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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.
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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