ISE 2014 Chapter 2 Section 2.4

Transcription

1 ISE 2014 Chapter 2 Section 2.4 Cost Driven Design Optimization - Problem 2-22 Set Up The cost of operating a large ship (Co) varies as the square of its velocity (v); specifically, Co = knv 2, where n = trip length in miles and k = 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 the perishable cargo (Cc ) which the customer has set at $1,500 per hour. At what velocity should the trip be planned to minimize the total cost (CT ), which is the sum of the cost of operating the ship and the cost of perishable cargo? 1

2 Pr Solution (per approach on p ) 1. The design variable is v, the velocity of the ship. 2. Our objective is to find the velocity that minimizes total cost. C T = C O + C C What is the value of Co? What is the value of Cc? 2

3 3. The first derivative of the cost model with respect to velocity is: dc T dv = 0 = 4. Solving the first derivative equation for v yields: 5. The second derivative is: d 2 C T dv 2 = 3

4 Section 2.4. Cost Driven Design Optimization Another Example Automobile doors can be reinforced for the added safety of passengers inside. Measured in terms of door weight (W), safety (S) can be described with the following linear function: S = 8W + 80 But added weight reduces the fuel economy of the automobile and its aesthetic appearance. The resultant penalty cost (C) for safety, expressed as a function of weight, is graphed below. Cost in Dollars (C) C = W Pounds of Weight (W) (a) If the automobile must have doors, how much should they weigh (i.e., what is the optimal weight of the door)? (b) Over what range of weight is the added safety economical (cost effective)? 4

5 Section 2.5. Present Economy Studies - Cost analyses where the influence of time on money is not a significant consideration. Examples: Material Selection (also tool selection) Make versus Purchase Studies Alternative Machine Speeds Optimizing Design Parameters - bridge span length - design tolerances and quality 5

6 ISE 2014 Present Economy Sec. 2.5 Rule 1 Maximize profit (revenues and/or profits are not constant) Input is identical Y' Alt. A 0.97Y Y' Alt. 0.90Y B Examples (1) mining operations where the body is fixed in size (2) equipment capacity (e.g., aircraft seating) is fixed 5.1

7 ISE 2014 Present Economy Sec. 2.5 Rule 2 Minimize costs/expenses (revenues are constant or non-existent) X/0.9 X/0.8 Alt. C Alt. D X' X' Output is identical Examples (1) hp or kw delivered to a process or workload (2) production to satisfy known demand (e.g., concrete at a building site) 5.2

8 Section 2.5. Present Economy The Cost of Drying Our Hands Have you ever wondered why so many electric warm-air hand dryers are replacing paper towel dispensers in restrooms nowadays? They re cheaper, that s why! Assume the cost of a single paper towel is $0.005 (many of us use two). If the electric hand-dryer is rated at 1000 watts, let s compute the cost of its use for a 15-second cycle. Given: Then: Or 1 kw = 1000 watts; electricity costs $0.07/kWh Cost for 1 blow-dry cycle equals 15seconds 1hour $0.07 1kW 60seconds / minute 60minutes kw hr $ kW hr $ kw hr What about the cost of the hand-dryer itself versus the towel dispenser? What about the comfort, convenience and time required to dry your hands? Are electric hand-dryers the way to go? YES? NO? 6

9 Section Material Selection Example After machining, the finished volume of a certain metal part is 0.17 cubic inch. Data for two types of metal being considered for manufacturing the part are given below: Brass Aluminum Machining time/piece (min) Cost of material ($/lb) Scrap value ($/lb) 0.24 none Cost of operator ($/hr) Density of material (lb/ in 3 ) Volume of raw material (in 3 ) Determine the cost per part for both types of material and recommend which material to use. 7

10 Example Continued Brass: Labor: (0.64 min/pc)($12.00/hr)(1 hr/60 min) = $0.128/pc Material: ($0.96/lb) (0.31 lb/ in 3 ) (0.3 in 3 ) -(0.3 in 3 / pc -.17 in 3 / pc)($0.24/lb)(0.31 lb/in 3 ) = $0.079/pc Total Cost = $0.207/pc Aluminum: Labor: (0.42 min/pc)($12.00/hr)(1 hr/60 min) = $0.084/pc Material: ($0.52/lb - $0.00/lb)(0.10 lb/in 3 ) (0.45 in 3 ) = $0.023/pc Total Cost = $0.107/pc (Choose Aluminum to minimize total cost) 8

11 Section Example - Make versus Purchase Studies Given: Find: The company is currently purchasing a part and is considering manufacturing the part in-house. This company is not operating at full capacity, and no other use for the excess capacity is contemplated. Unit costs are given below. Manufacture Direct Material $1.50 Direct Labor 1.35 Overhead - variable 0.90 Overhead - fixed Purchase Total Unit Cost $8.00 $7.50 Should the part be made in-house or purchased?

12 Example Continued Assumptions: - Utilizing the excess capacity has no opportunity costs. - This product will not change fixed overhead for the plant. Solution: Focus on differences - what really changes? Manufacture Purchase Difference Direct Material $1.50 $1.50 Direct Labor Overhead - variable Overhead - fixed Total Unit Cost $8.00 $7.50 $3.75 The incremental cost to make the product in-house is actually $3.75 per unit versus $7.50 to purchase. Q: What happens if fixed overhead changes as a result of producing the product in house? 10