • NP=-0.8t2+400t-10,000 • Set deriv = 0 and solve for t • Deriv of NP w/ respect to t =-1.6t+400 • t=250 tons • Substitute t into NP equation to get NP=$40,000 • Avg profit/ton=$40,000/250tons=$160 per ton
Breakeven (Algebra) • Set TC=TR and solve for t • -0.8t2+400t-10,000=0 • Must use quadratic equation • T=26 and 474 (if you sell within this range you’ll make a net profit)
Tips to solve any type of breakeven problem • TC=FC+VC (usually linear but could possibly be nonlinear) • TR=p*D (may be linear or nonlinear) • NP=TR-TC • Breakeven pt(s) occur when TC=TR • Maximize (or minimize) nonlinear equations by finding derivative and setting equal to zero • Maximize Profit • Maximize Revenues • Minimize Costs
Time Value of Money • Most of this course is based on the fundamental concept that money has a time value • Must take this concept into account since projects have different cash flow patterns at different times
Present Economy Problems • Time is not a significant factor
Conditions: • No investment of capital • Long-term costs are the same for all alternatives • Alternatives have identical results
Example-Present Economy • A metal part can be machined on an engine lathe (one at a time) or turret lathe (many at a time) Material costs are the same regardless of the machine used. • Parts are produced in batches according to the customer’s order • Based on the following cost data, what machine should be used for order sizes of 25, 100 and 500 units?
Summary • Turret lathe costs less per unit but has a high setup cost • Engine lathe costs more per unit but has no setup costs When do I use which machine?
Comparison Costs Switchover occurs somewhere between 25 and 100 units (n=1 setup)
Determine breakeven point-Math • 2.10x+48=3.45x • x=35 units (assumes N=1 setup)
Determine Breakeven Unit between 2 Machines with Math • Set Total Costs equal to each other between two machines: • Let Fixed Costs for Alternatives 1 and 2= FC1 & FC2 • Let Variable Costs for Alternatives 1 and 2=VC1 & VC2 • Set total costs equal to each other: FC1+VC1*D=FC2+VC2*D • Then D=(FC2-FC1)/(VC1-VC2) Note: embed this equation into the breakeven project
Next lecture • Estimates • Accounting Principles