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ECLT ECL T 5930 5930/SEE /SEEM M 5740 5740:: Engin Engineerin eering g Econo Economics mics 2013–14 Second Term Master of Science in ECLT & SEEM
Instr In struct uctor ors: s: Dr Dr.. An Antho thony ny Man Man–C –Cho ho So Dr.. Ma Dr Man n Ho Hong ng Ke Keit ith h Won ong g Department of Systems Engineering & Engineering Management The Chinese University of Hong Kong January 16, 2014
Course Personnel (Updated)
• Instructors: Dr. Anthony Man–Cho SO/Dr. Man Hong Keith WONG – Office: ERB 604 – Phone: 3943 8477 – Office Hours: By appointment – Email: [email protected] – Website: http://www.se.cuhk.edu.hk/~ manchoso
• Teaching Assistants: Name Ms. Xin LIU Mr. Ch Chong Man TANG Ms. Weijie WU Mr. Kairen ZHANG
Office ERB 814 ERB 905 ERB 905 ERB 514
– Office Hours: By appointment
Phone Email 3 4438 [email protected] 3 4241 [email protected] 3 4241 [email protected] 3 8319 [email protected]
Recap: Engineering Economic Analysis & Engineering Design Process 1. Pro Proble blem m defi definit nition ion 2. Prob Problem lem formulati formulation on and evaluation 3. Synt Synthesis hesis of possible solutions solutions (alternatives (alternatives)) 4. Analy Analysis, sis, optimizatio optimization, n, and evaluation 5. Specificati Specification on of preferred preferred alternativ alternativee 6. Comm Communicat unication ion via perfo performance rmance monitoring monitoring
This Lecture: Analysis of Short–Term Alternatives
• Focus on short term, hence time value of money is negligible • Identify various cost elements in an alternative • Perform economic breakeven analysis and cost–driven design optimization
Identifying Costs
• Cost elements differ in their frequency of occurrence, relative magnitude and degree of impact on the problem at hand
• Correctly identifying them is crucial in an engineering economic analysis • Broad Categories – Fixed, variable and incremental costs – Direct, indirect and standard costs – Cash and book costs – Sunk costs – Opportunity costs – Life-cycle costs
Fixed, Variable and Incremental Costs
• Fixed costs: costs that are unaffected by changes in activity level over a feasible range of operations for the capacity available
– e.g.: license fees, pipeline installation costs
• Variable costs: costs that vary in total with quantity of output or other measures of activity levels
– e.g.: costs of material and labor used in a product or service
• Incremental costs: additional cost that results from increasing the output of a system by one (or more) units
– depends on various factors, such as economies of scale, state of the production system, etc. – e.g.: incremental cost of producing a barrel of oil
Fixed, Variable and Incremental Costs (Cont’d)
cost
cost
units produced
(a) Fixed cost: cost constant over a range of production
units produced
(b) Variable cost: cost varies with amount of production
Figure 1: Graphs illustrating fixed, variable and incremental costs
Application: Highway Surfacing
• A contractor has to choose from one of two sites on which to set up asphalt– mixing plant equipment.
• The cost factors relating to the mixing sites are as follows: Cost Factor Average hauling distance Monthly rental of site Cost to set up and remove equipment Hauling expense Flagperson
Site A 6 miles $1,000 $15,000 $1.15/yd3–mile not required
Site B 4.3 miles $5,000 $25,000 $1.15/yd3–mile $96/day
Application: Highway Surfacing (Cont’d)
• The job requires 50,000 cubic yards of mixed asphalt paving material. • Also,
four months (17 weeks of five working days per week) are needed to complete the job.
• Assume that the cost of return trip is negligible. • Questions: Identify the costs. Which is the better site?
Application: Highway Surfacing (Cont’d) Cost Factor Rent Setup/Removal Flagperson Hauling
Fixed
√ √ √
Variable
√
Site A $4,000 $15,000 $0 $345,000
Site B $20,000 $25,000 $8,160 $247,250
• Total cost for site A: $364,000 • Total cost for site B: $300,410 • So site B is better. – Note that the higher fixed costs of site B are being traded off for reduced variable costs.
Direct, Indirect and Standard Costs
• Direct costs: costs that can be reasonably measured and allocated to a specific output or work activity
– e.g.: material and labor costs directly associated with an economic activity
• Indirect costs (aka overhead or burden ): costs that are difficult to attribute or allocate to a specific activity
– e.g.: costs of common tools, general supplies, equipment maintenance, electricity – typically allocated through a selected formula (e.g., proportional to direct labor hours, direct labor dollars, etc.).
• Standard costs: planned costs per unit of output that are established in advance of actual production or service delivery
– developed using anticipated level of production – play an important role in cost control and other management functions, such as estimating future manufacturing costs, measuring operating performance by comparing actual vs standard unit cost, etc.
Remark about the Terminologies
• The previously introduced categories are not necessarily mutually exclusive. • Can you think of a
fixed variable
direct indirect cost that is a standard cost?
Cash and Book Costs
• Cash costs: costs that involve payment of cash and result in cash flow • Noncash or book costs: costs that do not involve cash payments but rather represent the recovery of past expenditures over a fixed period of time
– e.g.: depreciation charged for the use of assets such as equipment
• In engineering economic analysis, only cash flows or potential cash flows matter – e.g.: Depreciation is not a cash flow, but it affects income taxes, which is a cash flow. More about this later.
Sunk Costs
• Sunk costs: costs that occurred in the past and have no relevance to estimates of future costs and revenues related to an alternative course of action
– e.g.: money spent on a passport, deposit used to secure a flat
• Sunk cost is not part of the prospective cash flows and can be disregarded in an engineering economic analysis
Sunk Costs: Example
• John finds a motorcyle he likes and pays $40 as down payment, which will be applied to the $1,300 purchase price, but will be forfeited if he does not take the motorcycle.
• Over the weekend, he finds another equally desirable motorcyle for a purchase price of $1,230.
• For the purpose of deciding which motorcycle to buy, the $40 is a sunk cost. It
should not enter into the decision, except that it lowers the remaining cost of the first motorcycle.
Sunk Costs: Another Example
• Tom bought a bad second hand mower machine for $100, hoping to spend an
additional $160 on accessories and repair it. Then he would be able to sell it for $500. However, after spending $200, he found that he would still need additional $250 to finish the repairing.
• Question: What is the sunk cost in this case? – Hint: Sunk costs are irretrievable consequences of past actions.
Opportunity Costs
• Opportunity costs: costs that are measured in terms of the value of the best alternative that is not chosen (i.e., foregone)
– one of the most important concepts in economics – difficult to define (what is the “best” alternative?) and is often hidden or implied
• Rule of thumb: avoided benefit = cost, avoided cost = benefit • Example
– A student who could earn $20,000 for working during a year, but chooses instead to go to school for a year and pay $5,000 in tuition. – His opportunity cost is $20,000 + $5,000 = $25,000.
• Question – By taking a plane Larry can travel from Hong Kong to Guangzhou in 1 hour. The same trip takes 5 hours by bus. Airfare is $600 and the bus fare is $200. – If Larry is not travelling, he can work and earn $200 per hour. – What is the opportunity cost for Larry if he travels by bus?
Life–Cycle Costs
• Life cycle: roughly, the life cycle of an economic activity consists of two phases: acquisition and operation
Acquisition Phase needs preliminary detailed design; assessment; design; production definition of advanced planning; requirements prototype resource testing acquisition Operation Phase production operation; retirement maintenance and disposal and support
• Life–cycle costs: summation of all costs related to an economic activity during its life span
Elements of Breakeven Analysis
• To perform breakeven analysis, we need to know our sources of revenue and expenditure.
• Typically, these depend on total cost, unit selling price and the actual demand. • These three elements are inter–related: – Higher the price, lower the demand – Higher the demand, higher the total cost of production Hence, we must specify the relationships among them.
Cost Function
• For simplicity, we assume that the total cost (C T ) is made up of fixed costs (C F ) and variable costs (C V ), i.e.,
C T = C F + C V .
• Since fixed costs essentially do not vary with the amount of activity,
we can
treat C F as a constant.
• On the other hand, let us assume that the variable costs depend linearly on the demand, i.e., C V = c × D, where D is the demand, and c > 0 is the per unit variable cost.
Cost Function (Cont’d)
total cost C T
= C F + c × D
slope = c
C F
demand
Figure 2: Graph of the total cost function, which is a sum of fixed and variable costs
Demand Function
• Typically, higher the price, lower the demand. • For simplicity, we assume that unit price ( p) and demand (D) are linearly related, i.e., p = a − b × D, where a, b > 0 and 0 ≤ D ≤ a/b (why?). • The coefficient b is related to the demand elasticity. Generally, the lower the b, the more elastic the demand.
– Question: What kind of goods have high (or low) demand elasticity?
Demand Function (Cont’d)
price
p = a − b × D
slope =
b
−
demand
Figure 3: Graph of the demand function
Total Revenue Function
• The total revenue (TR) is simply the product of unit selling price ( p) and number of units sold (D), i.e.,
2
TR = p
× D = (a − b × D) × D = aD − bD , where a, b > 0 and 0 ≤ D ≤ a/b. • We have expressed total revenue as a function of demand. In particular, we can ˆ find the demand D that maximizes the total revenue: dTR =a dD
− 2bD = 0 ⇐⇒
ˆ = a. D 2b
ˆ Just need to set the price right! • How to attain the demand D? ˆ = a. pˆ = a − b × D 2
• Question: Is maximizing total revenue the right thing to do?
Cost–Volume Relationships
cost/ revenue
C T = C F + c × D max profit
C F TR = aD
D
∗
ˆ D
−
demand
Figure 4: Cost–volume relationships
bD2
Profit Function
• By definition,
profit is simply the difference between total revenue and total
cost, i.e.,
profit = total revenue = TR = = where a, b, c > 0 and 0
− total cost
− C T (aD − bD ) − (C F + c × D) −bD + (a − c)D − C F , 2
2
≤ D ≤ a/b (a negative profit means a loss).
• From this identity, we can ask two fundamental economic questions: – Under what conditions would we achieve maximum profit? – Under what conditions would we breakeven?
Profit Maximization vs Breakeven
• To maximize profit, we take the first derivative of the profit function and solve d(profit) a−c = −2bD + (a − c) = 0 ⇐⇒ D = . ∗
dD
2b
• For this to make sense, we must have a > c to start with. • On the other hand, at a breakeven point, the total revenue equals total cost, i.e.,
aD
− bD
2
= C F + c
bD 2 + (c
× D ⇐⇒
− a)D + C F = 0.
Upon solving this quadratic equation, we get
−(c − a) ± (c − a) − 4bC 2
′
D =
2b
2
F
.
• For this to make sense, we must have (c − a) ≥ 4bC F to start with.
Breakeven Points
cost/ revenue
C T = C F + c × D
C F TR = aD
′
D1
′
D2
demand
Figure 5: Breakeven points
−
bD2
Profit Maximization vs Breakeven (Cont’d)
• Note that if the maximum profit is non–negative, then for any demand D that falls in the range D ≤ D ≤ D , i.e., ′
′
1
2
−(c − a) − (c − a) − 4bC 2
2b
we will be making a profit.
F
−(c − a) + (c − a) − 4bC 2
≤D≤
2b
F
,
Application: Finding Optimal Demand of Electronic Switch
• A company produces an electronic timing switch. • Running the production line costs $73,000 per month.
Moreover, it costs $83
to produce one unit.
• The price–demand relationship is determined as p = $180 − 0.02 × D. Questions: 1. Is there a demand level such that profit occurs? 2. What are the breakeven points? What is the range of profitable demand?
Application: Finding Optimal Demand of Electronic Switch
• In this problem, the fixed cost is $73,000 per month, and the variable cost is $83 per unit. Hence, the total cost function is given by C T = $73, 000 + $83
× D.
• Since a − c = 180 − 83 > 0, our previous result applies. The demand level that yields the maximum profit is given by ∗
D =
a
− c = 180 − 83 = 2, 425 units per month. 2b 2 × 0.02
The actual profit is given by
profit = total revenue = =
− total cost (aD − b(D ) ) − (C F + c × D ) (180 × 2, 425 − 0.02 × (2, 425) ) − (73, 000 + 83 × 2, 425) ∗
∗
2
= $44, 612 per month.
∗
2
Application: Finding Optimal Demand of Electronic Switch
• To find the breakeven points, we need to solve bD 0.02
×D
2
+ (83
2
+ (c
− a)D + C F = 0, or
− 180)D + 73, 000 = 0.
The solutions are ′
D1 = ′
D2 =
97
− 59.74 = 932 units per month,
0.04 97 + 59.74 = 3, 918 units per month. 0.04
• In particular, the range of profitable demand is 932
≤ D ≤ 3, 918.
Cost–Driven Design Optimization
• In many engineering problems, there is a tradeoff between cost and performance of a design.
– e.g.: building airplanes, building bridges, writing software
• How to optimize the design based on cost considerations?
Cost–Driven Design Optimization: An Example
• The cost of operating a jet can be given by C O = knv 3/2, where
– k is a constant of proportionality, – n is the trip length in miles, – v is velocity in miles per hour.
• At 400 miles per hour, the average cost of operation is $300 per mile. • The cost of passengers’ time (C C ) is set at $300,000 per hour. • Question: At what velocity should the trip be planned to minimize the total cost C T , which is defined as
C T = C O + C C ?
Cost–Driven Design Optimization: An Example (Cont’d)
• Let us first determine k, the constant of proportionality. From the data, we have 300 =
C O = k(400)3/2 n
=
⇒
k = 0.0375.
• Thus, the total cost is given by C T = C O + C C = 0.0375
× nv /
3 2
+ 300, 000
× nv .
• To minimize the total cost, we take the first derivative of C T with respect to v and solve i.e.,
dC T 3 = dv 2
1/2
× 0.0375 × nv − 300, 000 ×
0.05625
• Solving this equation yields v
∗
× v / − 300,v 000 = 0. 1 2
= 490.68 mph.
2
n = 0, v2
Application: To Produce or Not to Produce?
• Department A of a manufacturing plant occupies 100 square meters and produce, among other things, 576 pieces of product X per day.
• The average daily production costs for product X are summarized as follows: Direct labor
1 operator working 4 hours per day at $22.50 per hour; part–time manager at $30 per day
Direct material Overhead
$120.00 $86.40
at $0.82 per square meter Total cost per day
$82.00 $288.40
• One can also outsource the production of X to another company at a cost of $0.35 per piece. This results in a total purchase cost of 576 × $0.35 = $201.60. Question: Should the plant shut down the production line for X and purchase it from the other company?
Instr In struct uctor ors: s: Dr Dr.. An Antho thony ny Man Man–C –Cho ho So Dr.. Ma Dr Man n Ho Hong ng Ke Keit ith h Won ong g Department of Systems Engineering & Engineering Management The Chinese University of Hong Kong January 16, 2014
Course Personnel (Updated)
• Instructors: Dr. Anthony Man–Cho SO/Dr. Man Hong Keith WONG – Office: ERB 604 – Phone: 3943 8477 – Office Hours: By appointment – Email: [email protected] – Website: http://www.se.cuhk.edu.hk/~ manchoso
• Teaching Assistants: Name Ms. Xin LIU Mr. Ch Chong Man TANG Ms. Weijie WU Mr. Kairen ZHANG
Office ERB 814 ERB 905 ERB 905 ERB 514
– Office Hours: By appointment
Phone Email 3 4438 [email protected] 3 4241 [email protected] 3 4241 [email protected] 3 8319 [email protected]
Recap: Engineering Economic Analysis & Engineering Design Process 1. Pro Proble blem m defi definit nition ion 2. Prob Problem lem formulati formulation on and evaluation 3. Synt Synthesis hesis of possible solutions solutions (alternatives (alternatives)) 4. Analy Analysis, sis, optimizatio optimization, n, and evaluation 5. Specificati Specification on of preferred preferred alternativ alternativee 6. Comm Communicat unication ion via perfo performance rmance monitoring monitoring
This Lecture: Analysis of Short–Term Alternatives
• Focus on short term, hence time value of money is negligible • Identify various cost elements in an alternative • Perform economic breakeven analysis and cost–driven design optimization
Identifying Costs
• Cost elements differ in their frequency of occurrence, relative magnitude and degree of impact on the problem at hand
• Correctly identifying them is crucial in an engineering economic analysis • Broad Categories – Fixed, variable and incremental costs – Direct, indirect and standard costs – Cash and book costs – Sunk costs – Opportunity costs – Life-cycle costs
Fixed, Variable and Incremental Costs
• Fixed costs: costs that are unaffected by changes in activity level over a feasible range of operations for the capacity available
– e.g.: license fees, pipeline installation costs
• Variable costs: costs that vary in total with quantity of output or other measures of activity levels
– e.g.: costs of material and labor used in a product or service
• Incremental costs: additional cost that results from increasing the output of a system by one (or more) units
– depends on various factors, such as economies of scale, state of the production system, etc. – e.g.: incremental cost of producing a barrel of oil
Fixed, Variable and Incremental Costs (Cont’d)
cost
cost
units produced
(a) Fixed cost: cost constant over a range of production
units produced
(b) Variable cost: cost varies with amount of production
Figure 1: Graphs illustrating fixed, variable and incremental costs
Application: Highway Surfacing
• A contractor has to choose from one of two sites on which to set up asphalt– mixing plant equipment.
• The cost factors relating to the mixing sites are as follows: Cost Factor Average hauling distance Monthly rental of site Cost to set up and remove equipment Hauling expense Flagperson
Site A 6 miles $1,000 $15,000 $1.15/yd3–mile not required
Site B 4.3 miles $5,000 $25,000 $1.15/yd3–mile $96/day
Application: Highway Surfacing (Cont’d)
• The job requires 50,000 cubic yards of mixed asphalt paving material. • Also,
four months (17 weeks of five working days per week) are needed to complete the job.
• Assume that the cost of return trip is negligible. • Questions: Identify the costs. Which is the better site?
Application: Highway Surfacing (Cont’d) Cost Factor Rent Setup/Removal Flagperson Hauling
Fixed
√ √ √
Variable
√
Site A $4,000 $15,000 $0 $345,000
Site B $20,000 $25,000 $8,160 $247,250
• Total cost for site A: $364,000 • Total cost for site B: $300,410 • So site B is better. – Note that the higher fixed costs of site B are being traded off for reduced variable costs.
Direct, Indirect and Standard Costs
• Direct costs: costs that can be reasonably measured and allocated to a specific output or work activity
– e.g.: material and labor costs directly associated with an economic activity
• Indirect costs (aka overhead or burden ): costs that are difficult to attribute or allocate to a specific activity
– e.g.: costs of common tools, general supplies, equipment maintenance, electricity – typically allocated through a selected formula (e.g., proportional to direct labor hours, direct labor dollars, etc.).
• Standard costs: planned costs per unit of output that are established in advance of actual production or service delivery
– developed using anticipated level of production – play an important role in cost control and other management functions, such as estimating future manufacturing costs, measuring operating performance by comparing actual vs standard unit cost, etc.
Remark about the Terminologies
• The previously introduced categories are not necessarily mutually exclusive. • Can you think of a
fixed variable
direct indirect cost that is a standard cost?
Cash and Book Costs
• Cash costs: costs that involve payment of cash and result in cash flow • Noncash or book costs: costs that do not involve cash payments but rather represent the recovery of past expenditures over a fixed period of time
– e.g.: depreciation charged for the use of assets such as equipment
• In engineering economic analysis, only cash flows or potential cash flows matter – e.g.: Depreciation is not a cash flow, but it affects income taxes, which is a cash flow. More about this later.
Sunk Costs
• Sunk costs: costs that occurred in the past and have no relevance to estimates of future costs and revenues related to an alternative course of action
– e.g.: money spent on a passport, deposit used to secure a flat
• Sunk cost is not part of the prospective cash flows and can be disregarded in an engineering economic analysis
Sunk Costs: Example
• John finds a motorcyle he likes and pays $40 as down payment, which will be applied to the $1,300 purchase price, but will be forfeited if he does not take the motorcycle.
• Over the weekend, he finds another equally desirable motorcyle for a purchase price of $1,230.
• For the purpose of deciding which motorcycle to buy, the $40 is a sunk cost. It
should not enter into the decision, except that it lowers the remaining cost of the first motorcycle.
Sunk Costs: Another Example
• Tom bought a bad second hand mower machine for $100, hoping to spend an
additional $160 on accessories and repair it. Then he would be able to sell it for $500. However, after spending $200, he found that he would still need additional $250 to finish the repairing.
• Question: What is the sunk cost in this case? – Hint: Sunk costs are irretrievable consequences of past actions.
Opportunity Costs
• Opportunity costs: costs that are measured in terms of the value of the best alternative that is not chosen (i.e., foregone)
– one of the most important concepts in economics – difficult to define (what is the “best” alternative?) and is often hidden or implied
• Rule of thumb: avoided benefit = cost, avoided cost = benefit • Example
– A student who could earn $20,000 for working during a year, but chooses instead to go to school for a year and pay $5,000 in tuition. – His opportunity cost is $20,000 + $5,000 = $25,000.
• Question – By taking a plane Larry can travel from Hong Kong to Guangzhou in 1 hour. The same trip takes 5 hours by bus. Airfare is $600 and the bus fare is $200. – If Larry is not travelling, he can work and earn $200 per hour. – What is the opportunity cost for Larry if he travels by bus?
Life–Cycle Costs
• Life cycle: roughly, the life cycle of an economic activity consists of two phases: acquisition and operation
Acquisition Phase needs preliminary detailed design; assessment; design; production definition of advanced planning; requirements prototype resource testing acquisition Operation Phase production operation; retirement maintenance and disposal and support
• Life–cycle costs: summation of all costs related to an economic activity during its life span
Elements of Breakeven Analysis
• To perform breakeven analysis, we need to know our sources of revenue and expenditure.
• Typically, these depend on total cost, unit selling price and the actual demand. • These three elements are inter–related: – Higher the price, lower the demand – Higher the demand, higher the total cost of production Hence, we must specify the relationships among them.
Cost Function
• For simplicity, we assume that the total cost (C T ) is made up of fixed costs (C F ) and variable costs (C V ), i.e.,
C T = C F + C V .
• Since fixed costs essentially do not vary with the amount of activity,
we can
treat C F as a constant.
• On the other hand, let us assume that the variable costs depend linearly on the demand, i.e., C V = c × D, where D is the demand, and c > 0 is the per unit variable cost.
Cost Function (Cont’d)
total cost C T
= C F + c × D
slope = c
C F
demand
Figure 2: Graph of the total cost function, which is a sum of fixed and variable costs
Demand Function
• Typically, higher the price, lower the demand. • For simplicity, we assume that unit price ( p) and demand (D) are linearly related, i.e., p = a − b × D, where a, b > 0 and 0 ≤ D ≤ a/b (why?). • The coefficient b is related to the demand elasticity. Generally, the lower the b, the more elastic the demand.
– Question: What kind of goods have high (or low) demand elasticity?
Demand Function (Cont’d)
price
p = a − b × D
slope =
b
−
demand
Figure 3: Graph of the demand function
Total Revenue Function
• The total revenue (TR) is simply the product of unit selling price ( p) and number of units sold (D), i.e.,
2
TR = p
× D = (a − b × D) × D = aD − bD , where a, b > 0 and 0 ≤ D ≤ a/b. • We have expressed total revenue as a function of demand. In particular, we can ˆ find the demand D that maximizes the total revenue: dTR =a dD
− 2bD = 0 ⇐⇒
ˆ = a. D 2b
ˆ Just need to set the price right! • How to attain the demand D? ˆ = a. pˆ = a − b × D 2
• Question: Is maximizing total revenue the right thing to do?
Cost–Volume Relationships
cost/ revenue
C T = C F + c × D max profit
C F TR = aD
D
∗
ˆ D
−
demand
Figure 4: Cost–volume relationships
bD2
Profit Function
• By definition,
profit is simply the difference between total revenue and total
cost, i.e.,
profit = total revenue = TR = = where a, b, c > 0 and 0
− total cost
− C T (aD − bD ) − (C F + c × D) −bD + (a − c)D − C F , 2
2
≤ D ≤ a/b (a negative profit means a loss).
• From this identity, we can ask two fundamental economic questions: – Under what conditions would we achieve maximum profit? – Under what conditions would we breakeven?
Profit Maximization vs Breakeven
• To maximize profit, we take the first derivative of the profit function and solve d(profit) a−c = −2bD + (a − c) = 0 ⇐⇒ D = . ∗
dD
2b
• For this to make sense, we must have a > c to start with. • On the other hand, at a breakeven point, the total revenue equals total cost, i.e.,
aD
− bD
2
= C F + c
bD 2 + (c
× D ⇐⇒
− a)D + C F = 0.
Upon solving this quadratic equation, we get
−(c − a) ± (c − a) − 4bC 2
′
D =
2b
2
F
.
• For this to make sense, we must have (c − a) ≥ 4bC F to start with.
Breakeven Points
cost/ revenue
C T = C F + c × D
C F TR = aD
′
D1
′
D2
demand
Figure 5: Breakeven points
−
bD2
Profit Maximization vs Breakeven (Cont’d)
• Note that if the maximum profit is non–negative, then for any demand D that falls in the range D ≤ D ≤ D , i.e., ′
′
1
2
−(c − a) − (c − a) − 4bC 2
2b
we will be making a profit.
F
−(c − a) + (c − a) − 4bC 2
≤D≤
2b
F
,
Application: Finding Optimal Demand of Electronic Switch
• A company produces an electronic timing switch. • Running the production line costs $73,000 per month.
Moreover, it costs $83
to produce one unit.
• The price–demand relationship is determined as p = $180 − 0.02 × D. Questions: 1. Is there a demand level such that profit occurs? 2. What are the breakeven points? What is the range of profitable demand?
Application: Finding Optimal Demand of Electronic Switch
• In this problem, the fixed cost is $73,000 per month, and the variable cost is $83 per unit. Hence, the total cost function is given by C T = $73, 000 + $83
× D.
• Since a − c = 180 − 83 > 0, our previous result applies. The demand level that yields the maximum profit is given by ∗
D =
a
− c = 180 − 83 = 2, 425 units per month. 2b 2 × 0.02
The actual profit is given by
profit = total revenue = =
− total cost (aD − b(D ) ) − (C F + c × D ) (180 × 2, 425 − 0.02 × (2, 425) ) − (73, 000 + 83 × 2, 425) ∗
∗
2
= $44, 612 per month.
∗
2
Application: Finding Optimal Demand of Electronic Switch
• To find the breakeven points, we need to solve bD 0.02
×D
2
+ (83
2
+ (c
− a)D + C F = 0, or
− 180)D + 73, 000 = 0.
The solutions are ′
D1 = ′
D2 =
97
− 59.74 = 932 units per month,
0.04 97 + 59.74 = 3, 918 units per month. 0.04
• In particular, the range of profitable demand is 932
≤ D ≤ 3, 918.
Cost–Driven Design Optimization
• In many engineering problems, there is a tradeoff between cost and performance of a design.
– e.g.: building airplanes, building bridges, writing software
• How to optimize the design based on cost considerations?
Cost–Driven Design Optimization: An Example
• The cost of operating a jet can be given by C O = knv 3/2, where
– k is a constant of proportionality, – n is the trip length in miles, – v is velocity in miles per hour.
• At 400 miles per hour, the average cost of operation is $300 per mile. • The cost of passengers’ time (C C ) is set at $300,000 per hour. • Question: At what velocity should the trip be planned to minimize the total cost C T , which is defined as
C T = C O + C C ?
Cost–Driven Design Optimization: An Example (Cont’d)
• Let us first determine k, the constant of proportionality. From the data, we have 300 =
C O = k(400)3/2 n
=
⇒
k = 0.0375.
• Thus, the total cost is given by C T = C O + C C = 0.0375
× nv /
3 2
+ 300, 000
× nv .
• To minimize the total cost, we take the first derivative of C T with respect to v and solve i.e.,
dC T 3 = dv 2
1/2
× 0.0375 × nv − 300, 000 ×
0.05625
• Solving this equation yields v
∗
× v / − 300,v 000 = 0. 1 2
= 490.68 mph.
2
n = 0, v2
Application: To Produce or Not to Produce?
• Department A of a manufacturing plant occupies 100 square meters and produce, among other things, 576 pieces of product X per day.
• The average daily production costs for product X are summarized as follows: Direct labor
1 operator working 4 hours per day at $22.50 per hour; part–time manager at $30 per day
Direct material Overhead
$120.00 $86.40
at $0.82 per square meter Total cost per day
$82.00 $288.40
• One can also outsource the production of X to another company at a cost of $0.35 per piece. This results in a total purchase cost of 576 × $0.35 = $201.60. Question: Should the plant shut down the production line for X and purchase it from the other company?
