Chapter 2 - Cost Terminology and Design Economics
Cost terminology Fixed, variable, and incremental costs that remain constant over a specific range of Fixed costs that operating conditions. It is subject to change when larger changes in the operating conditions involved such as plant expansion or shutdown. that vary in total with the output units, e.g., Variable costs that material cost. are the additional costs that result from Incremental costs are increasing the output level by one or more units.
Recurring and nonrecurring costs are those that are repetitive and occur Recurring Costs are when an organization produces similar goods or services on a continuing basis, e.g., variable cost and also a periodic fixed cost like office rent. are those that are not repetitive, e.g., Nonrecurring Costs are construction construction cost of the manufac manufacturing turing plant.
Example 2.1
Raw mtl.s Direct labor Rent Admin. salaries Utilities Property tax Property set up
Fixed
Variable
√ √ √ √ √
√
√ √
Recurring
√ √ √ √ √ √
Nonrecurring
√
Cost terminology Direct, indirect (overhead), and standard costs Direct costs that can be attributed to a specific activity or output, e.g., labor and material costs directly associated with the a product. Indirect costs that cannot be attributed to a specific activity or output. Normally, they are allocated through a selected formula, such as by proportion, to the outputs or work activities. Cost of common tools is an example of the indirect costs. Standard costs are planned costs per unit of output that are established in advance of actual production. The standard costs play an important role in cost control and other management functions. One typical use is to measure operating performance by comparing actual cost per unit with the standard unit cost.
Cost terminology Cash and book costs Cash Costs that involve payment of cash and would result in a cash flow. Book Costs that do not involve cash transaction and is reflected in the accounting system as a noncash cost, e.g., depreciation.
Sunk costs Sunk Costs are those that have occurred in the past and irretrievable. Sunk costs are not relevant in the engineering economic analysis because they cannot be changed regardless of what decisions is made now or in the future. The original cost of an equipment is considered to be the sunk cost for a firm in deciding to replace it or not.
Cost terminology Opportunity cost The opportunity cost is incurred because there are only limited resources available. The opportunity cost of the selected alternative is the value of the next best alternative opportunity that is forgone (given up). The value of the next best alternative opportunity that is forgone can be obtained as the answer of the following question: "What would you gain if you selected the best alternative opportunity instead of the choice you are considering ?"
The opportunity cost is forward looking in that it measures the value that the decision maker sacrifices at the time the decision is made and beyond.
Cost terminology Opportunity cost - Example 1
Suppose you have $2,000 in your bank account, you can either keep it in the bank or invest it in stock market. What is the opportunity cost of investing in stock market? You will earn the accured interest of $2,000 if you choose to keep it in the bank instead of investing it in stock market. Therefore, The opportunity cost of investing in stock market is the accured interest of $2,000. Remark: By the forward looking nature of the opportunity cost, $2,000 is not counted into the opportunity cost since it has already been existed before the time of making decision.
Opportunity cost - Example 2
Option 1: Continue your own business with the annual expense of $180,000; Option 2: Stop the business and go out to earn $75,000 per year. What is the opportunity cost of continuing in business over the next year ? If you choose option 2, the amount of benefit you gain = earn $75,000 (the income you earn for choosing Option 2) + $180,000 (the amount you save for choosing Option 2) = $255,000. Therefore, the opportunity cost of continuing in business over the next year is $255,000.
Cost terminology
Life-cycle cost is the sum of all costs related to a product or service during its life span. Acquisition phase: Investment cost, capital investment, working capital. Operation phase: operation and maintenance cost, disposal cost.
Questions
If the capacity of an operation is significantly changed (e.g., a manufacturing plant), the fixed costs will also change. (T/F) A nonrefundable cash outlay (e.g., money spent on a passport) is an example of a sunk cost. (T/F) Stock you own from a bankrupt and out-of-business company is an example of a ... cost. (Choose one: opportunity/sunk/direct/variable)
Answers
If the capacity of an operation is significantly changed (e.g., a manufacturing plant), the fixed costs will also change. (T) A nonrefundable cash outlay (e.g., money spent on a passport) is an example of a sunk cost. (T) Stock you own from a bankrupt and out-of-business company is an example of a ... cost. (sunk)
General economics concepts Price-demand relationships Model 1: Selling price per unit ( p) is a linear function of demand ( D) a , and a > 0, b > 0 p = a bD for 0 D b 2 Model 2: Selling price per unit is independent of demand Maximizing total revenue (TR): (Use Model 1) TR = p. D = (a bD) D = aD bD2 TR is a concave function. (Check the second derivative: d 2 TR = 2b < 0) dD2 1
−
≤ ≤
−
−
−
ˆ from the first derivative: Find the maximizer D dTR ˆ= a a bD D 2 0 = = 2b dD Maximizing profit: (Use Model 1) Profit = TR TC , what is Total Cost (TC )? Assumption: TC = C F + cv D where C F is the fixed cost and cv is the variable cost per unit. Thus, Profit = aD bD2 (C F + cv D)
−
⇒
−
−
−
Maximizing profit cont’d
Maximizing profit: Profit = aD bD2 (C F + cv D) = bD2 + (a cv ) D C F In order for a positive profit to occur and to avoid negative demand, a cv > 0 or a > cv . From the first derivative, we have dProfit a−cv ∗ a c bD D 2 0 . = = = v 2b dD
−
−
−
−
−
−
− − ⇒ Profit Since d dD = −2b < 0 for all D, D 2
∗
2
maximizer of the Profit. Remark: Companies
=
a−cv is 2b
the
may also gain insights from Breakeven points other than the profit-maximizing points. At a breakeven point, total revenue is equal to total cost.
Breakeven analysis
Finding the breakeven points: (Model 1) Solve bD2 + (a cv ) D C F = 0. We obtain two roots:
D =
−
−
−
2 1 2 −(a−cv )±[(a−cv ) −4bC F ] /
−2b
Finding the breakeven points: (Model 2) p is independent of D. TR = p. D TC = C F + cv D F TR = TC D = pC (unique!) −cv
⇒
Example 2.2
A company produces circuit boards used to update outdated computer equipment. The fixed cost is $42,000 per month, and the variable cost is $53 per circuit board. The selling price per unit is p = 150 0.02 D. Maximum output of the plant is 4,000 units per month.
−
Determine the optimum demand that maximizes profit? What is the maximum profit per month? What is the breakeven point? What is the company’s range of profitable demand?
Solution:
p = 150 0.02 D, where a = 150 and b = 0.02; C F = 42, 000; cv = 53. cv = 2, 425. a) D∗ = a− 2b b) Profit = aD bD2 (C F + cv D) =
−
−
−
2
(42, 000 + 53 (2, 425)) 150(2, 425) 0.02(2, 425) −(150−53)±[(150−53)2 −4(0.02)(42,000)]1/2 c) D = −2(0.02) D1 = 481; D2 = 4, 369.
−
−
d) D is profitable in the range
481
− 4, 369.
= $75, 612.50.
Cost driven design optimization
The cost driven design optimization refers to how to design the product in order to minimize the cost. General approach: 1
2
3
Identify the design variable that is the primary cost driver (e.g., the thickness of the material). Write an expression for the cost model in terms of the design variable. Find the optimum value of the design variable to have the minimum cost value.
Example 2.3
The cost of operating a large ship (C O ) varies as the square of its velocity (v); specifically, C O = knv2 , where n is the trip length in miles and k is a constant of proportionality. It is known that at 12 miles/hour the average cost of operation is $100 per mile. The owner of the ship wants to minimize the cost of operation, but it must be balanced against the cost of perishable cargo (C C ), which the customer has set at $1,500 per hour. At what velocity should the trip be planned to minimize the total cost (C T ), which is the sum of the cost of operating the ship and the cost of perishable cargo?
Example 2.3
Solution:
We need to find k. Cost per mile is C nO = kv2 . At v = 12, cost per mile is given as $100. Hence, k = (100 = 0.694. 12) 2
We want to minimize C T where C T = C O + C C = knv2 + 1, 500 nv 1,500 1/3 dC T n ∗ ∗ knv v v = , = = ( ) = 10.25 0 2 1 500 0 2k dv v miles per hour. (Note: Check the second derivative to verify C T is convex, i.e., d C T > 0.) dv
⇒
2
2
−
2
⇒
⇒
Present economy studies
When the time horizon over which projects and alternatives are compared and evaluated is one year or less, and the influence of the time on money can be ignored, these analyses are called present economy studies .
Example 2.4 Ocean water contains 0.8 ounce of gold per ton. Method A: costs $210 per ton of water processed and will recover 80% of the metal. Method B: costs $150 per ton of water processed and will recover 60% of the metal. The two methods require the same investment and are capable of producing the same amount of gold every day. If the extracted gold can be sold for $380 per ounce. Assume that the supply of ocean water is unlimited. On the basis of profit per ounce of gold extracted, which method of extraction should be used?
Example 2.4 Solution:
Method A: Profit per ounce = $380 per ounce ($210 per ton of water)/(0.8 $51.88.
−
Method B: Profit per ounce = $380 per ounce ($150 per ton of water)/(0.8 $67.50.
−
× 80%) = × 60%) =
Select Method B Alternatively, since both methods have the same revenue, we could also choose the method with the minimal cost.
Example 2.5 Machine A and Machine B are being considered for the production of a part of a product. The capital investment associated with the machines are about the same and can be ignored. The important differences between the machines are their production capacities (product rate production hours);
× available
their reject rates (% of parts produced that cannot be sold).
Production rate Hours available for production Reject rate
Machine A 100 parts/hr 7 hrs/day 3%
Machine B 130 parts/hr 6 hrs/day 10 %
Example 2.5
The material cost is $6.00 per part, and all defect-free parts produced can be sold for $12 each. The rejected parts have no value. For either machine, the operator cost is $15.00 per hour and the variable overhead rate for traceable costs is $5.00 per hour. Assume that the daily demand for this part is large enough that all defect-free parts can be sold. Which machine should be selected? What would be the reject rate have to be for Machine B to be as profitable as Machine A?
Example 2.5 Solution:
Since both machines have different total daily revenue and total daily cost, we should base on the daily profit to select the machine. Profit per day
= Revenue per day
− Cost per day
= (Production rate)(Production hrs)($12/part) (1
− reject rate)
×
−(Production rate)(Production hrs)($6/part) −(Production hrs)($15/hr + $5/hr) Machine A: Profit per day = $3,808 per day. Machine B: Profit per day = $3,624 per day. Therefore, select Machine A.
Example 2.5
The breakeven reject rate, X , for Machine B, can be obtained by equating the profits from Machine A and Machine B.
3808
= (130)(6)(12)(1
X =
0.08.
− X ) − (130)(6)(6) − (6)(15 + 5)
So, the reject rate for Machine B can be no higher than 8% for it to be as profitable as Machine A.
