The price drivers pay for gasoline often varies a great deal across regions throughout the United States. The following data show the price per gallon for regular gasoline for a random sample of gasoline service stations for three major brands of gasoline (Shell, BP, and Marathon) located in 11 metropolitan areas across the upper Midwest region(OhioGasPrices.com website, March 18, 2012). Metropolitan Area Shell BP Marathon Akron, OH 3.77 3.83 3.78 Cincinnati, OH 3.72 3.83 3.87 Cleveland, OH 3.87 3.85 3.89 Columbus, OH 3.76 3.77 3.79 Ft. Wayne, IN 3.83 3.84 3.87 Indianapolis, IN 3.85 3.84 3.87 Lansing, MI 3.93 4.04 3.99 Lexington, KY 3.79 3.78 3.79 Louisville, KY 3.78 3.84 3.79 Muncie, IN 3.81 3.84 3.83 Toledo, OH 3.69 3.83 3.86 Write a SAS code to perform following tasks: 1. ANOVA Analysis, exploratory data analysis of the data and evaluate the tenability of the analysis of variance assumptions. 2. Perform at least three appropriate multiple comparisons tests for the significant main effects. 3. Create at least two pairs of suitable orthogonal contrasts and perform the appropriate
title1 'Single Factor CRD Experiment'; data Oilprice; input Price Block$ Company$ @@; datalines; 3.77 Akron Shell 3.72 Cincinnati Shell 3.87 Cleveland Shell 3.76 Columbus Shell 3.83 FtWaye Shell 3.85 Indianapolis Shell 3.93 Lansing Shell 3.79 Lexington Shell 3.78 Louisville Shell 3.81 Muncie Shell 3.69 Toledo Shell 3.83 Akron BP 3.83 Cincinnati BP 3.85 Cleveland BP 3.77 Columbus BP 3.84 FtWaye BP 3.84 Indianapolis BP 4.04 Lansing BP 3.78 Lexington BP 3.84 Louisville BP 3.84 Muncie BP 3.83 Toledo BP 3.78 Akron Marathon 3.87 Cincinnati Marathon 3.89 Cleveland Marathon 3.79 Columbus Marathon 3.87 FtWaye Marathon 3.87 Indianapolis Marathon 3.99 Lansing Marathon 3.79 Lexington Marathon 3.79 Louisville Marathon 3.83 Muncie Marathon 3.86 Toledo Marathon ; run; proc means data=WORK.Oilprice chartype mean std min max n vardef=df; var Price; class Company; run; proc glm data=WORK.Oilprice PLOTS=DIAGNOSTICS(UNPACK); class Block Company; model Price=Company Block; lsmeans Company / adjust=tukey pdiff alpha=.05; lsmeans Company/ adjust=bon pdiff alpha=.05; lsmeans Company / adjust=scheffe pdiff alpha=.05; run;
Trucks arrive at Chittagong for loading onto a ferry at a rate of 200 per 8-hour day, and the Port can process an average of 220 trucks per day. Arrivals follow a Poisson distribution and service times follow an exponential distribution. a. What is the average number of trucks in the queue? b. What is the average time a truck spends in the queue? c. What is the average number of trucks in the system? d. What is a truck’s average time in the system? e. What is the average number of trucks in the server? f. What is the average time a truck spends in the server? g. What is the probability that a truck will be loading when an arrival occurs? h. What is the probability that there is no one in the system?
Given, Arrivals: λ = 200 / 8hr or 25 every hour Services: μ= 220 trucks per day = 220/8=27.5 trucks per hour P=λ/μ=200/220=0.909<1 So, we can apply littles law in this case, a) Average number of trucks in the queue=L_q=λ^2/(μ(μ-λ))=〖25〗^2/(27.5(27.5-24))=6.49 per hour b) Average time a truck spends in the queue= W_q=λ/μ 1/((μ-λ))=0.909×1/3.5=0.259 hours c) Average number of trucks in the system=L=λ/((μ-λ))=25/((27.5-24))=7.143 per hour d) Average time a truck spends in the system= W=1/((μ-λ))=1/3.5=0.285 hours e) Average number of trucks in the server= μ=25 per hour f) Average time a truck spends in the server=1/μ=1/25=0.04 hour g) probability that a truck will be loading when an arrival occurs=λ/μ=200/220=0.909
You are manufacturing two different models of USB key. One stores 3GB of data and the other 1GB of data. The profit on the 3GB key is 80 cents and the profit on the 1GB key is 60 cents. The manufacture of the 3GB key takes 30 minutes, while the 1GB key takes 20 minutes. There are 300 hours of manufacturing time available. In addition, each key must be packaged, and it takes 12 minutes to package a 3GB key and 15 minutes to package a 1GB key. The packaging department has 200 hours available. Given the demand, it has been decided to manufacture at least twice as many 1GB keys as 3GB keys. Provide a linear programming formulation of this problem.
Let, Number of 3GB USB key = X Number of 1GB USB key = Y Objective Function: The objective of this linear programming is profit maximization and The profit on the 3GB key is 80 cents and the profit on the 1GB key is 60 cents. Maximize Z=0.80X+0.60Y Constraint Function: Total available time for manufacturing = 300 hours= 18000 minutes Total available time for packaging= 200 hours= 12000 minutes Manufacturing time constraints 30X+20Y≤18000 Packaging time constraint 12X+15Y≤12000 Demand Constraint 2X≤Y Non-Negativity Constraint X≥0 and Y≥0