DevinHarrisMidterm Assignment-V2_SOP-7200-nov16
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SOP-7200: Strategic Operations Management
Comprehensive Midterm Assignment – Version 2
Your Name: Devin Harris
Course Session: SOP-7200-MB900
Answer the following four questions. Be sure to answer all parts of the questions thoroughly. (25 points each)
1.
Alyssa’s Custom Cakes currently sells 5 birthday, 2 wedding, and 3 specialty cakes each month for $50, $150, and $100 each, respectively. It takes 90 minutes to produce a birthday cake, 240 minutes to produce a wedding cake, and 60 minutes to produce a specialty cake. Alyssa’s current multifactor productivity ratio is 1.25.
a)
Use the multifactor productivity ratio provided to calculate the average cost of the cakes produced. (5x50) + (2x150) + (3x100) = $850
5+2+3 / 1.25 = 8
($850/10) x 8 = $680
$680 / 10 = $68
average cost of cakes produced
b)
Calculate Alyssa’s labor productivity ratio in dollars per hour for each type of cake. (1 / 1.5) x $50 = $33.33
per hour for each type of birthday cake labor productivity ratio
(1 / 4) x $150 = $37.50
per hour for each type of wedding cake labor productivity ratio
(1 / 1) x $100 = $100.00
per hour for each type of specialty cake labor productivity ratio
c)
Based solely on the labor productivity ratio which cake should Alyssa try to sell the most. Specialty cakes, being that they produce the most amount of money per hour labor wise.
d)
Based on your answer in part (a), is there a type of cake Alyssa should stop selling?
$50 - $68 = -$18
$150 - $68 = $82
$100 - $68 = $32
Alyssa should stop selling birthday cakes
, due to the fact that she is taking a loss with every sale of this type of cake at -$18, whereas the other two types of cakes bring in great amounts of profits at $82 and $32.
SOP-7200: Strategic Operations Management
2.
Consider the following data for a project never before attempted by your company: a)
Draw the network diagram for this project.
b) Identify the critical path and estimate the project’s duration.
A-C-E = 5 + 2 + 4 = 13 weeks
B-D-E = 3 + 5 + 4 = 12 weeks
B-D-F = 3 + 5 + 7 = 15 weeks
Critical Path & the estimate of the project’s duration
c) Calculate the slack for each activity.
Activity
Expected Time (weeks)
Immediate Predecessor(s)
A
5
-----
B
3
-----
C
z2
A
D
5
B
E
4
C, D
F
7
D
B-D-F do not
have slack (
zero
)
Slack of activity E = Duration of B-D-F – Duration of B-D-E = 15 – 12 = 3 weeks
Slack of activity A, C = Duration of A-C-E - Duration of B-D-E = 15 – 13 = 2 weeks
Activity
Slack (weeks)
A
2
B
0
C
2
D
0
E
3
C
A
D
E
B
F
SOP-7200: Strategic Operations Management
F
0
3.
Suppose you are in charge of a large mailing to the alumni of your college, inviting them to contribute to a scholarship fund. The letters and envelopes have been individually addressed (mailing labels were not used). The letters are to be processed (matched with correct envelope, time estimated to be 0.2 minutes each), folded (0.12 minutes each), and stuffed into the correct envelope (0.10 minutes each). The envelopes are to be sealed (0.05 minutes each), and a large commemorative stamp is to be placed in the upper right hand corner of each envelope (0.10 minutes each). a)
Make a process chart for this activity, assuming that it is a one person operation.
b)
Estimate how long it will take to stuff, seal, and stamp 2,000 envelopes. Assume that the person doing this work is paid $8 per hour. How much will it cost
to process 2,000 letters?
Total processing time = 0.2 + 0.12 + 0.1 + 0.05 + 0.1 = 0.57 minutes
Time taken to process 2000 letters = 2000*0.57 = 1140 minutes or 19 hours
Cost to process 2000 letters = 19 x 8 = $
152
c)
Consider each of the following process changes. Which changes would reduce the time and cost of the current process?
●
Each letter has the same greeting “Dear Alumnus or Alumna,” instead of the person’s name. ●
Mailing labels are used and have to be put on the envelopes (0.10 minutes each).
●
Pre-stamped envelopes are used.
●
Envelopes are stamped by postage meter which can stamp 200 letter per minute.
●
Window envelopes are used.
●
A pre-addressed envelope is included with each letter for contributions (adds 0.05 minutes to stuffing step) REMOVE this last step in the process, as it is not needed
d)
Would any of these changes be likely to reduce the effectiveness of the mailing? If so, which ones? Why?
As a result of the first process change, the mail would not be as effective. Their goal is to gain funding from alumnae and by addressing each letter as "Dear Alumnus or Alumna," the letter becomes impersonal. In such a case, the donor might reconsider donating, which would not be worthwhile.
e)
Would the changes that increase time and cost be likely to increase the effectiveness of the mailing? Why or why not?
Match with correct envelope (0.20)
Folded (0.12)
Stuff into correct envelope
Sealed (0.05)
Place Commemorative Stamp (0.10)
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SOP-7200: Strategic Operations Management
The first and third selections provide the most effective solutions, combining a personal touch and a gentle nudge to contribute. While they increase cost, their returns are notably higher in comparison. Furthermore, by utilizing individually addressed letters and envelopes, a personal touch is added and the impersonal nature of mass mailing is eliminated. The third alternative, the addition of a pre-stamped envelope for contributions, makes it more challenging to ignore the impulse to
contribute. This serves as a reminder as well as a reminder that the task requires participation.
4.
Jim’s Outfitters, Inc., makes custom fancy shirts for cowboys. The shirts could be flawed in cautious ways, including flaws in the weave or color of the fabric, loose buttons or decorations, wrong dimensions, and uneven stitches. Jim randomly examined 10 shirts, with the following results:
Shirt
Defects
1
8
2
0
3
7
4
12
5
5
6
10
7
2
8
4
9
6
10
6
a)
Assuming that 10 observations are adequate for these purposes, determine the three sigma control limits for defects per shirt.
Defects per shirt = (8 + 0 + 7 + 12 + 5 + 10 + 2 + 4 + 6 + 6) / 10 c = 60 / 10 = 6 average defects per shirt
SD = 3 x 1.18 = 3.56
UCL = 17.31
LCL = 6 – (3 x sqrt(6))
6 - (3 x 1.18,0) = 0 defects per shirt
b)
Suppose that the next shirt has 13 flaws. What can you say about the process now?
Mean = (8 + 0 + 7 + 12 + 5 + 10 + 2 + 4 + 6 + 6 + 13) / 10 c = 73 / 11 = 6.64 average defects per shirt
SD = 3.98
UCL = 18.58
LCL = 0.00