Operations Management – 357 357 – Section Section 03 Team # 9 Caitlynn Dutton – Tim Tim Jankowski – Ryan Ryan Beck Assignment #2 February 25th, 2014
1. Question #1 (C.T.E) pg. 208 Computer repair service has a design capacity of 80 repairs per day. Its effective capacity, however, is 64 repairs per day, and its actual output is 62 repairs per day. Manager would like it increase repairs; which of the following factors would you recommend the manager investigate; quality problems, absenteeism, or scheduling and balancing?
Efficiency = Actual Output/Effective Capacity
Utilization = Actual Output/Design Capacity
62/64 = 0.968
62/80 = 0.775
Efficiency = 97%
Utilization = 77.5%
Efficiency is much higher than utilization due to the fact that the firm’s capacity could
use improvement to increase overall potential output. Out O ut of all the following factors recommended for the manager to investigate, the most appropriate answer would be “scheduling and balancing” since that is the category that correlates with optimizing capacity.
(Stevenson, 2012, pg. 187)
2. Question #4 (Problems) pg. 209 A.
Determine each alternative’s break -even
Break-Even Point
point in units
Q(BEP) = FC/Revenue-VC (Stevenson, 2012, pg. 202)
Firm
FC
Revenue
VC
Q(BEP)
A
40,000
$15/unit
$10/unit
8,000
B
30,000
$15/unit
$11/unit
7,500
Firm A :
Q(BEP) = 40,000/15-5 = 8,000 units
Firm B:
Q(BEP) = 30,000/15-11 = 7,500 units
B. At what volume of output would the two alternatives yield the same profit?
Profit (P) = Q (Revenue-VC) - FC
(Stevenson
Q(15 – 10) – 40,000 = Q(15 – 11) – 30,000 Q15 – Q10 – 40,000 = Q15 - Q11 – 30,000 Q5 – 40,000 = Q4 – 30,000 Q5 – Q4 = - 30,000 + 40,000 Q = 10,000 When volume = 10,000 units the two alternatives yield the same profit
C. If expected annual demand is 12,000 units, which alternative would yield higher profit?
Firm A:
P = 12,000(15 – 10) – 40,000 = 20,000
Firm B:
P = 12,000(15 – 11) – 30,000 = 18,000
If expected annual demand is 12,000 units then Firm A would yield higher profit
3. Question #3 (CTE) pg. 277 What are the risks of automating a production process? What are the risks for a service process?
Two risks that occur from automating a production process are the possibility of a shortage of inventory, or the possibility of over-capacity. The reason why these risk exist is due to the fact that if demand increases, the cost of altering the process to meet demand might exceed the benefits, or in the opposite case a decrease in demand may result to excess inventory. Excess inventory means minimal revenue to pay off the costs of the process application, which is also a risk. When dealing with a service process, the amount of risk is determined by the degree of variability of the service being provided. Services that contain a high degree of variability will be less likely to succeed in the automating process because offering high variety services comes with the opportunity cost of possibly not matching demand with expected consumers. Ultimately, the firms’ capability of correctly predicting demand determines whether automating the production or service process will be feasible. (Stevenson, 2012, pg. 243)
4. Question #1 pg 277 An assembly line with 17 tasks is to be balanced. The longest task is 2.4 minutes, and the total time for all tasks is 18 minutes. The line will operate for 450 minutes per day.
A.
What are the minimum and maximum cycle times?
Minimal Cycle Time = “the longest task time”
=2.4 minutes
Maximum Cycle Time = “the sum of the task times” =18 minutes Minimal Cycle Time = 2.4 minutes Maximal Cycle Time = 18 minutes
B.
Range of Output Possible:
450 min per day/ 2.4 min = 187.5 450 min per day/ 18 min = 25
C. What is the minimum number of workstations needed if max output rate is to be sought.
N(min) = Max(output) × Max(cycle-time) / Min. Per Day N(min) = 187.5(18) / 450 = 7.5, round to 8 OR
N(min) = Max(cycle-time) / Min(cycle-time) 18/2.4 = 7.5 , rounds to 8 is the minimum number of workstations needed
D.
What cycle time will provide an output rate of 125 units per day?
Output = Operation Time per day /Cycle Time 450/125 = 3.6 minutes per cycle
E.
Potential output using 9 min and 15 minutes as cycle time.
9 minutes ------ OT/CT = 450/ 9 = 50 units will be produced using 9 min. cycle time 15 minutes -----OT/CT = 450/15 = 30 units will be produced using 15 min. cycle time
Work Cited
Stevenson, W. J. (2012). Operations management. New York: McGraw-Hill Irwin.
