Final Exam Review

xlsx

School

Temple University *

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Course

3103

Subject

Economics

Date

Feb 20, 2024

Type

xlsx

Pages

39

Uploaded by tuq05086

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Mod C: Transportation Models Step 1: Enter the Unit Cost of Moving from Origin 'i' to Destination 'j', Supply at 'i', and Cost Table Newark St. Louis Denver Seattle Portland 101 85 75 45 New Orleans 87 65 85 92 Raleigh 65 77 83 115 Dummy Origin 5 Demand at Destination 330 375 405 265 Step 2: Open the Solver from 'Data' tab in the Top Ribbon and click 'Solve'. Volume Table Dest 1 Dest 2 Dest 3 Dest 4 Origin 1 0 0 135 265 Origin 2 0 350 0 0 Origin 3 300 0 0 0 Origin 4 30 25 270 0 Origin 5 0 0 0 0 Column Total 330 375 405 265 Demand at Destination 330 375 405 265 Total Transportation Cost 64300 1). Tunai enterprise manufacturers fishing equipment. Use the Neward, DE St. Louis, MO Denver, CO Portland, OR $ 101.00 $ 85.00 $ 75.00 New Orleans, LA $ 87.00 $ 65.00 $ 85.00 Raleigh, NC $ 65.00 $ 77.00 $ 83.00 Supply 330 375 405 a). Is the problem balanced No If so/not, which b). What is the total transportation cost? $ 64,300.00
d Demand at 'j' Dest 5 Supply at Origin 400 350 300 325 1375 \ 1375 Dest 5 Row Total Supply at Origin 0 400 400 0 350 350 0 300 300 0 325 325 0 0 0 0 0 following table to find the total shipping costs from each manufacturing facility to their 4 warehous Seattle, WA Capacity $ 45.00 400 $ 92.00 350 $ 115.00 300 265 h area needs to be balanced? Supply by 325
ses.
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Mod C: Transportation Models Step 1: Enter the Unit Cost of Moving from Origin 'i' to Destination 'j', Supply at 'i', and Cost Table Newark St. Louis Denver Seattle Portland 101 85 75 45 New Orleans 87 65 85 92 Raleigh 65 77 83 115 Dummy Origin 5 Demand at Destination 330 375 405 265 Step 2: Open the Solver from 'Data' tab in the Top Ribbon and click 'Solve'. Volume Table Dest 1 Dest 2 Dest 3 Dest 4 Origin 1 0 0 135 265 Origin 2 0 280 70 0 Origin 3 205 95 0 0 Origin 4 125 0 200 0 Origin 5 0 0 0 0 Column Total 330 375 405 265 Demand at Destination 330 375 405 265 Total Transportation Cost 66840 c). What if 95 units needs to be shipped from Raleigh to St. Lo d). Based on your answer to c)., were all demand levels met?
d Demand at 'j' Dest 5 Supply at Origin 400 350 300 325 1375 \ 1375 Dest 5 Row Total Supply at Origin 0 400 400 0 350 350 0 300 300 0 325 325 0 0 0 0 0 ouis, what would the total transportation cost be? $ 66,840.00 Yes If so/not, what warehouse(s) were backordered?
Mod C: Transportation Step 1: Enter the Unit Cost of Moving from Origin 'i' to Destination 'j', Supply at 'i', and Cost Table C P S G 17 22 27 H 23 21 19 V 16 25 18 Origin 5 Demand at Destination 275 225 300 Step 2: Open the Solver from 'Data' tab in the Top Ribbon and click 'Solve'. Volume Table Cranbury Plainfield SaddleBrook Gallway 225 45 Hendron 250 Vienna 275 5 Origin 4 Origin 5 Column Total 275 225 300 Demand at Destination 275 225 300 Total Transportation Cost 15405 2). Solve the problem below using the intuitive least cost metho Cranbury Plainfield Gallway $ 17.00 $ 22.00 Hendron $ 23.00 $ 21.00 Vienna $ 16.00 $ 25.00 Demand 275 225 a). What is the maximum quantity that can be shipped from Gal b). What is the maximum quantity that can be shipped from Vie c). What is the maximum quantity that can be shipped from Gal d). What will be the cost of this assignment?
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e). Were all demand levels met?All demand levels were met f). Was all supply exhausted? If so, which sites?
n Models d Demand at 'j' Dummy Dest 5 Supply at Origin 300 250 280 30 830 \ 830 Dummy Dest 5 Row Total Supply at Origin 30 300 300 250 250 280 280 0 0 0 0 30 0 30 0 od and answer the following: SaddleBrook Supply $ 27.00 300 $ 19.00 250 $ 18.00 280 300 llway to Plainfield? 225 enna to Cranbury? 275 llway to Saddlebrook? 45 $ 15,405.00
If so/not which site(s) was/were not? All supply was exhausted
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MOD B: LINEAR PROGRAMMING - 3 variable Step 1: Enter the LP model: nuggets mini-nuggets chips RHS Profits 5.50 4.25 6.75 pine bark 2.00 2.50 3.00 400 machine time 2.00 1.00 2.00 440 labor time 3.00 2.00 1.00 475 mini nug required 1.00 40 chips required 1.00 30 Step 2: Calculate the LHS in Constraints nuggets mini-nuggets chips LHS pine bark 2 2.5 3 400 machine time 2 1 2 310 labor time 3 2 1 425 mini nug required 0 1 0 40 chips required 0 0 1 30 Step 3: Run the Solver from the 'Data' tab in the Top Ribbon Optimal Solution x1 x2 x3 Profit 105.00 40.00 30.00 950.00 3). A garden store prepares various grades of pine bark for mulch: nugget The mulchingprocess requires pine bark (400 pounds available), machine Each bag of nuggets requires 2 pounds of bark, 2 minutes of machine tim Each bag of chips requires 3 pounds of bark, 2 minutes of machine time, a a). How many bags of each type of mulch should be made to maximize pr b). What is the optimal profit? 950.00 c). For this solution, what is the marginal value (shadow price) of pine bar d). What is the marginal value (shadow price) of machine time? e). What impact would decreasing labor hours by 30 have on the overall p
e Maximization problem Variable Cells Final Reduced Objective Cell Name Value Cost Coefficient Objective Function $B$22 x1 105 0 5.5 $C$22 x2 40 0 4.25 $D$22 x3 30 0 6.75 Constraints Final Shadow Constraint Cell Name Value Price R.H. Side $E$14 pine bark 400 2.75 400 RHS $E$15 machine t 310 0 440 400 $E$16 labor time 425 0 475 440 $E$17 mini nug r 40 -2.625 40 475 $E$18 chips requ 30 -1.5 30 40 30 ts, mini-nuggets, and chips, with profits of $5.50, $4.25, and $6.75 per bag respectively. e time (440 minutes available), labor time (475 minutes available). me, and 3 minutes of labor. Each bag of mini-nuggets requires 2.5 pounds of bark, 1 minute of machine time, and and 1 minute of labor. The garden store is commited to producing atleast 30 bags of chips and 40 bags of mini- rofit? Nuggets = 105.00 Mini-nuggets = 40.00 Chips = 30.00 rk? 2.75 0 profit? there would be no impact on profit
Allowable Allowable Increase Decrease 1E+030 1 2.625 1E+030 1.5 1E+030 Sensitivity report Allowable Allowable Increase Decrease 33.33333 210 1E+030 130 1E+030 50 84 28.57143 70 14.28571 d 2 minutes of labor. -nuggets.
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Microsoft Excel 16.0 Sensitivity Report Worksheet: [Final Exam Review.xlsx]3-variable_Max Report Created: 12/8/2023 2:25:56 PM Variable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$22:$D$22 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $E$14 Constraint 1 LHS 400 2.75 400 33.33333333 210 $E$15 Constraint 2 LHS 310 0 440 1E+030 130 $E$16 Constraint 3 LHS 425 0 475 1E+030 50 $E$17 Constraint 4 LHS 40 -2.625 40 84 28.57142857 $E$18 Constraint 5 LHS 30 -1.5 30 70 14.28571429
MOD B: LINEAR PROGRAMMING - 2 variable Minimizatio Step 1: Enter the LP model 1 2 Capacity (RHS) Costs 22,000.00 34,000.00 Objective Function HIGH 410.00 480.00 38000 MID 320.00 450.00 32000 LOW 220.00 390.00 29000 refinery 1 req 1.00 45 Constraint 5 Step 2: Calculate the LHS in Constraints (Current Usage) 1 2 Current Usage (LHS) Capacity (RHS) HIGH 410 480 41957.6923076923 38000 MID 320 450 36438.4615384615 32000 LOW 220 390 29000 29000 refinery 1 req 1 0 45 45 Constraint 5 0 0 0 0 Step 3: Run the Solver from the 'Data' tab in the Top Ribbon Optimal Solution x1 x2 Costs 45 48.97435897 $2,655,128.21 2,611,538.46 *cost for low but 500 *cost for e* 4). Noble Energy, located in Houston Texas, a small petroleum company, ow Refinery 1 costs $22,000 per day to operate, and it can produce 410 barrels o Refinery 1 (because of employee contractual commitments) needs to operate Refinery 2 is newer and more modern. It costs $34,000 per day to operate, a The company has orders totaling 38,000 barrels of high-grade oil, 32,000 ba How many days should it run each refinery to minimize its costs and still ref a). This is a minimization problem. b). How many days should it run each refinery to minimize its costs and still Refinery 1 45 Refinery 2 49 c). What is the optimal cost? $2,655,128.21 d). Are there any unused resources (capacity)? NO
e). What impact would there be if the demand for low-grade oil was decreas
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on problem 0 less* wns two refineries. of high-grade oil, 320 barrels of medium-grade oil, and 220 barrels of low-grade oil each day. e at least 45 days. and it can produce 480 barrels of high-grade oil, 450 barrels of medium-grade oil, and 390 barrels of low arrels of medium-grade oil, and 29,000 barrels of low-grade oil. fine enough oil to meet its orders? l refine enough oil to meet its orders during there 4-month planning period? If so/not, what constraint and how much or not? None, there are no unused resources
sed by 500 barrels? cost would decrease by $43,589.75
w-grade oil each day.
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MOD B: LINEAR PROGRAMMING - 2 variable Minimizatio Step 1: Enter the LP model 1 2 Capacity (RHS) Costs 22,000.00 34,000.00 Objective Function HIGH 410.00 480.00 38000 MID 320.00 450.00 32000 LOW 220.00 390.00 28500 refinery 1 1.00 45 Constraint 5 Step 2: Calculate the LHS in Constraints (Current Usage) x1 x2 Current Usage (LHS) Capacity (RHS) Constraint 1 410 480 41342.3076923077 38000 Constraint 2 320 450 35861.5384615385 32000 Constraint 3 220 390 28500 28500 Constraint 4 1 0 45 45 Constraint 5 0 0 0 0 Step 3: Run the Solver from the 'Data' tab in the Top Ribbon Optimal Solution x1 x2 Costs 45 47.69230769 $2,611,538.46 $ 2,655,128.21
on problem e). What impact would there be if the demand for low-grad
de oil was decreased by 500 barrels? cost would decrease by $ 43,589.74
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Mod D: Waiting Line Models Solving M/M/S problems Per hour Arrival rate 3 λ Lambda Service rate 7.5 μ Mu # servers (lambda/mu)^n/cumsum (n-1)n/(1-(lam/n*mu)) UtlizationP0 Lq Ls Ws Wq 0 1.00 1 0.40 1.00 0.67 0.40 0.60 0.267 0.667 0.222 0.089 2 0.08 1.40 0.10 0.20 0.67 0.017 0.417 0.139 0.006 3 0.01 1.48 0.01 0.13 0.67 0.001 0.401 0.134 0.000 4 0.00 1.49 0.00 0.10 0.67 0.000 0.400 0.133 0.000 5 8.5333333E-05 1.491733333 9.27536232E-05 0.08 0.67 5E-06 0.4 0.1333 2E-06 6 5.6888889E-06 1.491818667 6.0952381E-06 0.066667 0.67 3E-07 0.4 0.1333 1E-07 7 3.2507937E-07 1.491824356 3.44781145E-07 0.057143 0.67 1E-08 0.4 0.1333 5E-09 8 1.6253968E-08 1.491824681 1.71094403E-08 0.05 0.67 6E-10 0.4 0.1333 2E-10 9 7.2239859E-10 1.491824697 7.55998523E-10 0.044444 0.67 2E-11 0.4 0.1333 8E-12 10 2.8895944E-11 1.491824698 3.00999412E-11 0.04 0.67 8E-13 0.4 0.1333 3E-13 Utilization = how busy Po = idle or not busy Lq = customer waiting in line Ls = total customers in system (waiting in line and being serviced) Wq = Waiting time in line Ws = total time in system (waiting in line and being serviced)
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5). A team of technicians at XYZ inc. repair computer drives that break down at an average rate of λ It takes the team an average of μ = 8 minutes to complete repairs based on a negative exponential d The team cost per hour is $420. a). What is the utilization rate for the team? 40% b). What is the average time before the team can have the drives operational? 13.33 c). How much time is spent waiting for service? 5.33 d). How many computer drives are likely to be waiting for service at any one time? 0.27 e). What is the total wait cost per day based on a 9 hour operation? $ - Total daily cost $ 4,680.00 (# of servers*b*a)+(Lq*c*a) a Time (hr) per day 9 b team rate $ 420.00 c lost cost $ 375.00
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= 3 per hour (approximately Poisson in nature). distribution. Lost productivity due to breakdowns cost the company $375 per hour. minutes 0.266
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Waiting Line Models Solving M/M/1 problems Per hour Arrival rate 4.5 λ Lambda 3 40 Service rate 8.57143 μ Mu 9 120 Number # serves Utlization P0 Lq Ls Ws Wq 0 1 0.525 0.475 0.580263 1.105263 0.245614 0.128947 1 2 Utilization = how busy 3 Po = idle or not busy 4 5 Lq = customer waiting in line 6 Ls = total customers in system (waiting in line and being serviced) 7 8 Wq = Waiting time in line 9 Ws = total time in system (waiting in line and being serviced) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
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34 35
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6). Ashley runs a dog grooming service, and has asked you to analyz 3 dogs every 40 min She has gathered data and calculated that she can groom a dog in a 9 dogs every 120 min She also noted that about 3 dogs every 40 minutes arrive to her sho 4.5 dogs every 60 min a). How long on average, will a dog wait for grooming? 7.74 Probabilities Probability Cumulative Probability b). In total, how long will a dog spend in the shop waiting to be groo 0.475000 0.475000 0.249375 0.724375 c). Ashley's shop is fairly small, and he only has space for 4 dogs to b 0.130922 0.855297 What is the probability that there will be more than 4 dogs at any ti 0.068734 0.924031 0.036085 0.960116 0.018945 0.979061 0.009946 0.989007 0.005222 0.994229 0.002741 0.996970 0.001439 0.998409 0.000756 0.999165 0.000397 0.999562 0.000208 0.999770 0.000109 0.999879 0.000057 0.999937 0.000030 0.999967 0.000016 0.999983 0.000008 0.999991 0.000004 0.999995 0.000002 0.999997 0.000001 0.999999 0.000001 0.999999 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000 0.000000 1.000000
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0.000000 1.000000 0.000000 1.000000
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ze the current 'line situation'. about 7 minutes, based on a negative exponential distribution. op, during all operating hours, based on a Poisson distribution. minutes. omed? 14.74 minutes. be in the shop before it gets crowded. time? 4.0%
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Mod D: Waiting Line Models Solving M/M/S problems Per hour Arrival rate 4.5 λ Lambda Service rate 10.0000 μ Mu # servers (lambda/mu)^n/cumsum (n-1)n/(1-(lam/n*mu)) UtlizationP0 Lq Ls Ws Wq 0 1.00 1 0.45 1.00 0.82 0.45 0.55 0.368 0.818 0.182 0.082 2 0.10 1.45 0.13 0.23 0.63 0.024 0.474 0.105 0.005 3 0.02 1.55 0.02 0.15 0.64 0.002 0.452 0.100 0.000 4 0.00 1.57 0.00 0.11 0.64 0.000 0.450 0.100 0.000 5 0.00015377344 1.568146094 0.000168981799 0.09 0.638 1E-05 0.45 0.1 2E-06 6 1.1533008E-05 1.568299867 1.24681166E-05 0.075 0.638 6E-07 0.45 0.1 1E-07 7 7.4140765E-07 1.5683114 7.92344048E-07 0.064286 0.638 3E-08 0.45 0.1 8E-09 8 4.170418E-08 1.568312142 4.41898596E-08 0.05625 0.638 2E-09 0.45 0.1 4E-10 9 2.085209E-09 1.568312183 2.19495684E-09 0.05 0.638 7E-11 0.45 0.1 2E-11 10 9.3834405E-11 1.568312185 9.82559216E-11 0.045 0.638 3E-12 0.45 0.1 7E-13 Utilization = how busy Po = idle or not busy Lq = customer waiting in line Ls = total customers in system (waiting in line and being serviced) Wq = Waiting time in line Ws = total time in system (waiting in line and being serviced)
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Ashley wants to reduce the wait time, and is considering two options. d). What would the wait time be with a helper? 4.9 minutes The first would be to hire a helper, who would assist her, and reduce the grooming time from 7 to 6
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6 minutes. (NOTE: the helper only helps and does not work alone)
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Mod D: Waiting Line Models Per hour Arrival rate 4.5 λ Lambda The second option is Service rate 12.0000 μ Mu e). What would the w # servers Utlization P0 Lq Ls Wq Ws f). Assuming both op 1 0.375 0.625 0.11 0.49 0.03 0.11 which should Jill choo Utilization = how busy Po = idle or not busy Lq = customer waiting in line Ls = total customers in system (waiting in line and being serviced) Wq = Waiting time in line Ws = total time in system (waiting in line and being serviced) olving M/D/1 problems (Constant Time, car wash
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s to rent an Auto-Groomer, a new machine that would allow Ashley to groom a dog in a constant 5 minutes. wait time be with Auto-Groomer? 1.5 minutes ptions cost the same amount and all else is equal, ose to deliver the shortest wait time? Choose the Auto-Groomer
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7). Tommy Choi is the quality assurance director at Bimbo Bakery Corporate R&D Facility. They are investigating a moisture issue, where excessive moisture would create microbial gro Moisture (mean) for the bread is about 39.5% +/- 3.2%. Tommy conducted an extensive asses 3 models of Mettler Tolendo moisture analyzers are being evaluated: Model HC 103, priced at $ 4,500.00 Mean = 39.60% , standard deviation at Model HE 53, priced at $ 3,000.00 Mean = 38.50% , standard deviation at Model HX 204, priced at $ 12,000.00 Mean = 38.90% , standard deviation at a). What is the design tolerance (difference from Upper to Lower Specification Limits), from T b). What is the Cp, Cpk of the HC 103 1.255 1.216 c). Cp, Cpk of the HE 53 1.641 1.128 d). Cp, Cpk of the HX 204 1.939 1.576 e). Based on Cpk, which moisture analysis should Tommy be selectinHC 103 HC 103 Voice of consumer CP range 0.032 UCL-LCL 0.064 USL 0.427 6 * SD 0.051 Target 0.395 CP 1.255 LSL 0.363 CPK Voice of process UCL-avg 0.031 avg 0.396 3 * SD 0.0255 SD 0.0085 CPK 1 1.216 avg-LCL 0.033 3 * SD 0.0255 CPK 2 1.294 HX 204 Voice of consumer CP range 0.032 UCL-LCL 0.064 USL 0.427 6 * SD 0.033 Target 0.395 CP 1.939 LSL 0.363 CPK
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Voice of process UCL-avg 0.038 avg 0.389 3 * SD 0.0165 SD 0.0055 CPK 1 2.303 avg-LCL 0.026 3 * SD 0.0165 CPK 2 1.576
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owth negatively affecing the shelf life of their products. sment, and found that gives the optimal shelf life. 0.85% 0.65% 0.55% Tommy's specifications 6.4% HE 53 Voice of consumer CP range 0.032 UCL-LCL 0.064 USL 0.427 6 * SD 0.039 Target 0.395 CP 1.641 LSL 0.363 CPK Voice of process UCL-avg 0.042 avg 0.385 3 * SD 0.0195 SD 0.0065 CPK 1 2.154 avg-LCL 0.022 3 * SD 0.0195 CPK 2 1.128
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8). Stephanie Hill has been hand painting wooden Christmas ornaments for several years. Recently, she has hired some friends to help her increase the volume of her business. In checking the quality of the work, she notices that some slight blemishes occasionally are ap A sample of 22 pieces of work resulted in the following number of blemishes on each piece: 1,2,1,5,0,3,2,6,4,1,2,1,0,1,7,1,6,2,2,1,4,3. Develop upper and lower control limits for the number of blemishes on each piece based on a a). What are the UCL 5.66 and the LCL 0 ? b). Does the number of blemishes fall within an acceptable limit? No If not, whic pieces # of blemishes 1 1 z 2 2 2 n 22 3 1 avg 2.5 4 5 avg +- z * SQRT(avg) 5 0 6 3 UCL 5.662278 7 2 LCL -0.662278 0 8 6 9 4 10 1 11 2 12 1 13 0 14 1 15 7 16 1 17 6 18 2 19 2 20 1 21 4 22 3
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pparent. a confidence level of 95.45% (z=2). ch sample(s) are outside the calculated limit samples 8,15,17
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