Answers
Chapter 5
Q1
a) Mean demand (μ) = 60 cars
Variance (σ^2) = 36 cars
lead time = 3 months
Service level @ 95%
Z= 1.65
Standard deviation (σ) = √(variance) = √36 = 6 cars
Safety stock (SS) = z * σ = 1.65 * 6 = 10 cars
order quantity (Q) = μ + SS = 60 + 10 = 70 cars
b) Loss-of-goodwill cost = $100 per customer
Bookkeeping expenses = $50 per customer
Expected backorder cost (EBC) = Probability of stockout * Loss-of-goodwill +
Probability of stockout * Bookkeeping expenses
Service level @ 95%
z = 1.65
Probability of stockout = 1 - Probability of meeting demand = 1 - Φ(z)
Probability of stockout = 1 – Φ (1.65) = 1 - 0.9505 = 0.0495
Expected backorder cost (EBC) = 0.0495 * $100 + 0.0495 * $50
= $4.95 + $2.475
= $7.425
Q = μ + SS + Expected backorders
= 60 + 10 + EBC
= 60 + 10 + $7.425
= 77.425
Q ~ 78 cars
Hence, Happy Henry's should be buying 78 cars every 3 months, taking the cost of
backorders into account.
c) Average cost of an EX123 for Happy Henry's (CC) = $10,000
Average selling price of an EX123 (SP) = $13,500
Underage cost = 13500- 10000
= $ 3500
Cs = 125
Critical ratio = Cu / (Cu +Cs)
= 3500 / 3500 +125 = 0.9655
Z= 1.65
Q = 6 X 1.65 + 60 = 70.92 ~ 71
Purchasing Cars in every 3 months is 71 cars.
Q2
a) D = 320 (mean monthly demand)
S = $100 (order setup cost)
H = 0.28 * $1.50 = $0.42 (holding cost per unit per month, 28% annual interest rate)
L = 5 (lead time in months)
V = (81)^2 = 6561 (variance of lead time demand)
service level @ 95%
z= 1.65
EOQ = √ ( 2 X 320 X 100 / 042)
= 390.36
R = 5 X 320 + 1.65 X √ ( 5 X 6561 )
= 1897.945
Q = 390, R = 1898
b)
Holding cost per year = (Q / 2) * H
Setup cost per year = (D / Q) * S
Stock-out cost per year = (D - L * R) * C
C = $12.80 (cost per back-ordered demand)
Holding cost per year = ( 389 /2 ) X 0.42
=$ 81.69
Setup cost per year = (320 – 5 X 1898) X 12.80
= - $117,373
Avg annual cost = Holding cost per year + setup cost per year
= $81.69 + $ 82.26
= $163.95
c)
Avg annual cost with zero variance = Holding cost per year + Setup cost per year
= $ 81.69 + $ 82.26
= $ 163.95
Since the average annual cost with zero variance is the same as the average annual cost
obtained in part (b), the cost of uncertainty is zero.
d) service level @ 95%
z = 1.65
Q = √ ( ( 2 X 320 X 100 ) / 0.42)
= 390.36
R= 5 x 320 + 1645 x √ ( 5 x 6561 )
= 1897.945
Hence, the optimal values with a Type 1 service objective of 95 percent are Q = 390
and R = 1898.
e)
service level @ 95%
z= 1.65
C = $12.80 (cost per back-ordered demand)
Q = √ ( ( 2 X 320 X 100 ) / 0.42)
= 390.36
R = 5 * 320 + Z * √ (5 * 6561)
service level @ 95%
Type 2
z= 1 - 1.65 = -0.65
R = 5 X 320 + (- 0. 645) X √ ( 5 X 6561 )
= 1,483.177
Hence, the optimal values with a Type 2 service objective of 95 percent are Q = 390
and R = 1483. The imputed cost of shortage remains the same as in part (b): $12.80.
Chapter 7
Q1
a) time between arrivals = 5 minutes
=sevice rate=1/4.5
= 4.5 min
= arrival rate =1/5 = 0.2 per min
utilization =
= 0.2 / 0.222 = 0.9
Thus, the average utilization of the server is 0.9
b)
Time in queue = Wq = Lq /
Lq = 2 / ( 1 - )
= 0.9
time-in-system (bakery) = (waiting time +service time) = Ws = Wq + ( 1 /
Lq = 2 / ( 1 - )
= 0.92 / (1 - 0.9)
= 0.81 / 0.1
= 8.1
So, Lq = 8.1
)
Wq = Lq /
= 8.1 / 0.
Wq = 40.5 min
Ws = Wq + ( 1 /
= 40.5 + (4.5 )
Ws = 45 min
)
customers spend on average 45 min to complete their transactions at the bakery
(time in queue plus service time).
Wq = 40.5 min
Ws = 45 min
Wq * 100 / Ws
= 40.5 * 100 / 45
= 90 % of that time is spent queueing.
c) Average number-in-system (waiting + getting served)
= * Ws = 0.2 * 45
=9
Therefore, 9 customers are in the bakery on average.
d) t = 1 hour = 60 min
= 0.22222 per min
= 0.9
Probability that time-in-system / bakery is greater than t
= e- (1 - ) t
So, Probability that time-in-system / bakery is greater than 1 hour
= e-0.2222(1 - 0.9 ) 60
= e-1.33333 (e ~ 2.71828)
= 0.263598
e) We need to calculate P(0 customers) + P(1 customer) to find the probability that
there are fewer than two customers in the bakery.
Using the M/M/1 queuing model
=
/
P(n) =( 1-
= 0.9
)X
^n
For 0 customers
P(0) = (1 – 0.9 ) X 0.90 = 0.1
For 1 customers
P(0) = (1 – 0.9 ) X 0.91 = 0.09
P(0 customers) + P(1 customer) = 0.1 + 0.09
= 0.19
= 19%
f) The estimated wait times in the bakery system are substantially longer due to the
high level of traffic, indicating that the system is almost at capacity. This is due to the
fundamental assumptions of the M/M/1 queuing model. However, it's possible that the
model's idealized assumptions, such as exponential service durations, a single server,
and ad hoc customer arrivals, might not adequately reflect the intricate dynamics of a
genuine bakery environment. In truth, the distribution of service times might be more
different due to complexity, such as complicated orders and varying preparation times.
Additionally, the single-server premise disregards the implications of customer
interactions and potential simultaneous workloads. Also, the model's claim that no
customers will be ignored isn't always true. Since the M/M/1 model's assumptions are
skewed, it may produce wait times that are longer than anticipated and may not correctly
reflect actual bakery operations, even though it offers informative information.
Q2
a)
service rate is 2/80 = 0.025 customers per second per server’
average service time per client is 80 seconds
consumer arrival rate is 1/50 = 0.02 per second
Utilization = (arrival rate) / (service rate per server * number of servers)
Utilization = 0.02 / 0.025 X 2 = 0.4 = 40%
b) Average time in the system = Average time spent waiting in the queue + Average
service time
Average service time = 80seconds
Average time in the system = (average number of customers in the system) /
(arrival rate)
Average number of customers in the system = (utilization^2) / (1 - utilization)
= 0.4 ^ 2 / 1 – 0.4 = 0.64 / 0.6
= 1.067
Average time in the system = 1.067 / 0.02
= 53.35 seconds
Average time spent waiting in the queue = Average time in the system - Average service
time
Average time spent waiting in the queue = 53.35 – 80
= -26.65 seconds
Since the waiting time cannot be negative, the system must be modified in order to
reduce long wait periods and unhappy clients.
c)
Customers in the system on average = 0.02 X 53.35
= 1.067
So the average number of cars using the drive-through lane is 1.067.
d) The following are some ideas to increase customer satisfaction
* Wait times will be reduced and employee utilization will increase with more servers.
* Implementing a reservation system or mobile ordering to reduce wait times and
improve client satisfaction.
* Giving the workers greater training or refining the menu can reduce the variance in
service times by minimizing the time needed to produce each order.
* To improve the layout and improve traffic flow, you may add more lanes or rearrange
the menu display in the drive-through.
Chapter 8
Q1
a)
MPS for Computers
b) Planned order release for Motherboards assuming a lot for lot scheduling rule
c) Schedule of Outside orders for the Disk Drivers
Q2
a)
P = Average product cost of the product
D = Demand of the product
Q = Quantity to be ordered
Holding cost of the inventory, H = $0.3
Ordering cost of the product, S = $ 200
Q = Sqrt (2SD/H)
Avg Demand D = ( 335 + 200+ 140 + 440 + 300 + 200)/6
= 269.17
Q = √ ((2*200*269.17)/0.3) = 599
Week
1
2
3
4
5
6
Demand
335
200
140
440
300
200
Production
599
0
599
0
599
0
Inventory
264
64
523
83
382
182
b)
Silver Meal Method:
To calculate average cost per period.
C(1) = S
C(2) = (S + h*D)/2
C(3) = (S + h*D + 2h*D)/3
Period 1
C(1) =200
C(2) = (200 + 0.3*200)/2 = 130
C(3) = (130*2 + 2*0.3*140)/3 = 114.67
C(4) = (3*114.67 + 3*0.3*440)/4 = 185. Stop
Period 4:
C(1) = 200
C(2) = (200 + 300*0.3)/2 = 145
C(3) = (2*145 + 2*200*0.3)/3 = 136.67
Now y1 = 335 + 200 + 140 = 675
y4 = 440 + 300 + 200 = 940
c)
Least unit cost
Period
Demand
Beginning inventory
Ordering (setup)
cost/order
Carrying
cost/unit/period
Order quantity (lot
sizes)
Ending inventory
Ordering cost
Carrying cost
Total period cost
1
335
0
2
200
200
3
140
0
4
440
440
5
300
0
6 Totals
200
1615
200
$200.00 $200.00 $200.00 $200.00 $200.00 $200.00
$0.30
535
200
$200.00
$30.00
$230.00
$0.30
$0.30
0
580
0
440
$0.00 $200.00
$30.00 $66.00
$30.00 $266.00
$0.30
$0.30
$0.30
0
500
0
200
$0.00 $200.00
$66.00 $30.00
$66.00 $230.00
0
1615
0
$0.00 $600.00
$30.00 $252.00
$30.00 $852.00
4
440
0
0.5
0.1
22
103.75
2
6 Totals
200
1615
200
2.5
0.1
50
198.75
2
d) Part Period Balancing
Period
Demand
Beginning inventory
factor
Hodling cost per part
Holding cost
Cum Holding cost
Clubbing
Ordering (setup)
cost/order
Carrying
cost/unit/period
Order quantity (lotsizes)
Ending inventory
Ordering cost
Carrying cost
Total period cost
1
335
0
0.5
0.1
16.75
16.75
1
2
200
340
1.5
0.1
30
46.75
1
3
140
140
2.5
0.1
35
81.75
1
5
300
500
1.5
0.1
45
148.75
2
$200.00 $200.00 $200.00 $200.00 $200.00 $200.00
$0.30
675
340
$200.00
$51.00
$251.00
e)
Part period
EQO Balancing
$0.30
0
140
$0.00
$72.00
$72.00
$0.30
$0.30
$0.30
0
940
0
0
500
200
$0.00 $200.00
$0.00
$21.00 $75.00 $105.00
$21.00 $275.00 $105.00
$0.30
0
1615
0
$0.00 $400.00
$30.00 $354.00
$30.00 $754.00
Holding
Cost
Setup Cost
Total Cost
449
600
1049
354
400
754
Hence the least expensive method is Part Period Balancing with a total cost of $754.
Chapter 9
Q1
a) Network Diagram
12
3
7
1
14
2
6
3
6
3
5
6
12
8
4
3
b) Cycle time = 12 + 14+ 6+ 4+ 6+ 3+3 +12
=60 min
c)
Workstation 1: Activity 3 as it has 5 following tasks. Time left is 14 min.
Activity 2 as it has 4 following tasks.
Workstation 2: Activity 4 as it has 4 following tasks. Time left is 16 min.
Assigning activity 5 as it has 3 following tasks.
Time left is 10 min. Activity A cannot be assigned as it has higher time. So, activities
6 & 7 can be assigned.
Workstation 2 = 4,5,6,7
Workstation 3 = Activity 1 assigned. Time left is 8 min. So, activity 8 cannot be
assigned.
Workstation 4 = Activity 8
Q2
a)
Network Diagram
6(8)
4(12)
8(10)
5(10)
1(4)
7(12)
9(2)
10(10)
2(38)
F
S
b)
11(34)
Min cycle time = 45
No. of stations = 5
e) With a cycle time of 45 sec and 6 hours day
production = 6 X 60 X 60 / 45
= 480
d)
least cycle time with 4 stations = 68
Station
Activity club
Cycle time
1
2
3
4
2
1-4-6-8
5-7-9-10-11
3
38
34
68
45
If a company wants to increase production, task 3 must be improved in order to reduce
cycle time.
c)
Task
2
3
1
4
5
6
7
8
9
10
11
Time (seconds)
38
45
4
12
10
8
12
10
2
10
34
Immediate predecessors
1,2
2
4
5
6
7
8,9
3,10
RSW
136
79
78
74
68
62
58
54
46
44
34
Stations Cycle time
1
38
2
45
3
34
3
34
3
34
3
34
4
34
4
34
4
34
4
34
5
34