Problem Set 3 Hints

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Rutgers University, Newark *

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22:839:670

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Industrial Engineering

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Dec 6, 2023

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docx

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Boston University Metropolitan College AD 690 Problem Set 3 Transportation Manager’s Dilemmas Hints Hints for Question 1: Set up a Problem Solution Table. Below is a format that you can use. Problem Solution Table Box Trucks Shipment size for each supplier to each store = X cartons Number of shipments per year = X cartons per year/X per truck = X per year Annual shipping costs using box trucks = X shipments/year x ? stores x ? suppliers x $? = $? Average inventory at each store = X/2 = X cartons Annual inventory holding cost = X cartons x 0.40 x 10 x 5 = $? Total annual cost = $? + $? = $? Vans Shipment size for each supplier to each store = X cartons Number of shipments per year = X cartons per year/X per van= X per year Annual shipping costs using large trucks = X shipments/year x ? stores x ? suppliers x $? = $? Average inventory at each store =X/2 = X cartons Annual inventory holding cost = X cartons x 0.40 x 10 x 5 = $? Total annual cost = $? + $? = $? The solution is to use ? for the deliveries. The difference is $? – $? = $? in favor of ? . Hint: The SCM expects a written report outlining the logic of the recommendation. The recommendation might suggest that potential savings is sufficient to make a change. There may be other costs that have not been considered like training and operational aspects. For example, if you recommend vans, they are smaller and more compact than trucks, so it may be safer for drivers during extreme weather conditions. However, remember that vans do not carry as much load as box trucks, so the capacity of the vans is less. Hints for Question 2: Note that the Poisson distribution is a model that describes the probability of obtaining X successes of a Poisson process based on an expected value. The textbook suggests the following model: P ( X = x )= λ x e − λ x! ,x = 0,1,2,... The transportation manager’s situation per a Poisson Distribution (calculations) is as follows: The expected value of X is: µ (mean) = λ = 2 The standard deviation of X is: σ = λ ½ = 2 ½ =1.4 2σ = 2.8; µ ± 2σ=2 ± 2.8 or 0-4.8; This would include 95% of the possibilities. Therefore, you should consider 0 through 5 in your analysis.
Boston University Metropolitan College Hints for Question 2 (continued): Set up a Problem Solution Table. Below is a format that you can use. You will need to calculate values for the empty cells (Columns 2, 4, 6, 8, 10). ƛ = 2 Expected Fees based on Number of Trucks No. P(x≥) P(x) 0 2 4 6 8 10 0 0.1353 0.1353 0 1 0.4060 0.2707 54.14 2 0.6767 0.2707 108.28 3 0.8571 0.1804 108.24 4 0.9473 0.0902 72.16 5 0.9834 0.0361 36.1 0.9834 $379.92 Storage Fee $100.00 Truck Fee $200.00 Hint: Fees for storage are equal to P(x) times the number of trucks times $100. Fees for trucks are equal to P(x) times the number of trucks times $200. Hints for Question 3: The learning curve concept is interesting. It is important to note that the time to accomplish a task or a series of tasks declines as more experience (learning) is gained. The concept suggests that the required time decreases at a given rate as cumulative experience doubles. The phenomenon is attributable to the fact that it often takes less time to accomplish tasks as the workers gain more and more experience in performing their tasks. Less time means lower costs. Hint: You might consider using the following variable names and equations: Y = ax n x = the unit number Y = number of hours to achieve the x th unit a = number of hours to achieve the first unit n= log b/log 2, b = the learning rate (based on doubling) a = number of hours for the first day (trip) = 12 hours n= log b/log 2, b = 0.85 Y(x) = (12 hours) * (x) log 0.85/log2 ; Log 0.85/log 2= -? Y(x) = 12 hours * x -? Hint: Set up a Problem Solution Table. Below is a format that you can use. You will need to calculate values for the empty cells (Columns for Hours, Training Hours, Costs and Cost Sum). . Y(x) x x n Δ = Y(x) - 8 Hours Δ * 20 Drivers = Training Hours Overtime rate = $40 x 1.5=$60 Costs = $60 * Training Hours 12.0 1 1 4.0 80.0 $4800.00 10.2 2 0.8 5 9.2 3 0.7 7 8.6 4 0.7 2
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