Project #73869 - 4 short questions to answer just your view, no calculations needed and don't need to cite work. no format needed

 

4 short questions to answer just your view, no calculations needed and don't need to cite work. No format needed.

1.Alternative 2 would be the best option for me.  The anticipated worth of alternative two is near 100,000 dollars less than alternative 1.  In the final two demand instances the entire cost is less than alternative two than in one.  However in the 25000 units it would be lesser to go with alternative one since its entire cost is $200,000 lesser than two.  Generally the costs are lesser in alternative 2 than they are in alternative 1.

Capacity    $100,000(fixed cost)       $10 (variable cost)

Total cost    100,000 + 10(x)

100,000 + 10(25,000)=                    350,000  low

100,000 + 10(60,000)=                    700,000  medium

100,000 + 10(100,000)=                  1,100,000 high

 

EV       (350,000 x 30%) + (700,000 x 40%) +(1,100,000 x 30%) =  715,000

 

Capacity     $500,000(fixed cost)         $2 (variable cost)

Total cost   500,000 + 2(x)

500,000 + 2(25,000)=                      550,000  low

500,000 + 2(60,000)=                      620,000   medium

500,000 + 2(100,000)=                    700,000   high

EV       (550,000 x 30%) + (620,000 x 40%) + (700,000 x 30%)  =623,000

 

2

Option 1 Capacity $500,000 (fixed). $2 (variable) Total cost. 500,000 + 2(x) 500,000+ 2(25,000)= 550,000 Low 500,000+2(60,000)= 620,000 Med 500,000+2(100,000)=700,000 High EV 30%*(550,000)+40%*(620,000)+30%*(700,000)= 623,000 Option 2 Capacity $100,000 (fixed). $10 (variable) Total cost 100,000+10(x) 100,000+10(25,000)= 350,000 Low 100,000+10(60,000)= 700,000 Med 100,000+10(100,000)= 1,100,000 High EV 30%*(350,000)+40%*(700,000)+30%*(1,100,000)=715,000 I believe with a higher demand option two would work, but my preferred option is one. The fixed cost is a bit higher but the cost per unit is quite lower. While the overall cost is lower in option one as well
 
3

A decision tree provides a structured graphical technique that assists in the process of decision making (WikiHow, 2015). It is extremely useful in the cases where there are a series 

of decisions to choose from.

The decision tree shows that when the demand in units annually is smaller than 40000, it is advisable to choose the second option. The cost for 40000 units adds up to 400000 dollars and upon adding the fixed cost unit of 100000 dollars, the total price comes up to 500000 dollars which is the fixed cost pricing for the first option. Above that level however, the costs then turn in favour of the first option which turns out to be the cheaper option. As can be seen from the other two potions, the second option is considerably cheaper than the first option. However, in this case the managers must select one option before they are notified of the demand levels that are applicable. Therefore the managers must weigh all the relevant options before deciding on which option to take. Since the highest probability demand is for 60000 units, then the manager should select the first option with the fixed cost of 500000 dollars since the overall cost will be 80000 dollars cheaper.  There is an equal probability that it could be 25000 units or 100000 units meaning that there is a 70% probability that the demand will be above the desired 40000 units. It is therefore prudent to select the option with the 500000 fixed costs.

 

4Option 1 

Fixed Cost $500,000Variable Cost $2 per unit
 
Demand (Units per year)Probability
25,00030%
60,00040%
100,000                        30%

500k + 2(25k) = 500k + 50k = 550k(30%)
500k + 2(60k) = 500k + 120k = 620k(40%)
500k + 2(100k) = 500k + 200k = 700k(30%)
                                            
550k(.30) = 165k
620k(.40) = 248k
700k(.30) = 210k
EV = 623k

Option 2
Fixed Cost $100,000 Variable Cost $10 per unit
 
Demand (Units per year) Probability
25,000                         30%
60,000                                     40%
100,000                                   30%

100k + 10(25k) = 100k + 250k = 350k(30%)
100k + 10(60k) = 100k + 600k = 700k(40%)
100k + 10(100k) = 100k + 1M = 1,100,000(30%)
                                            
350k(.30) = 105k
700k(.40) = 280k
1,100,00(.30) = 330k
        EV = 715k

To compare: 
Option 1            Option 2
550k(.30) = 165k350k(.30) = 105k
620k(.40) = 248k             700k(.40) = 280k
700k(.30) = 210k             1,100,000(.30) = 330k

EV = 623kEV = 715k

I would prefer to go with the Option 1, where the fixed cost is high but the variable cost is low. At first glance it might seem like option 2 is the better value because of the lower fixed cost. Option 2 would be the better value only when demand is low. When demand is in middle or high range, option 2 will only cost Arktec more money. Option 1 is the better value overall because there is a combined higher probability that demand will be in middle or high range.

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