Answers to Problems at the End of Chapter 2 2-17
2.34a Shown below is the probability mass function for the number of good castings (x)
produced when Q castings are produced based on a probability of 0.85 that an individual
2.34b Shown below is the probability mass function for the number of good castings (x)
produced when Q castings are produced based on a probability of 0.98 that an individual
$64,315.
2.35 In Example 2.7, a = 2 min, b = 1 min, t = 6 min, n’ = 2.67, Co = $15/hr, and Cm = $50/hr.
Without other constraints the optimum number of machines to assign to an operator was
Answers to Problems at the End of Chapter 2 2-20
To calculate the cost per unit produced for each scenario, it is useful to evaluate each
alternative using a length of time equal to the least common multiple of the cycle times
2.36 For the optimum assignment in Example 2. 5 to remain unchanged, M < 1. Recall, in
Example 2.5, n’ = 2.67, Co = $15/hr, Cm = $50/hr, and , = Co/Cm = 15/Cm. Therefore, for
2.37 During 7 hours of work for the operator between 8:00 a.m. and 4:00 p.m., 136 units were
produced by the 3 machines. In steady state conditions, the repeating cycle is 9 minutes.
2.38 The problem statement was overly simplified, assuming sufficient demand exists to keep
5 machines busy and sales prices of the products are such that the calculation of cost per
unit produced can be performed by summing the units produced for both products. Also,
it is assumed that a machine is dedicated to producing either product 1 or product 2 and
cannot be assigned to produce a combination of the two products due to changeover
an operator and 3 machines producing product 2 to an operator; assign 2 machines
producing product 1 to an operator, 2 machines producing product 2 to an operator, and 1
machine producing product 2 to an operator; assign 2 machines producing product 1 to an
operator, 1 machine producing product 1 to an operator, and 2 machines producing
Answers to Problems at the End of Chapter 2 2-30
2.26a a = 6 + 4 = 10; b = 6; and t = 30. n’ = (10 + 30)/(10 + 6) = 2.5. No more than 2 mixers
2.26b Co = $12/hr; Cm = $25/hr; and = Co/Cm = 0.48.
2.40 The multiple activity chart is provided on the following page. The length of the repeating
Multiple Activity Chart
2.41a a = 5 min; b = 1 min; t = 20 min; Co = $12/hr, and Cm = $30/hr.
2.41b = Co/Cm = 0.4. Therefore,
2.41d For n* = 4, either Φ < 1 when n = 4 or Φ > 1 for n = 3.
Φ < 1 case: or
2.41e Consider the alternatives: {5, 5, 5}, {4, 4, 4, 3}, and {4, 5, 6}.
{5, 5, 5} case: Tc = 30 min.
TC{5, 5, 5} = [3($12) + 15($30)](30/60)(1/15) = $16.20/unit.
2.42 a = 4 min; b = 5 min + 3 min = 8 min; and t = 40 min. n’ = (4 + 40)/(4 + 8) or n’ = 3.67
2.43 The problem statement does not mention retrievals by the S/R; hence, no travel between
Answers to Problems at the End of Chapter 2 2-32
Multiple Activity Chart
2.44 a = 0.25 min; b = 0; and t = 1.0. Therefore, n’ = 1.25/0.25 = 5.0. An operator can tend 5
SECTION 2.5
2.45 a. Board for part shortages
2.46
2.47
2.48 – 2-50 The answers to these questions depend on choices made and course
Relevant process/machine
related issues
Training contents
definition
Relevant product related
issues
Other
concerns
Current skill
assessment
Participant
reaction
Operator
certification
program
Potential trainee
identification
Development of
knowledge material
Identification of key
process/machines
Process/machine
skills to be included
Product-specific
skills to be included
c
a
e
d
b
f
g