Reliability and Productivity Analysis
Professor’s Name
Student’s Name
Course Title
Date
Questions 1 Problem 9 page 848
Product A 12 min milling 5 min inspection 10 min drilling.
Product B 10 min milling 4 min inspection 8 min drilling
Product C 8 min milling 4 min inspection 16 min drilling
Available number of hours, 20 hours milling = 1200 minutes.
15 hours inspection = 900 minutes.
24 hours drilling = 1440 minutes.
A contributes $ 2.40/ unit to profit, produce B contributes $2.50/ unit to profit and product C contributes $3.00/ unit to profit. Let Z be the maximum profit.
The constraints equations are as follows;
2.4A + 2.5B + 3C = Z……………..1
12A + 10B + 8C = 1200…………..2
5A + 4B + 4C = 900………………….3
10A + 8 + 16C = 1440………………4
Using equation 2, A = (1200 - 10B – 8C)/12.
Then substituting the value of A to equation 3 we have; -B + 4C = 2400……………….5.
Also substituting the value of A in equation 4 we have; -B + 28C = 1320……………..6.
Solving equations 5 and 6 simultaneously, we have the value of C = -45 units and B = -2580 units. And from A = (1200 - 10B – 8C)/12, the value of A is calculated to be A = 2280 units.
Therefore substituting the values to equation 1 we have, Z = (2.4*2280) + (2.5 * -2580) + (3*-45) = -$1113. Therefore at profit maximization;
Quantity of product A = 2280 units
Quantity of product B = -2580 units
Quantity of product C = -45 units
Maximum profit = $(1113)
Question 2 problem 4(a) page 65
Labor productivity = quantity produced/ labor hours.
= 80/ (5*1) = 16 carts per worker per hour.
Question 3 problem 4(a) page 65
Labor productivity = quantity produced/ labor hours.
= 84/ (4*1) = 21 carts per worker per hour.
Question 4 problem 4(b) page 65
Multifactor = quantity produced/ (labor cost + machine cost)
= 80/ (10*5 + 45) = 0.8421 carts per dollar.
Question 5 problem 4(a) page 65
Multifactor = quantity produced/ (labor cost + machine cost)
= 84/ (4*10 + 55) = 0.8842 carts per dollar.
Question 6 problem 7(a) page 66
Hourly wage = $25
Overhead = $1.0 times labor cost
Material cost = $ 5 per customer
A: labor productivity = customers processed per employees.
= 36/4 = 9.00 customers per employee.
B: labor productivity = customers processed per employees.
= 40/5 = 8.00 customers per employee
C: labor productivity = customers processed per employees.
= 60/8 = 7.50 customers per employee
D: labor productivity = customers processed per employees.
= 20/3 = 6.67 customers per employee.
Question 7 problem 7(a) page 66
Multifactor productivity = customer processed/ (labor cost + material cost + overhead cost).
A: Multifactor productivity = customer processed/ (labor cost + material cost + overhead cost).
= 36/ (4*25) + (36*5) + (1*4*25) = 36/380 = 0.0947 customers per dollar input.
B: Multifactor productivity = customer processed/ (labor cost + material cost + overhead cost).
= 40/ (5*25) + (40*5) + (1*5*25) = 40/ 450 = 0.0889 customers per dollar unit.
C: Multifactor productivity = customer processed/ (labor cost + material cost + overhead cost).
= 60/ (8*25) + (60*5) + (1*8*25) = 60/ 700 = 0.0857 customers per dollar unit.
D: Multifactor productivity = customer processed/ (labor cost + material cost + overhead cost).
= 20/ (3*25) + (20*5) + (1*3*25) = 20/ 250 = 0.0800 customers per dollar unit.
Question 8 problem 6a page 181
Components reliabilities are; 0.98, 0.95, 0.94, and 0.90
The probability that the system doesn’t work = p
P = (1-0.98)* (1-0.95) * (1-0.94) *(1-0.90) = 0.000006.
Therefore the system reliability = 1-p = 1-0.000006 = 0.999994
Question 9 problem 6b page 181
The backup component should be 0.98 reliable.
Probability the system doesn’t work = p
P = 0.000006 * (1-0.98) = 0.00000012
Therefore the highest system reliability will be 1-p = 1- 0.00000012 = 0.99999988
Question 10 problem 7 page 181
Reliabilities for the components are; A =0.99 B= 0.96 C = 0.93
Original production reliability = 1- (0.01*0.04*0.07) = 0.999972.
Plan A reliability will be = 0.999972* 0.999972 = 0.999944.
Question 11 problem 7 page 181
Plan B = 1- (0.000028) ^2 = 1 – 0.00000000078 = 0.99999999922 ≈ 1.
Question 12 problem 14 page 182
MTBF = 20000 hours. T = 24,000 hours
Therefore, T/MTBF = 24000/20000 = 1.2
From table 4S.1, e^ - T/MTBF = 0.3012
For T = 4000 hours, T/MTBF = 4000/20000 = 0.2.
From table 4S.1, e^ - T/MTBF = 0.8187.
Therefore the probability of the bulb lasting between 4,000 hours to 24,000hours will be;
= (0.8187 – 0.3012) = 0.5175.
Question 13 problem 17 page 182
Mean = 6 years
Standard deviation = 1.5 years
Therefore after 7.5 years; z = 7.5-6/1.5 = +1.0
Thus from the appendix table B, P (T≥ 7.5) = 1-0.8413 = 0.1587
Question 14 problem 13 page 182
From the appendix table B, for a failure rate of approximately 10%, from tables exponential tables, z = -1.28
Therefore, -1.28 = T – 30/1
Hence, T = 30 -1.28 = 28.72 months.