HW1 452 1-7
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School
University of North Carolina, Charlotte *
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Course
MBAD-614
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
xlsx
Pages
14
Uploaded by SuperExplorationPorcupine30
1
a.
If units in inventory with cost of $1,000 per unit are sold for $2,000 per unit, how fast does the
company turn its inventory?
Inventory Turn Over = COGS/Average Inventory
Average Inventory= (beginning Inventory + Ending Inventory)/2
Beginning inventory wasn't provided, using $200000 as beginning
Average Inventory:
$ 200,000.00 COGS:
$ 400,000.00 Inventory turnover:
2
b.
The company uses a 25 percent per year cost of inventory. That is, for the hypothetical case that
one unit of $1,000 would sit exactly one year in inventory, the company charges its operations
division a $250 inventory cost. What---in absolute terms ($) ---is the per unit inventory cost for a
product that costs $1,000?
Per Unit Inventory Cost = (inventory Cost/ Number of units)
Per unit InventoryCost:
0.25
2
The following figures are taken from the 2003 financial statements of McDonald’s and Wendy’s. Figures a
McDonald's
Wendy's
Inventory
129.4
54.4
Revenue
17140.5
3148.9
COGS
11943.7
1634.6
Gross Profit
5196.8
1514.4
Average Inventory
64.7
27.2
Holding %
30%
30%
a.
In 2003, what were McDonald’s inventory turns? What were Wendy’s inventory turns?
Inventory Turnover = COGS/Average Inventory
McDonald's
184.601236476043
Wendy's
60.0955882352941
b.
A manufacturing company producing medical devices reported $
year. At the end of the same year, the company had $20,000,000 w
devices.
Suppose it costs both McDonald’s and Wendy’s $3 (COGS) per their value meal offerings, each sold at t
inventory for both companies is 30 percent a year. Approximately how much does McDonald’s save in in
of Wendy’s? You may assume that inventory turns are independent o
Inventory Cost per Value meal = (Inventory/Inventory Turnover) X (Holding Rate)
McDonald's:
0.210291115818381
Wendy's:
0.271567355928056
Wendy's - Mcdonald's
0.061276240109675
Mcdonald's save $0.06 per value of meal
3
a.
If the batch size is 10 donuts per batch, what is the capacity per day for each type of donut?
Capacity of the Machine =
1 dounut/2 minute
20+(10*2)
=
8X60
40 n=
480
n=
12 X10
b.
If the batch size is 20 donuts per batch, what is the capacity per day for each type of donut?
20+(20*2)
=
8X60
60 =
480
n=
8 X20
C.
What is the minimal batch size in order to satisfy demand for both types?
20+(2X)+20+(2y)
2X+2Y+40
<
6X80
2(X+Y)
<
440
X+Y
<
220
Demand for each batch is 80 donuts per day, so the Batch is passed since 220>160
d.
If the setup time is 8 minutes for either type, and batch size is 20 donuts per batch, what is the capacity p
n=
8+(20*2)
=
8*60
48 =
480
n=
10 X20
Each donuts batch is 20, so times 10 is 200
4
A donut store is open 8 hours a day and sells two types of donuts – blueberry cake donut and chocolate
donuts per day for EACH type. The store has only one machine which makes both types, and once it star
either type. The production alternates between the two types, i.e., a batch of one type is followed by a b
per batch for either type.
Furniture Face Lift refinishes old wood furniture. Their process for refinishing chairs has 8 workers and
station, then goes to Priming, then to Painting and finally to Inspection. Where there are multiple wo
independently on his/her own chair. Assume inventory buffers are allowed between each station and e
beginning of the day.
Station
Staffing
Stripping
3
2.5
Priming
2
1.5
Painting
2
1.75
Inspection
1
0.8
Total
8
6.55
a. What is the maximum number of chairs per hour that can be produced?
Maximum per chair
(station*worker)/Processing hours
Stripping:
1.2
Priming:
1.33333333333333
Painting:
1.14285714285714
Inspection:
1.25
Because Painting is the one has the lowest Capacity, so the maximum will be 1.1429 chair per hour
b.
if there is no bottle nect, so everyone will work 6.55 hours to get one cheir done.
Worker#
8
Total process hours:
6.55
Maximum Capacity:
1.22137404580153
The total will be 1.22 chair per hours
beginning of the day.
Processing Time (Hours per Chair Per Worker)
Suppose now that each worker is trained to do all tasks and each worker works on a chair from start to fi
Painting and Inspection. What is the maximum capacity of the process in c
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Sales:
$ 600,000.00 beginning Inventory:
$ 200,000.00 Ending Inventory:
$ 200,000.00 COGS:
$ 400,000.00 Unite/Year:
$250 Per unit Cost:
$1,000 are in million dollars.
$60,000,000 in sales over the last
worth of inventory of ready-to-ship
the same price of $4. Assume that the cost of nventory cost per value meal compared to that of the price
=
120 donuts
=
160 Donuts
per day for each type of donut?
=
200 Dounts
e frosted donut. On average the demand is 80 rts, it produces one donut every 2 minutes for batch of the other. The setup time is 20 minutes d 4 stations. Each chair starts at the Stripping orkers within a station, each worker works each station has inventory to work on at the
Capacity
Station:
4
1.2
4
1.333333
4
1.142857
4
1.25
4
finish, i.e., each worker does Stripping, Priming, chairs per hour?
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#5
A. Input
Output
Yield %
Dep A
2200
1760
80%
Dep B
1600
1440
90%
Dep C
1800
1800
100%
Dep D
1000
800
80%
Dep E
800
800
100%
We need 3 parts from department C and 1 part from department D to create one unit in departme
With C we can make 2400 units per day (800*3) and with D we can only make 800 units/day (800*
B. Where is the bottleneck?
What is the system capacity?
The bottleneck is in department D since it can only produce 800 units/day at 80% yield. Department D has the lowest capacity compared to other departments.
Output
max capacity
Utilization %
C. Department A
1760
2200
80%
Department B
1440
1600
#NAME?
Department C
1800
1800
#NAME?
Department D
800
1000
#NAME?
Department E
800
800
#NAME?
#6. A. The probability that no customers are in the system.
P0 = Prob (no waiting) = Prob(system is empty) = 1-p
p is server utilization is λ/μ
Arrival rate is
0.4 customers per minute
Service rate is 0.6 customers per minute
0.33 is the probability that no customers in the system
B. The average number of customers waiting in line
Lq = average number in line Lq = λ^2 / μ(μ-λ)
1.33 number of customers waiting in line C. The Average number of customers in the system
L
L=Lq+λ/μ
1.33+(0.4/0.6)
2 customers in the system D. The average time a customer spends waiting in line
Wq
Wq= Lq/λ
3.33 minutes
E. The average time a customer spends in the system W=Wq + 1/μ =L/λ
5 minutes
F. The probability that arriving customers will have to wait for service (hint: an arriving customer has
wait except when no customers are in the system upon his arrival).
1- P(0)
1-(1-(0.4/0.6))
0.67
#7.
Interarrival time is
6 min
λ
SD of interarival t
3.94 min
Avg service time
5 min
μ
SD of service time
2.83 min
# of servers
2 people
A. What is the arrival rate? What is the service rate per employee?
Arrival rate is 1/avg interarrival time
0.17 customers per min or 1/6
Service rate per employee
0.2 customers per minute or 1/5
B. What is the utilization of each employee?
utilization is λ/(Kμ)
K is number of servers 0.42
C. What is the average time a customer waits for service?
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Cva^2
0.66
Cvp^2
0.57
ANSWER:
equals 0.46 is Wq and in minutes D.
What is the average number of customers waiting to be served
Lq is lambda *Wq
Answer is 0.18
E. What is the averge number of customers being served?
=
λ/μ
1.2 average number of customers being served
F. Avg number of customers in the system
L = Lq +lambda/meu
1.38 avg number of customers in the system
G. Avg time a customer spends in the system
W
W= L / Lambda 0.23 minutes spent in queue
ent E. *1)
λ
μ
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s to
0.66
0.57