qam-amorefrozenfood - group8

QAM - II

Case Analysis: Amore Frozen Foods

------------------------------------------------------ Submitted To: Prof. Asok Chaudhuri ------------------------------------------------------

Submitted By:

Group # 8 (Section – A)

Aditaya Maini WMP7006 Aditya Shankar WMP7007 AkshayMathur WMP7008 Ankur Bansal WMP7013 Arun Kumar WMP7014 Ashish Kumar WMP7015

Amore Frozen Foods

Dear Mr. Jenkins,

As asked by the company we have done detailed cost benefit analysis of various options of

filling 8.0 oz, 8.22 oz and 8.44 oz of Macaroni and Cheese in our packing as to reach a

decision on our filling policy. This report elucidates the cost benefit analysis of three

possible quantities of Marconi & Cheese for various products of Amore Frozen foods.

Primary reason for shifting to 8.44 ounce of filling was the shortage of power supply at

stores which, no more exists, and hence the need to review our filling policy. The detailed

data, assumptions and calculations are attached in the exhibits for your reference.

For the cost benefit estimation on filling of Macaroni and Cheese in various products of

Amore Frozen foods on the basis of sample reports received from sampling department and

cost factors provided by the financial & manufacturing department (Exhibit 1).

Methodology adopted by us is that we have taken each possible case of 8.0oz, 8.22oz , and

present filling quantity of 8.44oz and based on sampling method suggested by FDA , what

could be possible fine and rejection cost associated with filling quantity and compared to

additional cost involved in filling the extra quantity over the required 8.oz. A test (Chi-

square, Exhibit 4) has been carried out to ensure that the samples received are independent

of each other and having normal distribution (Exhibit 5)

From the analysis, we found that the amount of 8.44 ounce of Macaroni & Cheese would

provide us the most optimum result for maximum annual net profit from the product line of

various products. Hence we would recommend this quantity of Macaroni and Cheese filling

in all the Amore frozen foods.

The analysis done is based on following assumptions:

1) Power: Primary reason for shifting to 8.44 oz of filling was the shortage of power

supply which no longer exists now.

2) Control chart: An analysis of R-chart & X-Bar chart for weight control is carried out

based on 8.44 oz of macaroni & cheese filling. (Exhibit 2)

3) The samples received from sampling department are normally distributed (Exhibit 3)

and are independent of each other (Chi-square test, Exhibit 4)

4) Standard deviation of sample is comparable to that standard deviation of population

implying analysis of sample can be extended to whole population (Exhibit 5).

5) ANOVA test & F-test are carried out across the samples shows that sample means &

variances are same & hence comparable (Exhibit 6, 7).

Following are the discussions on various quantities (8.0 oz, 8.22 oz & 8.44 oz):

1) For 8 oz filing 50% of our fillings will have the probability of being underweight and hence

this will be non-compliant of the FDA norms, thereby creating a bad brand name for the

company. Hence this option is completely ruled out (Exhibit 10).

2) For 8.22 oz filling about 1.25% of the fillings will be below the required 8oz , and as per

our production capacity 108000 𝑓𝑖𝑙𝑙𝑖𝑛𝑔𝑠 will be below 8oz. Loss from pies which cannot be

sold due to being underweight will be $211680 and possible fine by FDA on underweight

fillings will be $ 236850, making an overall extra loss of $448530 (Exhibit 9).

3) For 8.44 oz filling 0.000027% fillings will be below the required weight of 8.0 oz, and

considering the annual production capacity the number of pies having weight below 8 oz

will be negligible (Exhibit-8). The extra cost of filling 8.44 oz i.e. 0.22 extra cheese and

ingredients will be $ 420858 and probability of finding the under-weight samples by FDA is

negligible (0.000027%). This would also help in building company’s brand name.

Thus considering all the factors explained for the three filling options our team

recommends that we should continue with the existing policy of filling size of 8.44oz. The

ingredient cost saving obtained from shifting to 8.22 from the current 8.44 will be wiped

away by the FDA fines and the wastage of those fillings that cannot be sold anywhere

including the thrift store (way below the required weight). Therefore the 8.44 oz filling

seems to be the best solution in the given circumstances as the cost extra cost incurred in

ingredient cost can be compensated by the less wastage and significantly less percentage

of underweight pies and hence saving the cost and fine of FDA.

This option will also create a good brand image of our company as people are very cautious

and sensitive about food they consume.

Exhibit 1

AMORE FROZEN FOODS (A) MACRONI AND CHEESE FILL TARGETS

Assumptions

Production:

Production (Per Month) = 60000 dozen

Production Rate = 1000 dozen per 20 minutes

Annual Production = 60000 * 12 = 720000 dozens

Cost:

Selling Price = $ 4.5 per dozen

Standard Cost = $3.0

Contribution Margin (CM) = $ 4.5 - $ 3.0 = $1.5 per dozen

Thrift Store:

Sale Price = $3.60 per dozen

Demand = 60 dozen per week

Store Capacity = 1000 dozen pies

Annual Demand = 60 * 52 = 3120 dozens (52 weeks in a year)

• Energy crisis period is minimal and underweight frozen macaroni and cheeses are virtually non-existent in the industry.

• Underweight rejected batches are first sold at Thrift store and extra pies are donated to charity.

• FDA fines up to $ 15 for each 8 ounces package found to be substantially underweight. • Sample Data is normally distributed. (Refer Exhibit 3) • Sample Data is independent. (Refer Exhibit 4) • FDA is also using the same FDA approved process to detect the underweight pies that

already Amore Frozen Food is using.

Exhibit 2


Weight Control Chart

Here all the data points are within control limits. Therefore, no process stoppage or adjustment is required to eliminate any assignable cause of variation.

Exhibit 3


Normality Test

Here by analysing the Normal Probability Plot and Normal Distribution Histogram we can say that data is normally distributed with LeftSkewed.

Mean = 8.492 Median = 8.52 Mode = 9.0

Also Mean < Median < Mode in the sample data that also prove sample data is slightly Left Skewed.

7.88

8.28.48.68.8

9

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

Wei

ght

Z Value

Normal Probability Plot

0

5

10

15

20

1 2 3 4 5 6 7 8 9 10

Normal Distribution Histogram

Exhibit 4


Independent Test

Here we have analysed the sample data by Chi-Square Independence Test.

For a test of independence, the null and alternative hypothesis follows:

H0: Samples are independent H1: Samples are not independent.

Chi-Square Test Data Level of Significance 0.05 Number of Rows 5 Number of Columns 15 Degrees of Freedom 56

Results Critical Value 74.47 Chi-Square Test Statistic 0.283 p-Value 1 Do not reject the null hypothesis

Results indicate that we cannot reject the null hypothesis and there is no sufficient evidence to say that samples are different. So we can make the assumption of independent samples.

Exhibit 5


Population vs. Sample Standard Deviation

Standard Deviation of population: = 0.22

And by analysing the standard deviation of 15 samples taken in a day is:

Observations 8.64 8.54 8.5 8.36 8.52 8.56 8.52 8.34 8.66 8.42 8.28 8.6 8.6 8.6 8.2

8.6 8.76 8.6 8.34 8.44 8.54 8.44 8.84 8.82 8.5 8.56 8.8 8.5 8.8 8.7 8.7 8.68 8.2 8.52 8.38 8.5 8.82 8.78 8 8.4 8.58 8.7 8.8 8.4 8.1 8.5 8.54 8.3 8.68 8.5 8.7 8.18 8.06 8.94 8.16 8.32 8.4 8.2 8.6 8.5

8.52 8.72 8.6 8.2 8.52 8.44 8.36 8.38 8.12 8.32 8.38 8.4 8.6 8.2 8.7

Standard Deviation of Samples: S = 0.21

Here Sample SD is lower than Population SD.(0.21< 0.22)

It interprets data we used to work is not the complete population, but only a sample. When a sample is used we need to estimate the standard deviation from the sample. In this situation 1 is subtracted from the N number of cases. For small samples this N-1 tends to change the standard deviation slightly, making it a conservative estimate of the population's parameter. As sample size increases the effect of N-1 declines and the results of the two formulas converge towards the population parameter.

Exhibit 6


Difference among the Observed Sample’s Mean

Observations 8.64 8.54 8.5 8.36 8.52 8.56 8.52 8.34 8.66 8.42 8.28 8.6 8.6 8.6 8.2

8.6 8.76 8.6 8.34 8.44 8.54 8.44 8.84 8.82 8.5 8.56 8.8 8.5 8.8 8.7 8.7 8.68 8.2 8.52 8.38 8.5 8.82 8.78 8 8.4 8.58 8.7 8.8 8.4 8.1 8.5 8.54 8.3 8.68 8.5 8.7 8.18 8.06 8.94 8.16 8.32 8.4 8.2 8.6 8.5

8.52 8.72 8.6 8.2 8.52 8.44 8.36 8.38 8.12 8.32 8.38 8.4 8.6 8.2 8.7

H0: µ1 = µ1 =µ2 =µ3 =µ4 =µ5 =µ6 …………….. =µ15 H1: Not all µj are equal (where j = 1, 2, 3, 4……15)

ANOVA (Single Factor: Weight) Source of Variation SS df MS F P-value F critical

Between Groups 0.398592 14 0.028471 0.629812 0.829555 1.860242 Within Groups 2.71232 60 0.045205

Total 3.110912 74

Here we found that because computed t statistics F = 0.629812 is less than FU = 1.86, so reject the null hypothesis and concluded that there is no difference in the mean tensile strength among the fifteen samples. And also by p-value chances that 83 % times mean for fifteen samples are equal.

Exhibit 7


Difference among the Observed Sample’s Variance

Here we have used F test for differences in Variances and compare some samples as

H0: 12 = 2

2

H1: 1 2≠ 2

2

F Test for Differences in Two Variances

Data Level of Significance 0.05

Population 1 Sample Sample Size 5 Sample Standard Deviation 0.08

Population 2 Sample Sample Size 5 Sample Standard Deviation 0.1

Intermediate Calculations F Test Statistic 0.64 Population 1 Sample Degrees of Freedom 4 Population 2 Sample Degrees of Freedom 4

Two-Tail Test Lower Critical Value 0.104118 Upper Critical Value 9.60453 p-Value 0.67602

Do not reject the null hypothesis

And by interpreting we find out that we cannot reject the null hypothesis and samples are having same variance.

Similarly we have analysed for multiple samples and these results are we cannot reject the null hypothesis and samples are having same variance.

Exhibit 8


Fill Target at 8.44 oz.

8.44 oz. fill target with 2 standard deviation; 95.45% of package are with weight above 8.0 oz.

So the probability of package to be underweight is:

𝑍 = (8−8.493)0.22√5

= -5.01 (Using Central Limit Theorem)

And the p-value for -5.01 is 0.00000027 which mean probability of package to be underweight is less than 0.000027%. It mean package to be underweight is almost negligible or non-existent.

With higher fill targets company is able to save FDA fine but every pie is filled with additional 0.22 oz. of macaroni and cheese.

And annual impact with 0.22 oz. is been filled extra in each box is:

0.22 ∗ 60000 ∗ 12 ∗ 12 = 1900800 𝑜𝑧.

Annual cost saving = 1900800 ∗1.82 8.22

= $ 420858.3942 ($ 1.82 Cheese & Macaroni ingredient cost)

Approximately $ 420858 is a significant opportunity for company to save that amount as there is no energy crisis in U.S.

In the total population number of packages will be less than 8 oz. will be:

𝑍 = (8−8.44)0.22

= -2

And the p-value for -2 is 0.0228 which means probability of package to be less than 8 oz. is 2.28 %.

And it will be 720000∗12∗2.28

100 = 196992 pies

So with 0.000027% probability that FDA will find underweight package in market will be:

Then number of pies will be: 196992 ∗0.000027

100 = 0.05319pies (It’s approximately nil.)

Opportunity for company to save the amount is $420858

Exhibit 9



8.22 oz. fill target with 1 standard deviation; 84% of package are with weight above 8.0 oz.

So the probability of package to be underweight is:

𝑍 = (8−8.22)0.22√5

= -2.24 (Using Central Limit Theorem)

And the p-value for -2.23607 is 0.0125 which means probability of package to be less than 8 oz. is 1.25 %.

It’s a significant percentage and the number of pies will be lower than 8 oz. will be:

60000 ∗ 12 ∗ 12 ∗ 1.25100

= 108000 𝑝𝑖𝑒𝑠

And to have package to be less than 7.50 oz. probabilities will be:

𝑍 = (7.50−8.22)0.22√5

= -7.32

Its p-value will be 0.000000006 which mean it’s more than 6 SD of the mean and the probability to have pie less than 7.50 oz. is almost negligible.

As company is having Thrift Store also so company can sell underweight package there as 7 oz.

With annually 108000 underweight pies company can easily sell (3120 * 12 = 37440 pies) to meet Thrift store demand with margin of ($ 3.60 - $ 3.0 = $ 0.6) which recovers the standard cost of pies and give some negligible amount of profit also.

Marconi package that have not been utilized (I mean gone for Charity and company haven’t earned any revenue from them) = 108000 – 37440 = 70560 pies

So loss from those pies are = 70560 * $3 = $ 211680

Company is running 720 production cycles to produce 720000 dozen pies.

In the total population number of packages will be less than 8oz. will be:

𝑍 = (8−8.22)0.22

= -1

And the p-value for -1 is 0.1587 which means probability of package to be less than 8 oz. is 15.87 %.

And it will be 720000∗12∗15.87

100 = 1371168 pies

Company is able to catch only 108000 underweight pies in FDA approved inspection process; rest of the pies (1371168 – 108000 = 1263168 pies) will go in market.

So the probability will be 1.25 % that FDA will find underweight package in market.

Then number of pies will be: 1263168 ∗1.25

100 = 15789.6 pies~= 15790 pies

Total FDA fine on those pies will be: 15790 * $ 15 = $ 236850

Overall loss will be: $ 211680 + $ 236850 = $ 448530

Exhibit 10



With 8.0 oz. target Z value will be 0 and the probability of package to be underweight is 50% with p value 0.5.

By seeing the ThriftStore demandsit’s not recommended to set the target 8.0 as approximately 50 % underweight package standard cost will not been recovered. And it will be huge loss to company.

qam-amorefrozenfood - group8

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