Total productivity = output/input (50,000 x 3.5) / ((620 x 7.5) + 30,000 + 15,350) = 3.50
Labor Productivity - units/hour |
Model | Output in Units | Input in Labor Hours | Productivity
(Output/Input) | Deluxe Car | 4,000 | 20,000 | 0.20 | Limited Car | 6,000 | 30,000 | 0.20 | | Labor Productivity - dollars |
Model | Output in Dollars | Input in Dollars | Productivity
(Output/Input) | Deluxe Car | 4,000($8,000)=$32,000,000 | 20,000($12.00)=$240,000 | 133.33 | Limited Car | 6,000($9,500)=$57,000,000 | 30,000($14.00)=$420,000 | 135.71 | |
(b): The labor productivity measure is a conventional measure of productivity. However, as a partial measure, it may not provide all of the necessary information that is needed. For example, increases in productivity could result from decreases in quality, and/or increases in material cost. |
Multifactor productivity = Output / Labor + Capital
Percentage change in productivity: Month Output in Dollars Input in Hours Productivity (Output/Input) Percentage Change April $ 45,000 1,560 28.85 May $ 56,000 1,820 30.77 (30.77 − 28.85)/28.85 = 6.67%
Forcasting:
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Weighted moving avergae FJuly = .60(15) + .30(16) + .10(12) = 15.0 |
Three month average FJuly = (15 + 16 + 12) / 3 = 14.3 |
Single exponential smoothing FJuly = FJune + α(AJune – FJune) = 13 + .2(15-13) = 13.4 |
Simple linear regression analysis | x | y | xy | x2 | | 1 | 12 | 12 | 1 | | 2 | 11 | 22 | 4 | | 3 | 15 | 45 | 9 | | 4 | 12 | 48 | 16 | | 5 | 16 | 80 | 25 | | 6 | 15 | 90 | 36 | | | | | | Total | 21 | 81 | 297 | 91 | | | | | | | = 3.5 | = 13.5 | | a = =10.8 | Y = a + bx = 10.8 + .77x |
Regression equation FJuly, where July is the 7th month. | Y = a + bx = 10.8 + .77(7) = 16.2 |
Break even, profit per unit
3(a):
Profit per unit = (P–VC)*V – FC)/V | $.25 = (($5.50 – $4.50)*V - $900)/V | .25V = V – 900 | .75V = 900 | V = 1,200 units.
(d) At least one other student waiting in line is the same as at least two in the system. This probability is
1-(P0+P1). |
Probability of at least one in line is 1 − (.3333 + .2222) = .444 | |
What is the total cost per day for this option? Use model 1. | λ = 4/hour μ = 10/hour (a) Lq | = | ------------------------------------------------- λ2 | = | -------------------------------------------------
42 | = | .267 students | | | μ(μ − λ) | | 10(10 − 4) | | |
Waiting cost | = | number in line times goodwill loss per hour times number of hours per day. | | = | .267(10)8 | | = | $21.33 per day |
Total cost/day | = | waiting cost + additional service cost + Judy's Pay | | = | $21.33 + $99.50 + $75 | | = | $195.83 per day | Use model 1. | λ = 10/hour μ = 15/hour |
(a) Percentage of time ddesk is idle 1−ρ | = | 1− | -------------------------------------------------
10 | = | .3333 or 33.33% | | | | 15 | | |
(b) Probabitlity that both desks are busy P(both clerks busy) = P(one clerk busy) × P(one clerk busy) | |
(c) P(both clerks idle) = P(one clerk idle) × P(one clerk idle) | = (.3333)(.3333) = .1111 or 11.11% |
(d) Probability that both desks are idle Lq | = | ------------------------------------------------- λ2 | = | -------------------------------------------------
102 | = | 1.33 customers | | | μ(μ − λ) | | 15(15 − 10) | | |
(e) How much time does a customer spend at the service desk (waiting plus service time)? Ls | = | ------------------------------------------------- λ | = | -------------------------------------------------
10 | = | 2 | | | (μ − λ) | | (15 − 10) | | | WS | = | -------------------------------------------------
Ls | = | -------------------------------------------------
2 | = | .2 or 12 minutes | | | | | | | | |
Explanation: Use model 2. | λ = 60/50 per minute μ = 60/45 per minute | Average line length in time Lq | = | ------------------------------------------------- λ2 | = | -------------------------------------------------
(60/50)2 | = | 4.05 cars | | | 2μ(μ − λ) | | 2(60/45)(60/45− 60/50) | | | Average number of cars in the system. Ls = Lq + λ / μ = 4.05 + (60/50)/(60/45) = 4.95 cars |
Wq | = | -------------------------------------------------
Lq | = | -------------------------------------------------
4.05 | = | 3.375 | | | λ | | (60/50) | | | What is the expected average time in system Ws | = | -------------------------------------------------
Ls | = | -------------------------------------------------
4.95 | = | 4.125 minutes | | | | | | | | | Λ 60/50
Problem 5-12 A cafeteria serving line has a coffee urn from which customers serve themselves. Arrivals at the urn follow a Poisson distribution at the rate of three per minute. In serving themselves, customers take about 15 seconds, exponentially distributed. |
Required: |
(a) | How many customers would you expect to see on the average at the coffee urn? |
Average customers | |
(b) | How long would you expect it to take to get a cup of coffee? |
Expected Time | minute |
(c) | What percentage of time is the urn being used? (Omit the "%" sign in your response.) |
Percentage of time | % |
(d) | What is the probability that three or more people are in the cafeteria? (Round your answer to 3 decimal places.) |
Probability | |
(e) | If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 15 seconds, how does this change your answers to a and b? (Round your answers to 3 decimal places.) |
| | | Average number of people reduced by | | | Time reduced by | | minutes | |
Explanation:
(a)
Use model 1 | λ = 3 per minute μ = 4 per minute |
Ls | = | ------------------------------------------------- λ | = | -------------------------------------------------
3 | = 3 customers | | | μ − λ | | 4 − 3 | |
(b) Use model 1 | λ = 3 per minute μ = 4 per minute | Ws | = | -------------------------------------------------
Ls | = | -------------------------------------------------
3 | = 1 minute | | | λ | | 3 | |
(c) Use model 1 | λ = 3 per minute μ = 4 per minute |
ρ | = | ------------------------------------------------- λ | = | -------------------------------------------------
3 | = .75 or 75% | | | μ | | 4 | |
(d) Use model 1 | λ = 3 per minute μ = 4 per minute | Probability of 3 or more is equal to 1 – probability of 0, 1, 2 | ,
,
| Total of P0 + P1 + P2 = (.2500 + .1875 + .1406) = .5781 | Therefore, the probability of three or more is 1 − .5781 = .4219 |
(e) If a automatic vendor is installed, use model 2. | (a. revisited) | Lq | = | ------------------------------------------------- λ2 | = | -------------------------------------------------
32 | = 1.125 customers | | | 2μ(μ − λ) | | 2(4)(4 − 3) | | Ls = Lq + λ / μ = 1.125 + 3/4 =1.875 | | (b. revisited) | Wq | = | -------------------------------------------------
Lq | = | -------------------------------------------------
1.125 | = .375 minutes | | | λ | | 3 | | Ws | = | -------------------------------------------------
Ls | = | -------------------------------------------------
1.875 | = .625 minutes | | | λ | | 3 | | By converting to constant service time, the number in line is reduce from 3 to 1.875 people (a reduction of 1.125), and time in system is reduced from 1 minute to .625 minutes (a reduction of .375 minutes or 22.5 seconds). |
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Problem 5-18 Cathy Livingston, bartender at the Los Gactos Racquet Club, can serve drinks at the rate of one every 50 seconds. During a hot evening recently, the bar was particularly busy and every 55 seconds someone was at the bar asking for a drink. |
Required: |
(a) | Assuming that everyone in the bar drank at the same rate and that Cathy served people on a first-come, first-served basis, how long would you expect to have to wait for a drink? (Round your answer to 3 decimal places.) |
Expected time to wait for a drink | minutes | (b) | How many people would you expect to be waiting for drinks? |
Number of people waiting | | (c) | What is the probability that three or more people are waiting for drinks? (Round your answer to 4 decimal places.) |
Probability | | (d) | What is the utilization of the bartender (how busy is she)? (Round your answer to 2 decimal places. Omit the "%" sign in your response.) |
Utilization of the bartender | % | (e) | If the bartender is replaced with an automatic drink dispensing machine (with a constant service time), how long would you expect to have to wait for a drink? (Round your answer to the nearest whole number.) |
Expected time to wait for a drink | minutes |
Explanation: Use model 1. |
λ = 60/55 per minute μ = 60/50 per minute |
(a) Ls | = | ------------------------------------------------- λ | = | -------------------------------------------------
60/55 | = | 10.00 people | | | μ − λ | | 60/50−60/55 | | | Ws | = | -------------------------------------------------
Ls | = | -------------------------------------------------
10.00 | = | 9.167 minutes | | | λ | | 60/55 | | |
(b) Ls | = | ------------------------------------------------- λ | = | -------------------------------------------------
60/55 | = | 10.00 people | | | μ − λ | | 60/50−60/55 | | |
(c) Probability of more than 3 people is equal to 1 – probability of 0, 1, 2 | | | | Total of P0 + P1 + P2 = (.0909 + .0826 + .0751) = .2487 | Therefore, the probability of three or more is 1 − .2487 = .7513 |
(d) ρ | = | ------------------------------------------------- λ | = | -------------------------------------------------
60/55 | = | .9091 or 90.91% | | | μ | | 60/50 | | |
(e) Use model 2. | Ls | = | ------------------------------------------------- λ2 | = | -------------------------------------------------
(60/55)2 | = | 4.545 | | | 2μ(μ − λ) | | 2(60/50)(60/50−60/55) | | |
Ls = Lq + λ / μ = 4.545 + (60/55)/(60/50) = 5.455 minutes | Ws = Ls / λ = 5.455/(60/55) = 5.0 minutes | 