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3 years ago

14

The Action Toy Company has decided to manufacture a new train set, the production of which is broken into six steps. The demand

for the train is 4,500 units per 40-hour workweek:
Task Performance Time (seconds) Predecessors
A 22 -
B 30 A
C 15 A
D 14 A
E 12 B, C
F 27 D, E
Required:
a) Given the demand, the cycle time for the production of the new train set = ___________ (round your response to one decimal place).
b) The theoretical minimum number of work stations is __________? (round your response up to the next whole number)

1 answer:

horsena [70]3 years ago

5 0

Answer:

a. 32 secs

b. 3.75 or 4

Explanation:

The computation is shown below:

(a) Cycle Time is

= Operation time per week ÷ output per week

= (40 hours × 60 × 60) ÷ (4,500)

= 32 secs

(b) And, the Theoretical minimum number of workstations required is

= Sum of total task times ÷ cycle time

= (22 + 30 + 15 + 14 + 12 + 27) ÷ (32 sec)

= 3.75 or 4

By applying the above formulas we can get the cycle time and the theoretical minimum number of work stations

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Your friend is looking for investors in a risky business venture. To convince you to participate, she is offering you a 17% rate

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Answer:

B - $13,556.82

Explanation:

Amount to be invested is equal to the present value of future inflows

Present value = future value/(1+Interest rate)^Number of years

The actual amount at the end of the first year should be $3000 if there is an answer in the options

The amount at the end of the second year is $4000

The amount at the end of the third, fourth and fifth year is $5000

Hence, amount to be invested = 3000/(1.17) + 4000/(1.17)^2 + 5000/(1.17)^3 + 5000/(1.17)^4 + 5000/(1.17)^5

= $13,556.82

Hence, the answer is $13,556.82

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Compared to a perfectly competitive market, a natural monopolist will have a.) ______________ (higher / lower / either higher or

kipiarov [429]

Answer:

A). Lower

B). Lower

C). Higher

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2 years ago

Tidwell Industries has the following overhead costs and cost drivers. Direct labor hours are estimated at 100000 for the year. A

Darya [45]

Answer:

Overhead rate for ordering and receiving = $300 per order.

Explanation:

Given Activity costs and activity drivers

Activity Overhead Cost Driver Activity

Ordering & Receiving Order $150,000 500 orders

Machine Setup $324,000 450 setups

Machining $1,587,500 125,000 MH

Assembly Parts $1,260,000 1,000,000 parts

Inspection $330,000 500 inspections

Under activity based costing the rates are based on the nature of activity and the rates are based on number of those activities per driver. As for ordering and receiving there are total of 500 orders and total cost = $150,000 that is cost per order will be computed.

Overhead rate for ordering and receiving = $150,000/500 orders = $300 per order.

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In your own opinion, why do you think it is important to establish a daily job search routine?

Alenkinab [10]

The answer is: because every day there is more opportunities opening up

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Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of wa

Radda [10]

Answer:

$1170839.28 dyn/cm^2$

16.9816 psia

$117.083928 kN/m^2$

Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

$P_{abs}=P_{atm}+P_G$

And the gauge pressure is given by the contributions of columns of water ( $P_{w}$ ) and mercury( $P_{Hg}$ ), we can calculate the contribution of each column as:

$P= \rho g h$ (*)

where $\rho$ is the respective density, g gravity and h is height.

So we have all the data required to use the above equations ( $P_{atm}$ , height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( $P_{atm}$ , $P_{w}$ and $P_{Hg}$ ). Note that the units "x" mmHg means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

$mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}$

$mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O$

And pressure of water in mmHg is

$0.0737*56=4.1246 mmHg$

The absolute pressure is:

$P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246 = 881.1246 mmHg = 88.11246cmHg$

To pass to dyn/cm^2 units we need to use the equation (*)

$P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2}$

Note: We need to use cm Hg for units coherence

Now passing from dyn/ $cm^2$ to kN/ $m^2$ (or kPa) we need to consider that 1 dyn is $10^{-8}$ kN and 1 $cm^2$ is $10^{-4} m^2$ .

$1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2$

Now passing kN/ $m^2$ to psia. We need to consider that 1 psia is 6.89476.

$117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia$

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