Question

In: Operations Management

p .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .5000 .5040 .5080 .5120...

p .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998

A barista shop in NYC is open 200 days per year and sells an average of 77 pounds of Koko coffee beans a day. (Demand is assumed to be normally distributed with a standard deviation of 15 pounds per day). After ordering, with a fixed cost of $19 per order, beans are always delivered from Uganda in exactly 4 days. Annual holding costs for the beans are $3/pound.

Note: round all answers to two (2) decimal places and don't type commas or separators, only a period to indicate decimals.

a) The economic order quantity (EOQ) for Koko coffee beans is: Blank 1. Fill in the blank, read surrounding text. pounds per order.

b) The total annual holding costs of stock for Koko coffee beans are: $ Blank 2. Fill in the blank, read surrounding text. .

c) The total annual ordering costs for Koko coffee beans are: $ Blank 3. Fill in the blank, read surrounding text. .

d) Assume that management has specified that no more than a 1% risk of stockout during lead time is acceptable. The reorder point (ROP) is: Blank 4. Fill in the blank, read surrounding text. pounds.

e) The safety stock needed to attain a 1% risk of stockout during lead time is: Blank 5. Fill in the blank, read surrounding text. pounds.

Expert Solution

A barista shop in NYC is open 200 days per year

Per day demand =DD= 77 pounds

Annual Demand = D=200*77= 15400 pounds

standard deviation= σ= 15 pounds per day

Cost of Ordering =Co=$19 per order,

Lead time= L= 4 days

Annual holding cost = Ch= $3/pound

a.

EOQ is

=441.66 Pounds per order

a.The economic order quantity (EOQ) for Koko coffee beans is: 441.66 Pounds per order

b. total annual holding costs is

= (Q/2)* Ch

= (441.66/2)*3

=662.49

b.The total annual holding costs of stock for Koko coffee beans are $662.49

c. The total annual ordering costs is

=number of orders * Co

No. of orders = (D/Q) = 15400/441.66 = 34.87

= 34.87 *19

=662.53

c.The total annual ordering costs for Koko coffee beans are: $662.53

d.

ROP can be calculated as

ROP = SS + Lead time demand

= Z * σL + DD * Lead time

1% stockout is allowed so expected availability = 99% and Z for 99% CSL = 2.33

σ = Standard deviation = 15 pounds per day

σL = Standard deviation of demand over lead time = = =30

ROP = 2.33 *30 +77*4

=377.90

d.The reorder point (ROP) is:377.90 pounds

e. The safety stock needed to attain a 1% risk of stockout during lead time is

=Z * σL

= 2.33 * 30 = 69.9

e. The safety stock needed to attain a 1% risk of stockout during lead time is 69.9 pounds

======================================END==============================================

I tried best to answer your question with all my expertise. If you liked answer, please give an Up-vote.

This would be encouraging for me, Thank you !!!

keosha answered 11 months ago

Given the following program: Mailbox Contents 00 505 01 206 02 307 03 902 04 000...

Given the following program: Mailbox Contents 00 505 01 206 02 307 03 902 04 000 05 007 DAT 06 002 DAT 07 000 DAT What is the value in the calculator after the instruction in mailbox 04 is completed?

PSY 312 Power Analysis Homework h2 .02 .04 .06 .09 .12 .16 .20 .25 .30 ...

PSY 312 Power Analysis Homework h2 .02 .04 .06 .09 .12 .16 .20 .25 .30 .36 .42 .49 power .30 93 53 34 24 18 14 11 9 8 7 6 5 .40 132 74 47 33 24 19 15 12 10 8 7 6 .50 170 95 60 42 30 23 18 14 12 9 8 7 .60 257 143 90 62 45 34 24 20 16 13 11 9 .70 300 167 105 72 52 39 ...

observation_date FEDFUNDS 1954-07-01 0.80 1954-08-01 1.22 1954-09-01 1.06 1954-10-01 0.85 1954-11-01 0.83 1954-12-01 1.28 1955-01-01 1.39...

observation_date FEDFUNDS 1954-07-01 0.80 1954-08-01 1.22 1954-09-01 1.06 1954-10-01 0.85 1954-11-01 0.83 1954-12-01 1.28 1955-01-01 1.39 1955-02-01 1.29 1955-03-01 1.35 1955-04-01 1.43 1955-05-01 1.43 1955-06-01 1.64 1955-07-01 1.68 1955-08-01 1.96 1955-09-01 2.18 1955-10-01 2.24 1955-11-01 2.35 1955-12-01 2.48 1956-01-01 2.45 1956-02-01 2.50 1956-03-01 2.50 1956-04-01 2.62 1956-05-01 2.75 1956-06-01 2.71 1956-07-01 2.75 1956-08-01 2.73 1956-09-01 2.95 1956-10-01 2.96 1956-11-01 2.88 1956-12-01 2.94 1957-01-01 2.84 1957-02-01 3.00 1957-03-01 2.96 1957-04-01 3.00 1957-05-01 3.00 1957-06-01 3.00 1957-07-01 2.99 1957-08-01 3.24 1957-09-01 3.47...

car.txt 0001|Mercury|LeSabre|2005|Brown|11000|2019-09-01 0002|Chevrolet|Aveo|2013|Blue|12000|2019-09-02 0003|Datsun|240Z|1979|Orange|13000|2019-09-03 0004|Ford|Galaxie|1964|Black|14000|2019-09-04 0005|Porsche|Boxster|2014|Green|15000|2019-09-05 0006|Honda

car.txt 0001|Mercury|LeSabre|2005|Brown|11000|2019-09-01 0002|Chevrolet|Aveo|2013|Blue|12000|2019-09-02 0003|Datsun|240Z|1979|Orange|13000|2019-09-03 0004|Ford|Galaxie|1964|Black|14000|2019-09-04 0005|Porsche|Boxster|2014|Green|15000|2019-09-05 0006|Honda|Insight|2007|Silver|16000|2019-09-06 1) Design and implement a class of "car" with constructor and destructor, with member variable for each field, and method for get and set each field. (2) Read the file (car.txt) to instantiate each object for each record read from the file. (3) After the file input is done, then display each object to the console. Car List Record #1 (1) carID=0001 (2) carManufacturer=Mercury (3) carModel=LeSabre (4) carYear=2005 (5) carColor=Brown (6) carPrice=11000 (7)...

C++ UNIX car.txt 0001|Mercury|LeSabre|2005|Brown|11000|2019-09-01 0002|Chevrolet|Aveo|2013|Blue|12000|2019-09-02 0003|Datsun|240Z|1979|Orange|13000|2019-09-03 0004|Ford|Galaxie|1964|Black|14000|2019-09-04 0005|Porsche|Boxster|2014|Green|15000|2019-09-0 #i

C++ UNIX car.txt 0001|Mercury|LeSabre|2005|Brown|11000|2019-09-01 0002|Chevrolet|Aveo|2013|Blue|12000|2019-09-02 0003|Datsun|240Z|1979|Orange|13000|2019-09-03 0004|Ford|Galaxie|1964|Black|14000|2019-09-04 0005|Porsche|Boxster|2014|Green|15000|2019-09-0 #include <fstream> #include <iostream> #include <string> #include <sstream> using namespace std; int main() { const int RECORDS = 42; ifstream reader("car.txt"); if(!reader) { cout << "Error: cannot open input file" << endl; return -1; } string item[RECORDS]; int i = 0; while(!reader.eof()) { if((i+1) % 7 == 0) getline(reader,item[i++],'\n'); else ...

Q4. Suppose the character string “05/04/06” is assigned to an object, x. Which of the following...

Q4. Suppose the character string “05/04/06” is assigned to an object, x. Which of the following R commands would convert x to a date object that represents May 4, 2006? (Check all that apply) A. lubridate::mdy(x) B. lubridate::mdy(x, format = “m/d/y”) C. as.Date(x, format = ‘%m/%d/%y’) D. as.Date(x, format = “m/d/y”) E. as.Date(x, format = ‘m/d/y’) F. as.Date(x) Q5. Suppose you have a dataframe, df, with two datetime columns in them: firstPurchaseTime and lastPurchaseTime. These columns are a time stamp...

Questions 06, 07 and 08 refer to the same data American Airlines has 120 direct flights...

Questions 06, 07 and 08 refer to the same data American Airlines has 120 direct flights on a monthly basis from LAX Airport to Tokyo International Airport. The past experience from the previous year determine that 30% of its direct flights from LAX to Tokyo International Airport arrive late. Utilize the normal approximation to the binomial distribution to compute the following questions. Requirement Compute the probability 32 or fewer flights will arrive late Round your answer to three decimals the same as...

Instructions Scenario (information repeated for deliverable 01, 03, and 04) A major client of your company...

Instructions Scenario (information repeated for deliverable 01, 03, and 04) A major client of your company is interested in the salary distributions of jobs in the state of Minnesota that range from $30,000 to $200,000 per year. As a Business Analyst, your boss asks you to research and analyze the salary distributions. You are given a spreadsheet that contains the following information: A listing of the jobs by title The salary (in dollars) for each job You have previously explained...

Chapter 5 Homework Assignment (part 2) PE.05-03.ALGO PE.05-04.ALGO EX.05-04 EX.05-11 EX.05-12 EX.05-14.ALGO EX.05-15.ALGO EX.05-16.ALGO EX.05-17.ALGO EX.05-18.ALGO...

Chapter 5 Homework Assignment (part 2) PE.05-03.ALGO PE.05-04.ALGO EX.05-04 EX.05-11 EX.05-12 EX.05-14.ALGO EX.05-15.ALGO EX.05-16.ALGO EX.05-17.ALGO EX.05-18.ALGO Hide or show questions Progress:7/10 items eBook Show Me How Calculator Print Item Posting a Purchases Journal The purchases journal for Newmark Exterior Cleaners Inc. follows. The accounts payable account has a March 1, 20Y2, balance of $605 for an amount owed to Nicely Co. No payments were made on creditor invoices during March. PURCHASES JOURNAL Page 16 Date Account Credited Post. Ref. Accts. ...

x 0 1 2 3 4 P(X) .45 .3 .2 .04 .01 (c) Find the probability...

x 0 1 2 3 4 P(X) .45 .3 .2 .04 .01 (c) Find the probability that a person has 1 sibling given that they have less than 3 siblings. Hint: Use the conditional formula: P(A|B)=P(A and B)/P(B). In this case A: event of having 1 sibling and B: event of having less than 3 siblings. (d) Find the probability that a person has at least 1 sibling OR less than 2 siblings. Hint: Use the General Addition Rule: P(A...