jgt task 2

8/10/2019 JGT Task 2

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MEMORANDUM

O: CYNTHIA CROWNSHIELD

FROM:

SUBJECT: PRODUCTION ANALYSIS

DATE: MAY 29, 2014

CC: ALISTAIR WU, DIETER HANDEL,

Distribution Pattern

Alistair Wu has requested that I look at what the lowest shipping schedule and cost can be based ondata that he has provided. He wants to know what the lowest possible cost of shipping will can be.Mr. Wu is also considering increasing production at the Shanghai factory from 1,300 to 2,800 units,and wants to ensure that this growth will be an affordable choice. The first chart that was given liststhe factory capacity and what each warehouse demand is. The second chart lists the price of shippingfrom each factory to each warehouse. The chart looks at the demand in the warehouses as well as the

cost to ship there from each factory. It then generates a cost effective shipping plan to ship productsto the warehouse with least shipping expense and fills those warehouses first. It then allocatesadditional products to other warehouses that have higher shipping prices. This is evident withShuzworld H and Shuzworld F, the program ships the most products to the lower priced warehouses1 and 3. Warehouse 1 need 2,500 units but Shuzworld F only produces 2,200 units. That means

Warehouse 1 will take all off the units from Shuzworld F and still have a demand of 300 units. Warehouse needs 1,800 units and Shuzworld H produces 2,300 units. This means Warehouse 3 willreceive all of its units from Shuzworld H and Shuzworld H will still have an excess of 500 units.

After this, the lowest cost after the first is looked for. Warehouse 3 is already taken care of so noneof the costs need to be looked at in that column. The next most efficient cost is Shuzworld H to

Warehouse 1 and Shanghai to Warehouse 2. Shanghai can only produce 1,300 units so Warehouse 2

will take those units and still have a demand for 300 more units. Warehouse 1 only needs 300 moreunits to complete its demand so it will take those units from Shuzworld H leaving Shuzworld H withand excess of 200 units. At the end of this, Warehouse 2 still needs 200 units so the warehouse willget those units from Shuzworld H. The graph showing this is again on the excel sheet. Overall, whenthe costs are added up, the total cost of shipping the units this way will cost the company $13,600.

A1. Computer Analysis

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Shanghai Expansion

Alistair Wu is planning to expand production at the Shanghai plant from 1,300 to 2,800. Thequestion of what the price of the shipping will be after the increase and whether this would be afavorable move for the company. The computer program analysis shows that the increase inproduction will cause a more favorable transportation schedule and decrease transportation costs.

Instead of Warehouse 2 having to receive units from two different factories, it can receive all of theunits it need from the lower cost, Shanghai. Shanghai will still have 1,300 units left over andShuzworld H with have 200 extra units but it will decrease the transportation costs by 200 dollars.Below is the computer generated graph that shows these changes.

A1. Shanghai Expansion Computer Analysis

A1a. Decision Analysis Tool Justification

Excel OM/QM Transportation Model was used to make the analysis. Transportation Modeling waschosen because the tool finds the lowest optimal costs when shipping products from severallocations to several destinations. The program looks for the lowest cost in the shipping plan insteadof by route time or convenience. Mr. Wu was specifically concerned with finding the cheapestshipping schedule and this tool is made to do that.

B1. Machine Reliability

Shuzworld has three computer driven machines that are used to produce the deck shoes. At thismoment there is not a backup machine if one of the machines goes down. This means that the shoescannot be finished or production will be halted until a machine is replaced or repaired. If themachine is backed up the reliability will increase and the likelihood of delays in production willdecrease. By testing the reliability of each machine we can determine which machine should bebacked up to increase reliability. All three machines together with no backup have a reliability of75.68%. If machine one is backed up the reliability increases to 87.78%. If machine two is backed up

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the reliability increases to 82.49%. If machine three is back up the reliability increases to 76.43%. Ifany machine is backed up the reliability overall increases, however there is more reliability if machineone is backed up. Handel only wants to back up one machine so it must be decided which machine

with increased reliability the most when backed up. By using the formula {(Probability of firstcomponent working) + [(Probability of second component working) x (probability of needingsecond component)]. So the best machine to back up is machine one to increase the reliability from

75.78% to 87.78%.

B2. Computer Analysis

Machine 1 Analysis

Machine 2 Analysis

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Machine 3 Analysis

B2a. Decision Analysis Tool Justification

This decision model is used because it allows me to measure the overall system reliability comparedto the affect it would have if adding a back-up machine to any one of the current machines. This willdecrease delays in production should there be a malfunction of a machine. This tool shows the best

way to increase efficiency on the assembly of the computer driven shoe machines by enhancing thereliability of the least reliable machine.

C1. Shoelace Order

Angela Down from the Shuzworld accounting department wants help in minimizing their inventorycosts in regards to ordering shoelaces for a certain shoe. Shuzworld has been stuck with a higher

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inventory of shoelaces than they would like to have and would like to minimize their costs. They use300,000 shoelaces per year with an estimated cost of $125 for every order to the suppliers. Theestimated holding cost of each pair of laces is $.10.In order to help her with this; I used the economicorder quantity model (EOQ). The annual expenses for ordering and storing the inventory will be$1,369.31. The optimal inventory that they should keep on hand should be 27,386 and the averageinventory around 13,693. Ms. Down should order the optimal level of shoelaces 11 times a year to

avoid shortages.

C2. Computer Analysis

C2a. Decision Analysis Tool Justification

The economic order quantity is used for this situation because it shows the number of pairs of lacesthat can be added to the inventory each time an order is placed to minimize the holding and orderingcosts. It provides the information that can be used to control the inventory levels and costs.

D. Waiting Line System

Shuzworld aims to keep customer satisfaction at its highest. A customer approach that focuses onmaking the customer happy is of upmost importance. One way to achieve this is to ensure thatcustomer wait time is kept to a minimum. Specific questions have been posed for concern; howmany customers will be in the system; how many customers on average will be in line; and what theprobability is of one being in line or being served? The company is looking at staffing each store withonly one full-time cashier. The question of whether the need for an additional cashier has also been

presented. The pilot store showed that they had a sale every 10 minutes or 6 sales per hour and eachtransaction took an average of 5 minutes at the register or 12 transactions and hour. The waiting linesystem was used to analyze if one or two cahiers would be effective.

The first question posed is how many customers will be in the system (waiting in line or beingserved), on average? My data indicates that in the single cashier design the average number of

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customers in the system is 1. In the multiple cashier models with two cashiers, this number falls to0.53 customers.

Your next question asks what the average time a customer will spend in the system (waiting in lineand being served) will be is? In the single cashier model, my data shows that the average time in thesystem will be 0.1667 minutes or 10 minutes. When the multiple cashier models are applied utilizing

two cashiers the average time customers spend in the system falls to 0.0889 minutes or 5.33 minutes.

Your third question was, “how many customers will be in line on average”. There will be 0.5customers in line on average at any one time if you utilize the single cashier model. Adding a secondcashier reduces the average number of customers in line to 0.03.

Next you ask how long customers will have to wait in line, on average. The data shows that the singlecashier model averages a wait time in the queue of 0.083 min or 5 min. However utilization of thetwo cashier model will result in the average wait time falling to 0.0056 or 34 sec.

The fifth question posed regarded the probability of no one being in line or being served. Theprobability of no one being in line or being served is 0.5 or 50% if you use the single cashier system.

The probability of no one being served or in line if a second cashier is added is 0.6 or 60%.

D1. One Cashier Waiting Line

The analysis showed that a one cashier waiting line on average would have 1 customer in the systemand 0.5 customers in the queue. The average time a customer will spend in the system is 0.1667minutes or 10 minutes. The average wait time in the queue is 0.083 minutes or 5 minutes. Theprobability that no customer will be in line is 0.5. Server utilization is 0.5 or 50% in this model.

D2. Computer Analysis- One Cashier Line

D1. Two Cashier Waiting Line

The analysis showed that a two cashier waiting line on average would have 0.53 customers in thesystem and 0.03 customers in the queue. The average time a customer will spend in the system is0.0889 minutes or 5.33 minutes. The average wait time in the queue is 0.0056 or 34 seconds. Theprobability that no customer will be in line is 0.6 or 60%. Server utilization is 0.25 or 25% in thismodel

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