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Chapter 13: Breakeven Analysis
Breakeven analysis is performed to determine the value of a variable of a project that makes two elements equal, e.g. sales volume that will equate revenues and costs.
Single Project
The analysis is based on the relationship:
Profit = revenue โ total cost = R โ TC
At breakeven, there is no profit or loss, hence, revenue = total cost
or, R = TC
Note: It is to be noted that +ve sign is used for both the revenue and the costs. If we are to use โve sign for costs and +ve sign for revenue, then the above relationships become:
Profit = R + TC and R + TC = 0 at breakeven.
With revenue and costs given in terms of a decision variable, the solution yields the breakeven quantity for the decision variable.
Costs , which may be linear or non-linear, usually include two components:
Fixed costs (FC) โ Includes costs such as buildings, insurance, fixed overhead, equipment capital recovery, etc. These costs are essentially constant for all values of the decision variable.
Variable costs (VC) โ Includes costs such as direct labour, materials, contractors, marketing, advertisement, etc. These costs change linearly or non-linearly with the decision variable, e.g. production level, workforce size, etc. For the analysis to be followed here, the variation will generally be assumed to be linear.
Then, total cost, TC = FC + VC
Revenue also changes with the decision variable. Again, for the analysis, the variation will generally be assumed to be linear.
The following diagram illustrates the basics of the breakeven analysis.
Revenue, R
Revenue Total Cost, TC or Cost VC
FC
QBE, Breakeven quantity
Production, Q units/year
It can be seen that we have profit if the production level is above the breakeven quantity and loss if it is below.
Examples:
1. The fixed costs at Company X are $1 million annually. The main product has revenue of $8.90 per unit and $4.50 variable cost. (a) Determine the breakeven quantity per year, and (b) Annual profit if 200000 units are sold.
Let QBE be the breakeven quantity.
8.9QBE = 1,000,000 + 4.5QBE
QBE = 1,000,000/(8.90-4.50) = 227,272 units
(b) Profit = R โ TC
= 8.90Q โ 1,000,000 - 4.5Q
At 200,000 units: Profit = 8.90(200,000) โ 1,000,000 - 4.50(200,000) = $-120,000 (loss)
2. A product currently sells for $12 per unit. The variable costs are $4 per unit, and 10, units are sold annually and a profit of $30,000 is realized per year. A new design will increase the variable costs by %20 and Fixed Costs by %10 but sales will increase to 12,000 units per year. (a) At what selling price do we break even, and (b) If the selling price is to be kept same ($12/unit) what will the annual profit be?
Profit = revenue โ costs
30000 = 10000(12) โ [10000(4) + FC] FC = fixed costs
FC = 50000
(a) New variable cost = $4(1.2) = $4.8 per unit.
New fixed costs = 50000(1.1) = $
Let x = breakeven selling price per unit, then
12000x = 55000 + 12000(4.8) or, x = $9.38/unit
(b) Profit = 12000(12) โ 12000(4.8) - 55000 = $
5. An automobile company is planning to convert a plant from manufacturing economy cars to manufacturing sports cars. The initial cost for equipment conversion will be $200 million with a 20% salvage value anytime within a 5-year period. The cost of producing a car will be $21000, and it will be sold for $33000. The production capacity for the first year will be 4000 units. At an interest rate of 12% per year, by what uniform amount will production have to increase each year in order for the company to recover its investment in 3 years?
Let x = gradient increase per year.
Total costs = -200M(A/P,12%,3)+(0.20)(200M)(A/F,12%,3)-4000+ x(A/G,12%,3)
Revenue = 4000 + x(A/G,12%,3)
At breakeven, revenue + costs* = 0, then * -ve sign for costs is used.
[4000 + x(A/G,12%,3)](33,000 โ 21,000) = 200M(A/P,12%,3) - (0.20)(200M)(A/F,12%,3)
4000 + x(0.9246) = 200M(0.41635) - 40M(0.29635)
x = 2110 cars/year increase
6. Owners of a hotel chain are considering locating a new hotel in Karpaz. The complete cost of building a 150-room hotel (excluding furnishings) is $2million; the furnishings will cost $750 000 and will be replaced every 5 years for the same cost. Annual operating and maintenance cost for the facility is estimated to be $50 000. The average rate for a room is expected to be $15 per day. A 15-year planning period is used by the firm in evaluating new projects of this type; a terminal salvage value of 20% of the original building cost is anticipated; furnishings are estimated to have no salvage value at the end of each five-year replacement interval; land cost is not to be included. Determine the break-even value for the average number of rooms to be occupied daily based on a MARR of 10% (Assume the hotel will operate 365 days a year).
Annualizing Costs:
Building: AWB = -2M(A/P,10%,15) = -2M(0.13147) = -262940 / yr
Furnishings: AWF = -750000 โ 750000(P/F,10%,5) โ 750000(P/F,10%,10)
= -197836.06 / yr
Salvage Value = (0.2)2M = 400000
AWS = 400000(A/F,10%,15) = 400000(0.03147) = 12588
Total Annual Cost, AWC = -262940 โ197836 โ 50000 + 12588 = -
Revenue = 15(365)X, where X is number of rooms occupied.
At break-even,
15(365)X โ 498188 = 0 or X = 91 rooms per day on the average.
Two or more Alternatives
This is commonly applied to between alternatives that serve the same purpose. As a result, breakeven analysis is carried out between the costs of the alternatives. It involves the determination of a common variable between two or more alternatives. The procedure to follow for two alternatives is as follows:
- Define the common variable and its dimensional units.
- Use AW or PW analysis to express the total cost of each alternative as a function of the common variable. (Use AW values if lives are different).
- Equate the two relations and solve for the breakeven value of the variable.
- If the anticipated level is below the breakeven value, select the alternative with the higher variable cost (larger slope). If the level is above the breakeven point, select the alternative with the lower variable cost.
The same type of analysis can be performed for three or more alternatives. Then, compare the alternatives in pairs to find their respective breakeven points. The results are the ranges through which each alternative is more economical.
Examples:
7. A Textile company is evaluating the purchase of an automatic cloth-cutting machine. The machine will have a first cost of $22000, a life of 10 years, and a $500 salvage value. The annual maintenance cost of the machine is expected to be $2000 per year. The machine will require one operator at a total cost of $24 an hour. Approximately 1500 meters of material can be cut each hour with the machine. Alternatively, if human labor is used, five workers , each earning $10 an hour , can cut 1000 meters per hour. If the companyโs MARR is %8 per year , and 180,000 meters of material is to be cut every year should the company buy the automatic machine or use human labor instead? At how many meters cloth-cutting per year will the two alternatives breakeven?
Let x = meters of material to be cut
Automatic machine: Total annual cost, AWA = -22000(A/P,8%,10) + 500(A/F,8%,10) โ 2000 โ (x/1500)(24)
= -5244.15 โ x/62.
Manual: Total annual cost, AWM = -(x/1000)(5)(10) = -x/
At breakeven, AWA = AWM
-5244.15 โ x/62.5 = x/
or, x = 154240 m
Therefore, at 180000 m, select the automatic machine. (we make profit if quantity is above the breakeven).
10. The ABC Company is faced with three proposed methods for making one of their products. Method A involves the purchase of a machine for $5000. It will have a seven-year life, with a zero salvage value at that time. Using Method A involves additional costs of $0. per unit of product produced per year. Method B involves the purchase of a machine for $10000. It will also have a seven-year life, with $2000 salvage value at that time. Using Method B involves additional costs of $0.15 per unit of product produced per year. Method C involves the purchase of a machine for $8000. It will have a $2000 salvage value when disposed of in seven years. Additional costs of $0.25 per unit of product per year arise when Method C is used. An 8% interest rate is used by the ABC Company in evaluating investment alternatives. For what range of annual production volume values is each method preferred?
Let X = number of units per year
AWA = -5000.(A/P,8%,7) โ 0.2X = -5000.(0.19207) โ 0.2X = -960.35 โ 0.2X
AWB = -10000.(A/P,8%,7) + 2000.(A/F,8%,7) โ 0.15X = -10000.(0.19207) + 2000.(0.11207) โ 0.15X = -1696.56 โ 0.15X
AWC = -8000.(A/P,8%,7) + 2000.(A/F,8%,7) โ 0.25X = -1312.42 โ 0.25X
A vs B: -960.35 โ 0.2X = -1696.56 โ 0.15X XAB = 14724.
B vs C: -1696.56 โ 0.15X = -1312.42 โ 0.25X XBC = 3841.
A vs C: No intersection for X > 0.
C A B
Total Cost 4000
X โค 14724, select A X > 14724, select B
2000
X
Number of units per year
11. Three types of design proposals for a commercial one - storey building is to be evaluated details given below: STEEL CONCRETE BRICK First cost $72/ft^2 $76/ft^2 $81/ft^2 Annual maintenance $14000 $9000 $ Annual heating cost $3/ft^2 $3.4/ft^2 $3.9/ft^2 SV (%of first cost) %80 %100 % Life (years) 20 20 20
For what range of building area (ft^2 ) which type of design is the most suitable (cheapest) to select? Carry out breakeven analysis using an interest rate of %18 per year and plot your ranges to illustrate.
Let X = area in ft^2
First cost of steel = 72x and its salvage value = (0.8)72x
Using AW values:
AWS = -(72x)(A/P,18%,20) + (0.8)(72x) โ 14000 โ 3x = -13.45x + 0.393x โ 14000 โ 3x = -16.057x โ 14000
Similarly,
AWC = -(76x)(A/P,18%,20) + (1.0)(76x) โ 9000 โ 3.4x = -17.082x โ 9000
AWB = -(81x)(A/P,18%,20) + (1.1)(81x) โ 6000 โ 3.9x = -18.423x โ 6000
Breakevens:
Steel vs Concrete: -16.057x โ 14000 = -17.082x โ 9000 x = 4878 ft^2
Steel vs Brick: -16.057x โ 14000 = -18.423x โ 6000 x = 3381 ft^2
Concrete vs Brick: -17.082x โ 9000 = -18.423x โ 6000 x = 2237 ft^2
B 0 < x โค 2237 Select Brick C 2237 < x โค 4878 Select Concrete Total cost x > 4878 Select Steel S
Area, ft^2
