Suad Alwan, the purchasing agent for Dubai Airlines, is interested in determining what he can expect to pay for airplane number
4 if the third plane took 20,000 hours to produce. What would Alwan expect to pay for plane number 5? Number 6? Use an 85% learning curve and a $40-per-hour labor charge.
1 answer:
Answer:
Alwan expect to pay for airplane 4= $747818.48
Explanation:
given data
expect to pay airplane = 4
3rd plane produce = 20,000 hours
learning curve = 85%
solution
As here logarithmic approach allow get labor for any unit, TN, as
TN = T1(Nb)
here TN is time for the Nth unit and T1 is hours to produce the first unit
so
b = (log of the learning rate) ÷ (log 2) = slope of the learning curve
so
T3 = T1(3log(0.85)÷log2)
so we get
So Alwan expect to pay for airplane 4 = $747818.48
You might be interested in
Answer:
9644
Explanation:
cost of paper on which a card is printed = $0.40 per card
cost of printing = $0.10 per card
profit made per card sold = $3.75
number of areas where customers are located (n)= 4
mean of customers from each region = 2300
standard deviation for each region = 200
note : each region is independent
The optimal production quantity for the card can be calculated going through these steps
first we determine
the cost of card = $0.10 + $0.40 = $0.50
selling value = $3.75
salvage value = 0
next we calculate for the z value
= ( selling value - cost of card) / ( selling price - salvage value )
= ( 3.75 - 0.50 ) / 3.75 = 0.8667
Z( 0.8667 ) = 1.110926 ( using excel formula : NORMSINV ( 0.8667 )
next we calculate
<em>u</em> = n * mean demand
= 4 * 2300 = 9200
б = = 200 * 2
= 400
Hence optimal production quantity for the card
= <em>u</em> + Z (0.8667 ) * б
= 9200 + 1.110926 * 400
= 9644.3704
≈ 9644