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January 18, 2022

The purpose of this assignment is to use analytics techniques to analyze a case problem. Part 1 Read Case Study Case 15.2 Ebo

Business Finance

Home>Business & Finance homework help

 

*** I ONLY NEED PART 2 DONE OF THIS ASSIGNMENT (500-750-word summary to company management). I HAVE ATTACHED THE ORIGINAL ASSIGNMENT THAT HAS BOTH PARTS COMPLETED BUT PART 2 WAS NOT ANSWERED CORRECTLY AND I NEED A BETTER SUMMARY FORTHE INSTRUCTOR***

The purpose of this assignment is to use analytics techniques to analyze a case problem.

Part 1

Read Case Study Case 15.2 “Ebony Bath Soap” from the textbook, and then complete the following items.

  1. For Questions 1 and 2 of the case, use the Palisade DecisionTools Excel software to set up a simulation model and run a simulation with 500 trials for the case. Ensure that all Palisade software output is included in your files and that only one Excel file is open when running a simulation. Use the "Topic 3 Case Study Template" file as a starting point. Hint: The RiskSimtable function was be helpful for running the simulations.
  2. Respond to Question 3 as written in the problem. Ignore the confidence interval portion of the question.
  3. Respond to Question 4 as written in the problem.

To receive full credit on the assignment, complete the following.

  1. Ensure that the Palisade software output is included with your submission.
  2. Ensure that Excel files include the associated cell functions and/or formulas if functions and/or formulas are used.
  3. Include a written response to all narrative questions presented in the problem by placing it in the associated Excel file.
  4. Include screenshots of all simulation distribution results for output variables.
  5. Place each problem in its own Excel file. Ensure that your first and last name are in your Excel file names.

Part 2

In a 500-750-word summary to company management, address the following. Include relevant charts and graphs within your summary, as needed.

  1. Describe the case specific business requirements and how they can be communicated across all levels of the organization.
  2. Based on the simulation results, discuss the Annual Cost output statistical distributions. Assume that your audience as minimal background in statistics.
  3. Discuss which Annual Cost output probability distribution has the most dispersion, and explain why this is so.
  4. Explain the descriptive, predictive, and prescriptive analytics that have been used to formulate the solutions to the business needs.
  5. Based on the Annual Cost output statistical distributions and other information gleaned from your analysis, discuss the specific prescribed course of action you would recommend to company management and justify your recommendations. Include discussion of how the proposed analytics solutions can optimize organizational performance and effectiveness.

*** I ONLY NEED PART 2 DONE OF THIS ASSIGNMENT. I HAVE ATTACHED THE ORIGINAL ASSIGNMENT THAT HAS BOTH PARTS COMPLETED BUT PART 2 WAS NOT ANSWERED CORRECTLY AND I NEED A BETTER SUMMARY FORTHE INSTRUCTOR***

  • attachment

    attachment_2.xlsx

  • attachment

    attachment_1.xlsx

Tom Salmons – Part I _Answer

Ebony Bath Soap Solution
Question 1)
Question 2)
Risk is used to simulate 500 iterations of each of 6 values of U (those in the range J13:J18), using a RiskSimtable function in ce
The Summary Report shows the results, some of which are copied to the Simulation sheet.
Question 3)
The mean, standard deviation and confidence intervals for each value of U is tabulated in the Simulation sheet. The smallest yearly value occurs with
U= 60, although this could change if the simulation were done with different random numbers. A plot of the mean annual cost is shown below the tabulated results
Question 4)
Other values of U and L could be tested. Note that the policy as stated never returns to a production level of 120 onve the production level changes. Other policies could
be investigated which return to a 120 production level. For example, another policy would be to produce 120 units if inventory crosses the midpoint of the range [L, U].

See the Simulation sheet, columns A-H, for the solution to the 52-week simulation. The major components to calculating the cost for a given week are the demand generation, inventory calculation and production level setting. Column D tracks the inventory which is calculated as last week's inventory plus this week's production (which was set last week) minus this week's demand or zero, whichever is larger. This insures that inventory cannot be negative, and thus, no backorders. Column E indicates the production level to be set for the following week. A nested IF statement is used. The first check is to see whether this week's inventory is less than l (here, 30). If so, next week's production level is set to 130. If not, the inventory level is checked to whether is greater than u (here, 80). If so, next week's production level is set to 110. Otherwise, the production level is unchanged. All that remains is to calculate the inventory and production change cost. The inventory cost calculated in column F is simply the per unit inventory cost (here, 30) multiplied by this week's inventory. Calculating the production change cost in column G is more challenging. The production change cost (here, 3000) is multiplied by the reusult of an IF statement is used to check if the production level has been changed. If there was a change, 3000 is multiplied by one, otherwise, if no change occured, then 3000 is multiplied by 0. Column H is simply the sum of the two costs for the week. Cell H9 totals the costs over the year.

Mean Annual Cost

Mean Profit 30 40 50 60 70 80 114141.48 106346.82 103071.24 102708.66 105158.58 108302.7

U

Tom Salmons – Simulation

Ebony Bath Soap Simulation
Inputs Production policy:
Average demand 120 If inventory < 30 then produce 130
Stdev of demand 15 If inventory > ERROR:#NAME? then produce 110
Unit holding cost $30 Otherwise, don't change production level.
Prod change cost $3,000
Initial inventory 60
Current prod level 120 Annual cost ERROR:#NAME?
Simulation of 52 weeks Next week Sensitivity to U (cell E5) – see next sheet for more @Risk results
Week Normal Demand
Chris Albright: Make sure demand is an integer and not negative
Inventory Production Holding cost Change cost Weekly cost U Min Max Mean Stdev Low
Chris Albright: Lower limit of 95% confidence interval
High
Chris Albright: Upper limit of 95% confidence interval
Chris Albright: Make sure demand is an integer and not negative 0 =B8 =B9 30 $82,200 $147,180 $114,141 $11,768 $113,089 $115,194
1 =RiskNormal($B$4,$B$5 =ROUND(MAX(B14,0),0) =MAX(D13+E13-C14,0) =IF(D14<$E$4,$G$4,IF(D14>$E$5,$G$5,E13)) =D14*$B$6 =$B$7*IF(E14<>E13,1,0) =F14+G14 40 $72,930 $130,650 $106,347 $10,119 $105,442 $107,252
2 =RiskNormal($B$4,$B$5) =ROUND(MAX(B15,0),0) =MAX(D14+E14-C15,0) =IF(D15<$E$4,$G$4,IF(D15>$E$5,$G$5,E14)) =D15*$B$6 =$B$7*IF(E15<>E14,1,0) =F15+G15 50 $72,270 $126,930 $103,071 $9,776 $102,197 $103,946
3 =RiskNormal($B$4,$B$5) =ROUND(MAX(B16,0),0) =MAX(D15+E15-C16,0) =IF(D16<$E$4,$G$4,IF(D16>$E$5,$G$5,E15)) =D16*$B$6 =$B$7*IF(E16<>E15,1,0) =F16+G16 60 $65,490 $130,560 $102,709 $9,674 $101,843 $103,574
4 =RiskNormal($B$4,$B$5) =ROUND(MAX(B17,0),0) =MAX(D16+E16-C17,0) =IF(D17<$E$4,$G$4,IF(D17>$E$5,$G$5,E16)) =D17*$B$6 =$B$7*IF(E17<>E16,1,0) =F17+G17 70 $71,040 $131,820 $105,159 $10,306 $104,237 $106,080
5 =RiskNormal($B$4,$B$5) =ROUND(MAX(B18,0),0) =MAX(D16D17+E17-C18,0) =IF(D18<$E$4,$G$4,IF(D18>$E$5,$G$5,E17)) =D18*$B$6 =$B$7*IF(E18<>E17,1,0) =F18+G18 80 $71,040 $142,140 $108,303 $10,964 $107,322 $109,283
6 =RiskNormal($B$4,$B$5) =ROUND(MAX(B19,0),0) =MAX(D18+E18-C19,0) =IF(D19<$E$4,$G$4,IF(D19>$E$5,$G$5,E18)) =D19*$B$6 =$B$7*IF(E19<>E18,1,0) =F19+G19
7 =RiskNormal($B$4,$B$5) =ROUND(MAX(B20,0),0) =MAX(D19+E19-C20,0) =IF(D20<$E$4,$G$4,IF(D20>$E$5,$G$5,E19)) =D20*$B$6 =$B$7*IF(E20<>E19,1,0) =F20+G20
8 =RiskNormal($B$4,$B$5) =ROUND(MAX(B21,0),0) =MAX(D20+E20-C21,0) =IF(D21<$E$4,$G$4,IF(D21>$E$5,$G$5,E20)) =D21*$B$6 =$B$7*IF(E21<>E20,1,0) =F21+G21
9 =RiskNormal($B$4,$B$5) =ROUND(MAX(B22,0),0) =MAX(D21+E21-C22,0) =IF(D22<$E$4,$G$4,IF(D22>$E$5,$G$5,E21)) =D22*$B$6 =$B$7*IF(E22<>E21,1,0) =F22+G22
10 =RiskNormal($B$4,$B$5) =ROUND(MAX(B23,0),0) =MAX(D22+E22-C23,0) =IF(D23<$E$4,$G$4,IF(D23>$E$5,$G$5,E22)) =D23*$B$6 =$B$7*IF(E23<>E22,1,0) =F23+G23
11 =RiskNormal($B$4,$B$5) =ROUND(MAX(B24,0),0) =MAX(D23+E23-C24,0) =IF(D24<$E$4,$G$4,IF(D24>$E$5,$G$5,E23)) =D24*$B$6 =$B$7*IF(E24<>E23,1,0) =F24+G24
12 =RiskNormal($B$4,$B$5) =ROUND(MAX(B25,0),0) =MAX(D24+E24-C25,0) =IF(D25<$E$4,$G$4,IF(D25>$E$5,$G$5,E24)) =D25*$B$6 =$B$7*IF(E25<>E24,1,0) =F25+G25
13 =RiskNormal($B$4,$B$5) =ROUND(MAX(B26,0),0) =MAX(D25+E25-C26,0) =IF(D26<$E$4,$G$4,IF(D26>$E$5,$G$5,E25)) =D26*$B$6 =$B$7*IF(E26<>E25,1,0) =F26+G26
14 =RiskNormal($B$4,$B$5) =ROUND(CMAX(B27,0),0) =MAX(D26+E26-C27,0) =IF(D27<$E$4,$G$4,IF(D27>$E$5,$G$5,E26)) =D27*$B$6 =$B$7*IF(E27<>E26,1,0) =F27+G27
15 =RiskNormal($B$4,$B$5) =ROUND(MAX(B28,0),0) =MAX(D27+E27-C28,0) =IF(D28<$E$4,$G$4,IF(D28>$E$5,$G$5,E27)) =D28*$B$6 =$B$7*IF(E28<>E27,1,0) =F28+G28
16 =RiskNormal($B$4,$B$5) =ROUND(MAX(B29,0),0) =MAX(D28+E28-C29,0) =IF(D29<$E$4,$G$4,IF(D29>$E$5,$G$5,E28)) =D29*$B$6 =$B$7*IF(E29<>E28,1,0) =F29+G29
17 =RiskNormal($B$4,$B$5) =ROUND(MAX(B30,0),0) =MAX(D29+E29-C30,0) =IF(D30<$E$4,$G$4,IF(D30>$E$5,$G$5,E29)) =D30*$B$6 =$B$7*IF(E30<>E29,1,0) =F30+G30
18 =RiskNormal($B$4,$B$5) =ROUND(MAX(B31,0),0) =MAX(D30+E30-C31,0) =IF(D31<$E$4,$G$4,IF(D31>$E$5,$G$5,E30)) =D31*$B$6 =$B$7*IF(E31<>E30,1,0) =F31+G31
19 =RiskNormal($B$4,$B$5) =ROUND(MAX(B32,0),0) =MAX(D31+E31-C32,0) =IF(D32<$E$4,$G$4,IF(D32>$E$5,$G$5,E31)) =D32*$B$6 =$B$7*IF(E32<>E31,1,0) =F32+G32
20 =RiskNormal($B$4,$B$5) =ROUND(CMAX(B33,0),0) =MAX(D32+E32-C33,0) =IF(D33<$E$4,$G$4,IF(D33>$E$5,$G$5,E32)) =D33*$B$6 =$B$7*IF(E33<>E32,1,0) =F33+G33
21 =RiskNormal($B$4,$B$5) =ROUND(MAX(B34,0),0) =MAX(D33+E33-C34,0) =IF(D34<$E$4,$G$4,IF(D34>$E$5,$G$5,E33)) =D34*$B$6 =$B$7*IF(E34<>E33,1,0) =F34+G34
22 =RiskNormal($B$4,$B$5) =ROUND(MAX(B35,0),0) =MAX(D34+E34-C35,0) =IF(D35<$E$4,$G$4,IF(D35>$E$5,$G$5,E34)) =D35*$B$6 =$B$7*IF(E35<>E34,1,0) =F35+G35
23 =RiskNormal($B$4,$B$5) =ROUND(MAX(B36,0),0) =MAX(D35+E35-C36,0) =IF(D36<$E$4,$G$4,IF(D36>$E$5,$G$5,E35)) =D36*$B$6 =$B$7*IF(E36<>E35,1,0) =F36+G36
24 =RiskNormal($B$4,$B$5) =ROUND(MAX(B37,0),0) =MAX(D36+E36-C37,0) =IF(D37<$E$4,$G$4,IF(D37>$E$5,$G$5,E36)) =D37*$B$6 =$B$7*IF(E37<>E36,1,0) =F37+G37
25 =RiskNormal($B$4,$B$5) =ROUND(MAX(B38,0),0) =MAX(D37+E37-C38,0) =IF(D38<$E$4,$G$4,IF(D38>$E$5,$G$5,E37)) =D38*$B$6 =$B$7*IF(E38<>E37,1,0) =F38+G38
26 =RiskNormal($B$4,$B$5) =ROUND(MAX(B39,0),0) =MAX(D38+E38-C39,0) =IF(D39<$E$4,$G$4,IF(D39>$E$5,$G$5,E38)) =D39*$B$6 =$B$7*IF(E39<>E38,1,0) =F39+G39
27 =RiskNormal($B$4,$B$5) =ROUND(MAX(B40,0),0) =MAX(D39+E39-C40,0) =IF(D40<$E$4,$G$4,IF(D40>$E$5,$G$5,E39)) =D40*$B$6 =$B$7*IF(E40<>E39,1,0) =F40+G40
28 =RiskNormal($B$4,$B$5) =ROUND(MAX(B41,0),0) =MAX(D40+E40-C41,0) =IF(D41<$E$4,$G$4,IF(D41>$E$5,$G$5,E40)) =D41*$B$6 =$B$7*IF(E41<>E40,1,0) =F41+G41
29 =RiskNormal($B$4,$B$5) =ROUND(MAX(B42,0),0) =MAX(D41+E41-C42,0) =IF(D42<$E$4,$G$4,IF(D42>$E$5,$G$5,E41)) =D42*$B$6 =$B$7*IF(E42<>E41,1,0) =F42+G42
30 =RiskNormal($B$4,$B$5) =ROUND(MAX(B43,0),0) =MAX(D42+E42-C43,0) =IF(D43<$E$4,$G$4,IF(D43>$E$5,$G$5,E42)) =D43*$B$6 =$B$7*IF(E43<>E42,1,0) =F43+G43
31 =RiskNormal($B$4,$B$5) =ROUND(MAX(B44,0),0) =MAX(D43+E43-C44,0) =IF(D44<$E$4,$G$4,IF(D44>$E$5,$G$5,E43)) =D44*$B$6 =$B$7*IF(E44<>E43,1,0) =F44+G44
32 =RiskNormal($B$4,$B$5) =ROUND(CMAX(B45,0),0) =MAX(D44+E44-C45,0) =IF(D45<$E$4,$G$4,IF(D45>$E$5,$G$5,E44)) =D45*$B$6 =$B$7*IF(E45<>E44,1,0) =F45+G45
33 =RiskNormal($B$4,$B$5) =ROUND(MAX(B46,0),0) =MAX(D45+E45-C46,0) =IF(D46<$E$4,$G$4,IF(D46>$E$5,$G$5,E45)) =D46*$B$6 =$B$7*IF(E46<>E45,1,0) =F46+G46
34 =RiskNormal($B$4,$B$5) =ROUND(MAX(B47,0),0) =MAX(D46+E46-C47,0) =IF(D47<$E$4,$G$4,IF(D47>$E$5,$G$5,E46)) =D47*$B$6 =$B$7*IF(E47<>E46,1,0) =F47+G47
35 =RiskNormal($B$4,$B$5) =ROUND(MAX(B48,0),0) =MAX(D47+E47-C48,0) =IF(D48<$E$4,$G$4,IF(D48>$E$5,$G$5,E47)) =D48*$B$6 =$B$7*IF(E48<>E47,1,0) =F48+G48
36 =RiskNormal($B$4,$B$5) =ROUND(MAX(B49,0),0) =MAX(D48+E48-C49,0) =IF(D49<$E$4,$G$4,IF(D49>$E$5,$G$5,E48)) =D49*$B$6 =$B$7*IF(E49<>E48,1,0) =F49+G49
37 =RiskNormal($B$4,$B$5) =ROUND(MAX(B50,0),0) =MAX(D49+E49-C50,0) =IF(D50<$E$4,$G$4,IF(D50>$E$5,$G$5,E49)) =D50*$B$6 =$B$7*IF(E50<>E49,1,0) =F50+G50
38 =RiskNormal($B$4,$B$5) =ROUND(MAX(B51,0),0) =MAX(D50+E50-C51,0) =IF(D51<$E$4,$G$4,IF(D51>$E$5,$G$5,E50)) =D51*$B$6 =$B$7*IF(E51<>E50,1,0) =F51+G51
39 =RiskNormal($B$4,$B$5) =ROUND(MAX(B52,0),0) =MAX(D51+E51-C52,0) =IF(D52<$E$4,$G$4,IF(D52>$E$5,$G$5,E51)) =D52*$B$6 =$B$7*IF(E52<>E51,1,0) =F52+G52
40 =RiskNormal($B$4,$B$5) =ROUND(MAX(B53,0),0) =MAX(D52+E52-C53,0) =IF(D53<$E$4,$G$4,IF(D53>$E$5,$G$5,E52)) =D53*$B$6 =$B$7*IF(E53<>E52,1,0) =F53+G53
41 =RiskNormal($B$4,$B$5) =ROUND(MAX(B54,0),0) =MAX(D53+E53-C54,0) =IF(D54<$E$4,$G$4,IF(D54>$E$5,$G$5,E53)) =D54*$B$6 =$B$7*IF(E54<>E53,1,0) =F54+G54
42 =RiskNormal($B$4,$B$5) =ROUND(MAX(B55,0),0) =MAX(D54+E54-C55,0) =IF(D55<$E$4,$G$4,IF(D55>$E$5,$G$5,E54)) =D55*$B$6 =$B$7*IF(E55<>E54,1,0) =F55+G55
43 =RiskNormal($B$4,$B$5) =ROUND(MAX(B56,0),0) =MAX(D55+E55-C56,D0) =IF(D56<$E$4,$G$4,IF(D56>$E$5,$G$5,E55)) =D56*$B$6 =$B$7*IF(E56<>E55,1,0) =F56+G56
44 =RiskNormal($B$4,$B$5) =ROUND(MAX(B57,0),0) =MAX(D56+E56-C57,0) =IF(D57<$E$4,$G$4,IF(D57>$E$5,$G$5,E56)) =D57*$B$6 =$B$7*IF(E57<>E56,1,0) =F57+G57
45 =RiskNormal($B$4,$B$5) =ROUND(MAX(B58,0),0) =MAX(D57+E57-C58,0) =IF(D58<$E$4,$G$4,IF(D58>$E$5,$G$5,E57)) =D58*$B$6 =$B$7*IF(E58<>E57,1,0) =F58+G58
46 =RiskNormal($B$4,$B$5) =ROUND(MAX(B59,0),0) =MAX(D58+E58-C59,0) =IF(D59<$E$4,$G$4,IF(D59>$E$5,$G$5,E58)) =D59*$B$6 =$B$7*IF(E59<>E58,1,0) =F59+G59
47 =RiskNormal($B$4,$B$5) =ROUND(MAX(B60,0),0) =MAX(D59+E59-C60,0) =IF(D60<$E$4,$G$4,IF(D60>$E$5,$G$5,E59)) =D60*$B$6 =$B$7*IF(E60<>E59,1,0) =F60+G60
48 =RiskNormal($B$4,$B$5) =ROUND(MAX(B61,0),0) =MAX(D60+E60-C61,0) =IF(D61<$E$4,$G$4,IF(D61>$E$5,$G$5,E60)) =D61*$B$6 =$B$7*IF(E61<>E60,1,0) =F61+G61
49 =RiskNormal($B$4,$B$5) =ROUND(MAX(B62,0),0) =MAX(D61+E61-C62,0) =IF(D62<$E$4,$G$4,IF(D62>$E$5,$G$5,E61)) =D62*$B$6 =$B$7*IF(E62<>E61,1,0) =F62+G62
50 =RiskNormal($B$4,$B$5) =ROUND(MAX(B63,0),0) =MAX(D62+E62-C63,0) =IF(D63<$E$4,$G$4,IF(D63>$E$5,$G$5,E62)) =D63*$B$6 =$B$7*IF(E63<>E62,1,0) =F63+G63
51 =RiskNormal($B$4,$B$5) =ROUND(MAX(B64,0),0) =MAX(D63+E63-C64,0) =IF(D64<$E$4,$G$4,IF(D64>$E$5,$G$5,E63)) =D64*$B$6 =$B$7*IF(E64<>E63,1,0) =F64+G64
52 =RiskNormal($B$4,$B$5) =ROUND(MAX(B65,0),0) =MAX(D64+E64-C65,0) =IF(D65<$E$4,$G$4,IF(D65>$E$5,$G$5,E64)) =D65*$B$6 =$B$7*IF(E65<>E64,1,0) =F65+G65

Mean Annual Cost

Mean Profit 30 40 50 60 70 80 114141.48 106346.82 103071.24 102708.66 105158.58 108302.7

U

Tom – Part II_Summary Report

@RISK Summary Reports
This case has trying to evaluate its inventory. This can be comunicated across the all levels by having the
evalaution of good performces of its annual costs, asstets, and revenues.
The production for each week is found such that if the inventory for the previous week is less than 30 units, then the
production level is 130 units; if the previous week’s inventory is greater than 80 units, then the production level will be 110 units; otherwise,
the production level is kept at the same level as the previous week. The inventory level for each week is also found
by summing the previous week’s inventory level with the production level, then subtracting the demand. The total
cost for each week is found by multiplying the inventory level by the holding cost of $30 per unit. If the production level was switched,
then there is a cost of $3,000 in addition. Finally, the total average annual cost was found by adding all of the weekly costs together.
@Risk is used to run 500 iterations of the simulations with a range of upper limit (U) inventory values. First, we constructed a table with six simulations with U values ranging from 30 units to 80
units, each having an incremental increase of 10 units from the previous simulation. Second, we used =RiskSimtable() formula to pull the U values from the table that we have just created. Third, we
added =RiskOutput() to our average total annual cost. Finally, we are able to find the value of U that gives us the lowest average cost
1.      The inventory level over the span of 52 weeks is shown below. The corresponding total cost for the 52-week period is $104,451.58. The model is shown in Exhibit 1.
1.      With the values of U ranging from 30 to 80 units in increments of 10 units (L = 30 units throughout), the average total average annual cost of each upper inventory level is:
a.       The average total cost of $114,375.37 when U=30 units
b.      The average total cost of $106,614.36 when U=40 units
c.       The average total cost of $103,177.16 when U=50 units
d.      The average total cost of $102,633.32 when U=60 units
e.       The average total cost of $104,305.55 when U=70 units
f.       The average total cost of $108,073.62 when U=80 units
Using the simulated results, the average 52-week cost for each value is shown in the table below, along with their standard deviations and 95% confidence intervals
The 95% Confidence Interval can be generate using “RiskPercentile”. The graph of total cost versus U is also shown below
From the simulated results, the best upper inventory level(U) is 60 units when the lower inventory level (L) = 30 units, because it results in the lowest average total cost.
The simulation is shown in Exhibit 1
Simulation Decision: Upper Inventory Level [units] Average Total Annual Cost [$] Standard deviation [$] Lower Limit Confidence Interval (95%) [$] Upper Limit Confidence Interval (95%) [$]
1 30 114,354.37 12,237.20 89,789.51 138,169.19
2 40 106,614.36 10,952.92 83,491.46 126,169.19
3 50 103,177.16 10,101.97 82,704.06 121,533.62
4 60 102,633.32 10,023.55 82,318.04 120,876.08
5 70 104,305.55 10,328.10 83,833.10 124,279.20
6 80 108,073.62 10,502.45 85,917.21 129,116.30
Other than finding the best upper inventory limit, Ebony can also conduct the same analysis to find optimum lower limits of the inventory before switching the production level.
If the lower limit is to low, the company may not be able to react to the sudden surge in demand. As a result, the company will lose
sales from the supply shortage. Therefore, the company need to find the optimize level of inventory that will minimize the total cost, which included the cost of loss of sales.
Furthermore, they should investigate the optimum production level to maintain over the 52-week period. If it is possible for them to adjust their production level to more consistent value, then they
will not have lower the switching cost. Lastly, if the production capacity is not limited to the incremented of 10, then the company may need to find the new optimize production level which will
minimize the total cost while meeting most of the demand.
General Information
Workbook Name C16_3.xls
Number of Simulations 6
Number of Iterations 500
Number of Inputs 53
Number of Outputs 1
Sampling Type Latin Hypercube
Simulation Start Time 7/7/00 9:29:28
Simulation Stop Time 7/7/00 9:29:41
Simulation Duration 0:00:13
Random Seed 144998495
Output and Input Summary Statistics
Output Name Output Cell Simulation Minimum Maximum Mean Std Dev 5% 95%
Annual cost $H$9 1 82200 147180 114141.48 11768.2144078998 94590 133440
2 72930 130650 106346.82 10118.5800293926 88890 121950
3 72270 126930 103071.24 9776.486887124 85560 117990
4 65490 130560 102708.66 9673.5833709577 86280 117780
5 71040 131820 105158.58 10306.1854777509 88290 122010
6 71040 142140 108302.7 10963.5784219217 89370 125940
Input Name Input Cell Simulation Minimum Maximum Mean Std Dev 5% 95%
If inventory > $E$5 1 30 30 30 0 30 30
2 40 40 40 0 40 40
3 50 50 50 0 50 50
4 60 60 60 0 60 60
5 70 70 70 0 70 70
6 80 80 80 0 80 80
Normal $B$14 1 69.962928772 172.092666626 120.0113503265 15.0638986751 95.0924758911 144.4532928467
2 69.962928772 172.092666626 120.0113503265 15.0638986751 95.0924758911 144.4532928467
3 69.962928772 172.092666626 120.0113503265 15.0638986751 95.0924758911 144.4532928467
4 69.962928772 172.092666626 120.0113503265 15.0638986751 95.0924758911 144.4532928467
5 69.962928772 172.092666626 120.0113503265 15.0638986751 95.0924758911 144.4532928467
6 69.962928772 172.092666626 120.0113503265 15.0638986751 95.0924758911 144.4532928467
Normal $B$15 1 71.2089691162 163.7991485596 119.9849081116 14.9976170327 95.3269577026 144.6170654297
2 71.2089691162 163.7991485596 119.9849081116 14.9976170327 95.3269577026 144.6170654297
3 71.2089691162 163.7991485596 119.9849081116 14.9976170327 95.3269577026 144.6170654297
4 71.2089691162 163.7991485596 119.9849081116 14.9976170327 95.3269577026 144.6170654297
5 71.2089691162 163.7991485596 119.9849081116 14.9976170327 95.3269577026 144.6170654297
6 71.2089691162 163.7991485596 119.9849081116 14.9976170327 95.3269577026 144.6170654297
Normal $B$16 1 71.1585159302 167.9855041504 119.9977834778 15.0205687364 95.1532897949 144.4892272949
2 71.1585159302 167.9855041504 119.9977834778 15.0205687364 95.1532897949 144.4892272949
3 71.1585159302 167.9855041504 119.9977834778 15.0205687364 95.1532897949 144.4892272949
4 71.1585159302 167.9855041504 119.9977834778 15.0205687364 95.1532897949 144.4892272949
5 71.1585159302 167.9855041504 119.9977834778 15.0205687364 95.1532897949 144.4892272949
6 71.1585159302 167.9855041504 119.9977834778 15.0205687364 95.1532897949 144.4892272949
Normal $B$17 1 76.2275390625 168.0211791992 120.0103268585 14.9868392666 95.0752334595 144.6616973877
2 76.2275390625 168.0211791992 120.0103268585 14.9868392666 95.0752334595 144.6616973877
3 76.2275390625 168.0211791992 120.0103268585 14.9868392666 95.0752334595 144.6616973877
4 76.2275390625 168.0211791992 120.0103268585 14.9868392666 95.0752334595 144.6616973877
5 76.2275390625 168.0211791992 120.0103268585 14.9868392666 95.0752334595 144.6616973877
6 76.2275390625 168.0211791992 120.0103268585 14.9868392666 95.0752334595 144.6616973877
Normal $B$18 1 69.7681884766 163.5778045654 119.9828871307 15.0067829976 95.3220596313 144.5030212402
2 69.7681884766 163.5778045654 119.9828871307 15.0067829976 95.3220596313 144.5030212402
3 69.7681884766 163.5778045654 119.9828871307 15.0067829976 95.3220596313 144.5030212402
4 69.7681884766 163.5778045654 119.9828871307 15.0067829976 95.3220596313 144.5030212402
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