Iscom 305 week 2 individual assignement
Week Two Individual Assignment
Lewis Mack
Iscom/305 - Systems operations management
May 11, 2015
Ryan Derrickson
Week Two Individual Assignment
1-1
Problem solutions
Annandale= $ 40,000 in 250 hours, per hour production is 40000/250 =160
Blacksburg = $12,000 in 60 hours, per hour productivity is 12000/60= 200
Charlottesville = $60,000 in 500 hours per hour productivity 60000/500=120
Danville = $25000 in 200 hours per hour productivity is 25000/200= 50
The location with the highest productivity
Using the per hour production, Blacksburg has the highest labor productivity as stated with the number of sales per hour.
Problem 1-2
Average wage
Annandale at $6.75 an hour, and rent $1800, labor 250
Blacksburg pays $6.50 an hour, rent $1200, labor hours 60
Charlottesville $6, rent $2000, labor hour 500
Danville $5.50, rent $800, labor hour 200
Therefore for cost of production would be
Annandale= (6.75 X250) + 1800= 3487.50
Blacksburg = (6.50 X 60) + 2000= 2390
Charlottesville = (6 X 500) + 1200= 4200
Daville = (5.5 X 200) +800= 1900
The most productive site is will be gotten by finding the ratio of sales and cost of production.
Annandale =$ 40,000/3487.50= 11.45
Blacksburg= 12,000/2390 = 5.02
Charlottesville =$60,000/4200= 14.29
Daville = $25000/1900= 13.16
Blacksburg remains the one with the biggest productivity due to the lowest ratio
b.
To know the store to close using the cost multifactor will depend on the ratio of the cost of labor and the total cost of productivity.
Therefore
Cost of labor
Annandale= (6.75 X250)/3487.50= 0.48
Blacksburg = (6.50 X 60)/2390 = 0.16
Charlottesville = (6 X 500)/4200= 0.71
Daville = (5.5 X 200)/1900= 0.58
The company can close the Charlottesville location, but deliberation of other things such as consumer satisfaction, the need to serve the market, the market effectiveness as well as the raw materials required in the region.
Problem 2-1
(Please see excel file given)
a. Looking at the percentage failure cost as a percentage of
Lewis Mack
Iscom/305 - Systems operations management
May 11, 2015
Ryan Derrickson
Week Two Individual Assignment
1-1
Problem solutions
Annandale= $ 40,000 in 250 hours, per hour production is 40000/250 =160
Blacksburg = $12,000 in 60 hours, per hour productivity is 12000/60= 200
Charlottesville = $60,000 in 500 hours per hour productivity 60000/500=120
Danville = $25000 in 200 hours per hour productivity is 25000/200= 50
The location with the highest productivity
Using the per hour production, Blacksburg has the highest labor productivity as stated with the number of sales per hour.
Problem 1-2
Average wage
Annandale at $6.75 an hour, and rent $1800, labor 250
Blacksburg pays $6.50 an hour, rent $1200, labor hours 60
Charlottesville $6, rent $2000, labor hour 500
Danville $5.50, rent $800, labor hour 200
Therefore for cost of production would be
Annandale= (6.75 X250) + 1800= 3487.50
Blacksburg = (6.50 X 60) + 2000= 2390
Charlottesville = (6 X 500) + 1200= 4200
Daville = (5.5 X 200) +800= 1900
The most productive site is will be gotten by finding the ratio of sales and cost of production.
Annandale =$ 40,000/3487.50= 11.45
Blacksburg= 12,000/2390 = 5.02
Charlottesville =$60,000/4200= 14.29
Daville = $25000/1900= 13.16
Blacksburg remains the one with the biggest productivity due to the lowest ratio
b.
To know the store to close using the cost multifactor will depend on the ratio of the cost of labor and the total cost of productivity.
Therefore
Cost of labor
Annandale= (6.75 X250)/3487.50= 0.48
Blacksburg = (6.50 X 60)/2390 = 0.16
Charlottesville = (6 X 500)/4200= 0.71
Daville = (5.5 X 200)/1900= 0.58
The company can close the Charlottesville location, but deliberation of other things such as consumer satisfaction, the need to serve the market, the market effectiveness as well as the raw materials required in the region.
Problem 2-1
(Please see excel file given)
a. Looking at the percentage failure cost as a percentage of