Sales and Inventory System Essay Sample

9 September 2017

The aim of this chapter is to place cardinal operational steps that may be used to analyze procedure flows. They are linked together utilizing Little’s jurisprudence. We so present a series of illustrations that show how procedure flow analysis may be used to analyze public presentation. The aim is to analyze current public presentation every bit good as identify mark countries for betterment.

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We besides link the operational steps of public presentation to fiscal steps.

In a category of 100 proceedingss we start by discoursing the importance of constructing a clip based capableness in today’s competitory environment. We so set up Little’s jurisprudence to put up other operational steps – viz. stock list and throughput that impact flow clip. Several illustrations from the chapter are discussed to do this relationship clear. We so link these operational steps to fiscal steps to place what signifier betterments may take. We so discuss the Kellogg CRU Rental instance to show how such an analysis may be used to place cardinal countries for betterment. 3. 2 Additional Suggested Readings

We assign a short instance as auxiliary reading for the analysis of procedure flows. The instance is used to make a thorough analysis of flows and place cardinal drivers of cost and gross in a procedure. This apprehension is so used to place actions that improve public presentation. * “CRU Computer Rentals” . Kellogg Case. Writer: Sunil Chopra. Available from: hypertext transfer protocol: //www. Kellogg. northwesterly. edu/cases/index. htm. Suggested assignment inquiries are contained in the instance.

3. 3 Solutions to the Chapter Questions

Discussion Question 3. 1
The antonym of looking at norm is looking at a specific flow unit’s flow clip. and the stock list position and instantaneous flow rate at a specific point in clip. Because flow times change from flow unit to flux unit. it is better to look at the norm over all flow units during a period of clip. Similar for stock list and throughput.

Discussion Question 3. 2
In pattern. one frequently tracks stock list position sporadically ( each twenty-four hours. hebdomad. or month ) . Flow rate is typically besides tracked sporadically ( even more often than stock list position because it straight relates to gross revenues ) . It so is easy to cipher the norm of those Numberss to obtain mean stock list and throughput during a period.

In contrast. few companies track the flow clip of each flow unit. which must be done to cipher the mean flow clip ( over all flow units during a given period ) .

Discussion Question 3. 3
First. pull a procedure flow chart.
Second. cipher all operational flows: throughput. stock list. and flow clip for each activity. Third. cipher the fiscal flow associated with each activity. If the activity incurs a cost ( or earns a gross ) . the cost or gross rate is merely the throughput times the unit cost or gross. If the stock list incurs a keeping cost. the stock list cost rate is merely the mean stock list times the unit keeping cost. Fourth. summing all gross rates and subtracting all cost rates yields the net income rate. straight broken down in footings of the relevant throughputs and stock list Numberss. The latter therefore are the minimum set of operational steps to foretell fiscal public presentation.

Discussion Question 3. 4
For the section of revenue enhancement ordinances we have
Average stock list I = 588 undertakings.
Throughput R = 300 projects/yr ( we assume a stable system ) . Thus. Average flow clip T = I / R = 588 / 300 = 1. 96 year.
This is larger than six months. So we should differ with the section head’s statement.



Discussion Question 3. 5
If GM and Toyota have same bends. and we know that
bends = 1/flow clip = 1/T.
it follows that their norm flow times are the same. We besides know that Toyota’s throughput is twice that of GM. Therefore. from I=RT
it follows that Toyota has twice the stock list of GM. Thus. the statements are inconsistent. both companies have the same flowtime but Toyota has higher stock list than GM.



Discussion Question 3. 6
Yes. low stock lists means few flow units are held in the buffer. In contrast. fast stock list bends means short flow times ; i. e. . flux units do non pass a long clip in the procedure. As such. one can hold high bends with high or low stock lists ( it all depends on what the throughput is ) .

Discussion Question 3. 7
A short cost-to-cash rhythm means that it does non take long to change over an input into a sold end product. Clearly. this is good because we do non necessitate to finance the input for a long clip before it earns gross ( i. e. . lower working capital demands ) . Short cost-to-cash rhythm requires short flow times. which imply low stock lists ( for a given throughput ) . or high throughput ( for a given stock list ) .

Exercise 3. 1 ( Bank )
For the bank we have
Average stock list I = 10 people.
Throughput R = 2 people/min ( we assume a stable system ) .
Therefore.
Average flow clip T = I / R = 10/2 min = 5 min.




Exercise 3. 2 ( Fast-Food )
For the fast nutrient mercantile establishment we have
Average stock list I = 10 autos.
Throughput rating is as follows: Cars try to come in the thrust through country at a rate of 2 cars/min. However 25 % of autos leave when they see a long waiting line. Therefore. autos enter the thrust through at a flow rate R = 75 % * 2 cars/min = 1. 5 cars/min. Thus Average flow clip T = I / R = 10/1. 5 min =
6. 67 min.



Exercise 3. 3 ( Checking Histories )
For a checking history we have
Average stock list I = mean balance = $ 3. 000
Turns = 6 per twelvemonth.
Average flow clip T = 1 / turns = 1/6 twelvemonth = 2 months.
Therefore
Throughput R = I / T = 3. 000/2 = $ 1. 500 / month.





Exercise 3. 4 ( ER )
First pull the flow chart with all the informations given:

We assume a stable system. This implies that mean inflow peers mean outflow at every phase. In this instance you are given inventory Numberss I and flux rate R = 55 patients/hr. There are two flow units:

( 1 ) Those that are possible admits: flow rate = 55*10 % = 5. 5/hr. ( 2 ) Those that get a simple prescription: flow rate = 55*90 % = 49. 5/hr.

To happen the norm flow times. we use Little’s jurisprudence at each activity for which the flow clip is unknown:

( 1 ) Buffer 1: Roentgen = 55/hr ( both flow units go through at that place ) . I = 7. so that waiting clip in buffer 1 = T = I/R = 7/55 hour = 0. 127 hours = 7. 6 proceedingss.

( 2 ) Registration: flow clip T = 2 min = 2/60 hour. All flow units flow through this phase. Therefore flow rate through this phase is R = 55 / hour. Average stock list at enrollment is given by I = RT = 55*2/60 = 1. 83 patients.

( 3 ) Buffer 2: Roentgen = 55/hr ( both flow units go through at that place ) . I = 34. so that waiting clip in buffer 2 = T = I/R = 34/55 hour = 0. 62 hours = 37. 1 proceedingss.

( 4 ) Doctor clip: depends on the flow unit:
4a: possible admits: T = 30 proceedingss
4b: prescription folks: T = 5 proceedingss

OK. now we have everything to happen the entire norm flow times: happen the critical way for each flow unit. In this instance. each flow unit merely has one way. so that is the critical way. We find its flow clip by adding the activity times on the way:

( a ) For a possible admit. mean flow clip ( buffer 1 + enrollment + buffer 2 + physician ) = 7. 6 + 2 + 37. 1 + 30 = 76. 7 proceedingss ( B ) For a individual stoping up with a prescription. mean flow clip ( buffer 1 + enrollment + buffer 2 + physician ) = 7. 6 + 2 + 37. 1 + 5 = 51. 7 proceedingss.

The reply to the other inquiries is found as follows:

1. On norm. how long does a patient spend in the exigency room? We know the flow clip of each flow unit. The mean flow clip over all flow units is the leaden norm: 10 % of entire flow units spend 76. 7 proceedingss while 90 % spend 51. 7 proceedingss. Therefore. the expansive norm is:

T = 10 % * 76. 7 + 90 % *51. 7 = 54. 2 proceedingss.

2. On norm. how many patients are being examined by a physician? This inquiry asks for the mean stock list at the doctor’s activity. Again. first calculate stock list of each type of flow unit:
( a ) Potential admits: R = 5. 5 patients/hr. T = 30 min = 0. 5 hour. therefore. I = RT = 5. 5/hr*0. 5 hr = 2. 75 patients
( B ) Simple prescription: R = 49. 5 patients/hr. T = 5 min = ( 5/60 ) hour. therefore I = RT = 49. 5* ( 5/60 ) = 4. 125 patients Therefore. entire stock list at the physician is 2. 75 + 4. 125 = 6. 865 patients.

3. On norm. how many patients are in the ER?
This inquiry asks for entire stock list in ER = stock list in buffer 1 + stock list in enrollment + stock list in buffer 2 + stock list with physicians = 7 + 1. 83 + 34 + 6. 865 = 49. 695 patients.

Exercise 3. 5 ( ER. triage )
The procedure flow map with the triage system is as follows:

The stock list. and clip spent in assorted locations are as follows. In each instance the deliberate measure is italicized.

Throughput through ER. R = 55 patients / hr = . 9167/min.
Average stock list in exigency room. I = amount of stock list in all phases = 50. 63 patients
Average clip spent in the exigency room = I/R = 50. 63/ . 9167 = 55. 23 proceedingss.
For patients that are finally admitted. mean clip spent in the exigency room = clip in buffer 1 + enrollment + buffer 2 + triage nurse + buffer 3 + physician ( possible admit ) = 71. 18 proceedingss.


Exercise 3. 6 ( ER. triage with misclassification )
In this instance the procedure flow map is altered slightly since there are some patients sent from simple prescriptions to buffer 3.

( We will presume that the physician “instantaneously” recognizes misclassification so that a misclassified patient does non pass 5 proceedingss with the physician. However. if you assume such individual besides spends 5 proceedingss. the full methodological analysis below follows. merely increase the relevant flow clip by 5 minutes. ) The stock lists. throughputs and flow times are as follows:

Throughput through ER. R = 55 patients / hr
Average stock list in exigency room. I = amount of stock lists in all phases = 37. 46
Average clip spent in the exigency room T = I/R = 37. 46/ . 9167 = 40. 86 proceedingss.

To cipher flow times. we should separate three types of flow units: ( 1 ) those that are right identified as possible admits the first clip: flow rate = 55*9 % = 4. 95/hr. Average flow clip = clip in buffer 1 + enrollment + buffer 2 + triage nurse + buffer 3 + physician ( possible admit ) = 56. 82 proceedingss ( 2 ) those that are first mis-identified as simple prescription and subsequently corrected and redirected to possible admits: flow rate = 55*1 % = 0. 55/hr. Average flow clip = clip in buffer 1 + enrollment + buffer 2 + triage nurse + buffer 4 + buffer 3 + physician ( possible admit ) = 74. 80 proceedingss ( 3 ) those that are right identified to acquire a simple prescription the first clip: flow rate = 55*90 % = 49. 5/hr. Average flow clip = clip in buffer 1 + enrollment + buffer 2 + triage nurse + buffer 4 + buffer 3 + physician ( simple prescription ) = 38. 89 proceedingss

For patients that are finally admitted. that is types ( 1 ) and ( 2 ) . mean clip spent in the exigency room is the leaden norm of their flow times. Type ( 1 ) is fraction 4. 95/ ( 4. 95+0. 55 ) = 90 % of those admitted and type ( 2 ) is 0. 55/ ( 4. 95+0. 55 ) = 10 % of those admitted. Therefore. mean flow clip for patients that are finally admitted is 90 % *56. 82min + 10 % *74. 80. 8min = 58. 62 min.

Note that the overall mean flow clip over all patients is: 9 % *56. 82min + 1 % *74. 80min + 90 % *38. 89. 8min = 40. 86 min. in understanding with the figure derived straight from Little’s Law above.

Exercise 3. 7 ( Orange Juice Inc )

First let us detect that there are two periods in the twenty-four hours:

1. From 7am-6pm. oranges come in at a rate of 10. 000kg/hr and are processed. and therefore go forth the works. at 8000kg/hr. Because influxs exceed escapes. stock list will construct up at a rate of ?R = 10. 000-8. 000kg/hr = +2. 000 kg/hr.

Therefore. because we can non hold oranges stored overnight. we start with an empty works so that stock list at 7am is zero: I ( 7 am ) = 0. Because stock list builds up linearly at 2. 000kg/hr. the stock list at 6pm is I ( 6pm ) = 2. 000 kg/hr * 11 hour = 22. 000kg.

2. After 6pm. no more oranges come in. yet treating continues at 8000 kg/hr until the works is empty. Therefore. influxs is less than escapes so that stock list is depleted at a rate of ?R = 0 – 8. 000 kg/hr = – 8. 000 kg/hr.

Therefore. since we have that I ( 6pm ) = 22. 000kg. we know that stock list depletes linearly from that degree at a rate of -8. 000 kg/hr. Therefore. to empty the works. stock list must make zero and this will take an sum of clip ?t where: 22. 000 kilogram – 8. 000 kg/hr ?t = 0.

or
?t = 22. 000/8. 000 hr = 2. 75 hr = 2 hour 45min.
Therefore. the works must run until 6pm + 2hr 45min = 8:45pm.

This can all be diagrammatically summarized in the stock list construct up diagram shown above.

3. Truck kineticss: for this the stock list diagram is truly utile. Notice that we have taken a entire procedure position of the works. including the truck waiting waiting line. Therefore. stock list is entire stock list in the bins + stock list in the trucks ( if any are waiting ) . So. let’s draw the thick line on the stock list build-up diagram. stand foring the bin storage capacity. First stock list builds up in the bins. When the bin is full. so the trucks must wait. This happens at: 2. 000 kg/hr ?t = 6. 000kg.

so that the first truck will wait after ?t = 6. 000/2. 000 hr = 3 hour. which is at 10am. Now. the last truck that arrives ( at 6pm ) joins the longest waiting line. and therefore will wait the longest. That “unfortunate” truck will be able to get down dumping its contents in the bins when the bins start consuming. This is at 22. 000 kilogram – 8. 000 kg/hr ?t = 6. 000.

or after ?t = ( 22. 000-6. 000 ) /8. 000 hr = 2 hour. after 6pm. Therefore. the last truck departs at 8pm and the maximal truck waiting clip is hence 2 hours.

Now. among all the trucks that do wait ( i. e. . those geting after 10am ) . the first truck delaies practically zero proceedingss. and the last truck waits 2 hours. climaxing in an norm of ( 0 + 2 ) hrs/2 = 1 hr.

Notice that the trucks geting before 10am do non wait. Thus. the overall mean truck waiting clip is ( # trucks geting before 10am * 0 + # trucks geting after 10am * 1hr ) / ( entire # of trucks ) . Because input rate is 10. 000kg/hr and each truck carries 1. 000 kg/truck. the truck input rate is 10 trucks/hr. so that the overall mean truck waiting clip is: ( 10 trucks/hr * 3hrs * 0 + 10 trucks/hr * 8hrs * 1hr ) / ( 10 trucks/hr * 11 hour ) = 8/11 hour = 43. 63min. Average waiting clip can besides be calculated by detecting that the country of the upper trigon in the build-up diagram represents the entire sum of hours waited by all trucks: Area = ( 22. 000 – 6. 000 ) kilogram * ( 8pm – 10 am ) /2 = 16. 000 kilogram * 10 waiting hour /2 = 80. 000 kilogram waiting hrs = 80. 000 kilogram waiting hour / ( 1. 000 kg/truck ) = 80 truck waiting hour. Now. we merely calculated that there are 80 trucks that do wait. hence the mean waiting clip among those trucks that do wait is 80 truck waiting hrs/ 80 trucks = 1 hr.

Exercise 3. 8 ( Jasper Valley Motors )
Part a.

TURNStotal = 1/Ttotal so Ttotal = 1/8 old ages = 1. 5 months

Itotal = RtotalTtotal = 160 vehicles/month * 1. 5 months = 240 vehicles. which is the reply.

Typical mistakes: incorrect units and saying that “I = 160* ( 1/8 ) = 20 vehicles. ”

Part B.

Similar to portion a. we have Tnew = 1/7. 2 old ages = 1. 667 months and Tused = 1/9. 6 old ages = 1. 25 months.

Inew = 0. 6 * 160 vehicles/month * 1. 667 months = 160 new vehicles

Iused = 0. 4 * 160 vehicles/month * 1. 25 months = 80 new vehicles

Entire monthly funding costs so 160* $ 175 + 80* $ 145 = 28. 000 + 11. 600 = $ 39. 600/month.

Cost per vehicle are so $ 39. 600/month ( 160+80 ) = $ 165 per vehicle per month. which is the reply.

Typical mistakes:
1. Not recognizing that the cost driver is stock list. non throughput. ( Taking a throughput-weighted norm would give 60 % * $ 175 + 40 % * $ 145 = 163. alternatively of the right inventory-weighted. ) 2. Not taking a leaden norm. Clearly. the reply must fall between $ 145 and $ 175. 3. Giving entire monthly costs alternatively of per vehicle.

Part degree Celsius.

From Little’s Law. cutting clip 20 % while keeping R unchanged will cut down stock list by 20 % . From portion b. mean monthly funding costs for new vehicles is 160* $ 175 = $ 28. 000/month. A 20 % bead gives $ 5. 600 per month. which is the reply.

Typical mistakes:
1. Assuming the service works besides on used autos. taking to 20 % $ 39. 600/month= $ 7920/mo. 2. Merely saying the value per auto per month: We cut down Tnew from Tnew = 1/7. 2 old ages = 1. 667 months by 20 % *1. 667mo = 1/3 minute. This saves 1/3mo * $ 175/new auto. minute = $ 58. 33/new auto. ( Multiplying by 96new cars/mo would hold yield the right $ 5600/mo. ) 3. Reducing the flow clip by 20 % does non intend that turnover is increased by 20 % . ( On the contrary. really. turnover additions here from 7. 2 to 9. which is 25 % . )

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