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zzz [600]
2 years ago
10

A construction company distributes its products by trucks loaded at its loading station. A backacter in conjunction with trucks

are used for this purpose. If it was found out that on an average of 12 trucks per hour arrived and the average loading time was 3 minutes for each truck. A truck must queue until it is loaded. The backacter’s daily all-in rate is GH¢ 1000 and that of the truck is GH¢ 400.
a) Compute the operating characteristics: L, Lq, W, Wq, and P.

b) The company is considering replacing the backacter with a bigger one which will have an average service rate of 1.5 minutes to serve trucks waiting to have their schedules improved. As a manager, would you recommend the new backacter if the daily all-in rate is GH¢ 1300.

c) The site management is considering whether to deploy an extra backwater to assist the existing one. The daily all-in-rate and efficiency of the new backwater is assumed to be the same as that of the existing backwater. Should the additional backwater be deployed?
Engineering
1 answer:
blagie [28]2 years ago
6 0

Answer:

a) L = 1.5

L_q = 0.9

W =  \dfrac{1 }{8 } \, hour

W_q =  \dfrac{3}{40 } \, hour

P = \dfrac{3}{5 }

b) The new backacter should be recommended

c) The additional backacter should not be deployed

Explanation:

a) The required parameters are;

L = The number of customers available

L = \dfrac{\lambda }{\mu -\lambda }

μ = Service rate

L_q = The number of customers waiting in line

L_q = p\times L

W = The time spent waiting including being served

W = \dfrac{1 }{\mu -\lambda }

W_q = The time spent waiting in line

W_q = P \times W

P = The system utilization

P = \dfrac{\lambda }{\mu  }

From the information given;

λ = 12 trucks/hour

μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour

Plugging in the above values, we have;

L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5

P = \dfrac{12 }{20 } = \dfrac{3}{5 }

L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9

W = \dfrac{1 }{20 -12 } =  \dfrac{1 }{8 } \ hour

W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour

(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;

μ = 60/1.5 trucks/hour = 40 trucks/hour

P = \dfrac{12 }{40  } = \dfrac{3}{10}

W = \dfrac{1 }{40 -12 } =  \dfrac{1 }{38 } \, hour

W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour

λ = 12 trucks/hour

Total cost = mC_s + \lambda WC_w

m = 1

C_s = GH¢ = 1300

C_w = 400

Total cost with the old backacter is given as follows;

1 \times 1000 + 12 \times \dfrac{1}{8}  \times 400 = \$ 1,600.00

Total cost with the new backacter is given as follows;

1 \times 1300 + 12 \times \dfrac{1}{38}  \times 400 = \$ 1,426.32

The new backacter will reduce the total costs, therefore, the new backacter is recommended.

c)

Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour

\therefore W = \dfrac{1 }{40 -12 } =  \dfrac{1 }{38 } \, hour

Total cost with the one backacter is given as follows;

1 \times 1000 + 12 \times \dfrac{1}{8}  \times 400 = \$ 1,600.00

Total cost with two backacters is given as follows;

2 \times 1000 + 12 \times \dfrac{1}{38}  \times 400 = \$ 2,126.32

The additional backacter will increase the total costs, therefore, it should not be deployed.

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Explanation:

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2 years ago
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8 0
1 year ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30
shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0
2 years ago
Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons
mixas84 [53]

Answer:

work done = 48.88 × 10^{9} J

Explanation:

given data

mass = 100 kN

velocity =  310 m/s

time = 30 min = 1800 s

drag force = 12 kN

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to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time  - force ×velocity ×time  - mass ×descends )  10³ J

work done = ( 100 × 310 × 1800  - 12×310 ×1800  - 100 × 2200 )  10³ J

work done = 48.88 × 10^{9} J

6 0
2 years ago
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