Question

Problem 1.[15 points] Suppose that a metal rod is supposed to be cut 74.0 mm. in length.The upper spec limit is set at 78.5 mm., and the lower spec limit set at 69.5 mm. A metalrod that is out of spec must be scrapped at a cost of$42.00. The company uses theTaguchiquality loss functionto estimate quality costs. Suppose that a sample of 5 units was taken,and the rod measurements were: 75.6, 78.2, 68.7, 71.0 and 73.6 mm., respectively.What is the totalTaguchi quality costof that sample of 5 units? Please round all costs tothe nearest penny.Problem 2.[10 points] The manager of an apartment complex feels overwhelmed by thenumber of complaints she is receiving. Below is the check sheet she has kept for the past 12weeks. Develop aPareto chartusing this information.What recommendations would you make?WeekGroundsParkingPoolTenantIssuesElectrical/Plumbing1√√√√√√√√√2√√√√√√√√√3√√√√√√√√√4√√√√√√√√√5√√√√√√√√√√√6√√√√√√√7√√√√√√√8√√√√√√√√√√√9√√√√10√√√√√√√√√11√√√√√√12√√√√√√√√√1

Answer 1

1) The total Taguchi Quality Cost is;

L_total = $119.2

2) The recommendation for Mary Beth Mars because;

She should focus more on the parking/drive complaints, pool complaints  and grounds complaints.

Problem 1;

We are given that;

Dimension of length to be cut; D_i = 74 mm

Dimension of Length of upper spec limit; D_u = 78.5 mm

Dimension of Length of lower spec limit; D_l = 69.5 mm

Cost at which metal must be scrapped; L = $42

Let's find the unit formula to find cost for each of the 5 units sample.

T = L/(D_i - D_l)²

Plugging in the relevant values, we have;

T = 42/(74 - 69.5)²

T = 42/20.25

T = 2.074

Thus, the cost for each of the 5 units would be gotten from the formula;

L = 2.074(D_i - D_l)²

The measurement of each sample was given as;

75.6mm, 78.2mm, 68.7mm, 71.0mm and 73.6 mm.

Thus;

L₁ = 2.074(D_i - D_1)²

L₁ = 2.074(74 - 75.6)²

L₁ = $5.3094

Problem 1;

We are given that;

Dimension of length to be cut; D_i = 74 mm

Dimension of Length of upper spec limit; D_u = 78.5 mm

Dimension of Length of lower spec limit; D_l = 69.5 mm

Cost at which metal must be scrapped; L = $42

Let's find the unit formula to find cost for each of the 5 units sample.

T = L/(D_i - D_l)²

Plugging in the relevant values, we have;

T = 42/(74 - 69.5)²

T = 42/20.25

T = 2.074

Thus, the cost for each of the 5 units would be gotten from the formula;

L = 2.074(D_i - D_l)²

The measurement of each sample was given as;

75.6mm, 78.2mm, 68.7mm, 71.0mm and 73.6 mm.

Thus;

L₁ = 2.074(D_i - D_1)²

L₁ = 2.074(74 - 75.6)²

L₁ =  $5.30944

L₂ =  2.074(74 - 78.2)²

L₂ = $36.58536

L₃ = 2.074(74 - 68.7)²

L₃ = $58.25866

L₄ = 2.074(74 - 71)²

L₄ = $18.666

L₅ = 2.074(74 - 73.6)²

L₅ = $0.33184

Total Taguchi Quality Cost = L₁ + L₂ + L₃ + L₄ + L₅

Total Taguchi Quality Cost = $5.30944 + $36.58536 + $58.25866 + $18.666 + $0.33184

Total Taguchi Quality Cost = $119.1513

Approximating to the nearest penny gives;

Total Taguchi Quality Cost = $119.2

Problem 2;

The check sheet she has kept for the past 12 weeks is missing and so i have attached it.

From the check sheet attached we can see that;

Frequency of parking/drive complaints = 30

Frequency of pool complaints = 20

Frequency of grounds complaints = 17

Frequency of Tenant issues complaints = 14

Frequency of Electric/plumbing complaints = 5

With those frequencies, i have drawn a pareto's chart and attached it.

Pareto's chart is a bar graph that helps us to understand what is the priority per time. We can see from the chart that the complaint with the highest frequency is that of parking/drive which makes up around 34.88%. Then pool complaints which is 23.26% and then ground complaints which consists of 19.77%.

This 3 complaints add up to about approximately 78% which is close to 80% and thus Mary Beth Marrs is recommended to focus on these 3 issues primarily according to pareto's principle.

Read more at; brainly.com/question/17989104