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Solution Manual for Matching Supply with Demand An Introduction to Operations Management, 5th Edition Cachon. All Chapters A+ The Process View of the Organization Q2.1 Dell The following steps refer directly to Exhibit 2.1. #1: For 2001, we find in Dell’s 10-k: Inventory = $400 (in million) #2: F...

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Solution Manual for
Matching Supply with Demand An Introduction to Operations Management, 5th Edition Cachon
Chapter 2-19

Chapter 2
The Process View of the Organization

Q2.1 Dell
The following steps refer directly to Exhibit 2.1.
#1: For 2001, we find in Dell’s 10-k: Inventory = $400 (in million)
#2: For 2001, we find in Dell’s 10-k: COGS = $26,442 (in million)
26,442$ / year
#3: Inventory turns   66.105 turns per year
400$
40% per year
#4: Per unit Inventory cost   0.605% per year
66.105 per year

Q2.2. Airline
We use Little’s law to compute the flow time, since we know both the flow rate as well
as the inventory level:
Flow Time  Inventory / Flow Rate  35 passengers / 255 passengers per hour  0.137 hours
 8.24 minutes

Q2.3 Inventory Cost
(a) Sales  $60,000,000 per year / $2000 per unit  30,000 units sold per year
Inventory  $20,000,000/ $1000 per unit  20,000 units in inventory

Flow Time  Inventory / Flow Rate  20,000/ 30,000 per year  2/ 3 year  8 months
Turns  1/ Flow Time  1/ (2/ 3 year)  1.5 turns per year

Note: we can also get this number directly by writing: Inventory turns  COGS / Inventory

(b) Cost of Inventory: 25% per year /1.5 turns  16.66% . For a $1000 product, this would
make an absolute inventory cost of $166.66 .

Q2.4. Apparel Retailing

(a) Revenue of $100M implies COGS of $50M (because of the 100% markup).
Turns  COGS/ Inventory  $50M / $5M  10 .
(b) The inventory cost, given 10 turns, is 40%/10  4% . For a 30$ item, the inventory
cost is 0.4  $30  $1.20 per unit .

Q2.5. La Villa
(a) Flow Rate  Inventory / Flow Time  1200 skiers /10 days  120 skiers per day
(b) Last year: on any given day, 10% (1 of 10) of skiers are on their first day of skiing

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

, This year: on any given day, 20% (1 of 5) of skiers are on their first day of skiing

Average amount spent in local restaurants (per skier)
Last year  0.1 $50  0.9  $30  $32
This year  0.2  $50  0.8  $30  $34
% change  ($34  $32) / $32  6.25% increase

Q2.6. Highway
We look at 1 mile of highway as our process. Since the speed is 60 miles per hour, it
takes a car 1 minute to travel through the process (flow time).
There are 24 cars on ¼ of a mile, i.e. there are 96 cars on the 1 mile stretch (inventory).
Inventory = Flow Rate * Flow Time: 96 cars = Flow Rate * 1 minute
Thus, the Flow Rate is 96 cars per minute, corresponding to 96*60 = 5760 cars per hour.

Q2.7. Strohrmann Baking
The bread needs to be in the oven for 12 minutes (flow time). We want to produce at a
flow rate of 4000 breads per hour, or 4000/60 = 66.66 breads per minute.

Inventory = Flow Rate * Flow Time: Inventory = 66.66 breads per minute* 12 minutes
Thus, Inventory = 800 breads, which is the required size of the oven.

Q2.8. Mt Kinley Consulting

We have the following information available from the question:

Level Inventory (number of consultants at Flow Time (time spent at that
that level) level)
Associate 200 4 years
Manager 60 6 years
Partner 20 10 years

(a) We can use Little’s law to find the flow rate for associate consultants: Inventory =
Flow Rate * Flow Time; 200 consultants = Flow Rate * 4 years; thus, the flow rate is
50 consultants per year, which need to be recruited to keep the firm in its current size
(note: while there are also 50 consultants leaving the associate level, this says nothing
about how many of them are dismissed vs how many of them are promoted to
Manager level).

(b) We can perform a similar analysis at the manager level, which indicates that the flow
rate there is 10 consultants. In order to have 10 consultants as a flow rate at the
manager level, we need to promote 10 associates to manager level (remember, the
firm is not recruiting to the higher ranks from the outside). Hence, every year, we
dismiss 40 associates and promote 10 associates to the manager level (the odds at that
level are 20%)

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

, Now, consider the partner level. The flow rate there is 2 consultants per year (obtained
via the same calculations as before). Thus, from the 10 manager cases we evaluate every
year, 8 are dismissed and 2 are promoted to partner (the odds at that level are thereby also
20%).

In order to find the odds of a new hire to become partner, we need to multiply the
promotion probabilities: 0.2*0.2 = 0.04. Thus, a new hire has a 4% chance of making it to
partner.

Q2.9. Major US Retailers
a. Product stays on average for 31.9 days in Costco’s inventory
b. Costco has for a $5 product an inventory cost of $0.1311 which compares to a
$0.2049 at Wal-Mart

Q2.10. McDonald’s
a. Inventory turns for McDonald’s were 92.3. They were 30.05 for Wendy’s.
b. McDonald’s has per unit inventory costs of 0.32%, which for a 3$ meal about
$0.00975 . That compares to 0.998% at Wendy’s where the cost per meal is $0.0299 .

Q2.11. BCH
I = 400 associates, T = 2 years. R  I / T  400 associates / 2 yrs  200 associates / yr .

Q2.12. Kroger
Turns  R / I   12.3

Matching Supply with Demand: An Introduction to Operations Management
5e
Solutions to Chapter Problems

Chapter 3
Understanding the Supply Process: Evaluating Process Capacity

Q3.1 Process Analysis with One Flow Unit
(a) Capacity of the three resources in units per hour are 60  2 /10  12 , 60 1/ 6  10 ;
60  3 /16  11.25 . The bottleneck is the resource with the lowest capacity, which is
resource 2.
(b) The process capacity is the capacity of the bottleneck, which is 10 units/hr .
(c) If demand  8 units / hr , then the process is demand constrained and the flow rate is
8 units/hr
(d) Utilization = Flow Rate / Capacity . For the three resources they are 8 /12 , 8 /10 , and
8 /11.25 .

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

,Q3.2 Process Analysis with Multiple Flow Units
a) Bottleneck is resource 3 because it has the highest implied utilization of 125%. The
demands per hour of the three products are  5 ,  6.25 and  7.5 . The
total minutes of work demanded per hour at resource 1 is 5 × 5 + 6.25 * 5 + 7.5 * 5 =
93.75. Two workers at resource 1 produce 2 * 60 = 120 min of work per hour. So
resource 1’s utilization is 93.75 /120  0.78 . Utilization at the other resources are
similarly evaluated.
b) The capacity of resource 3 is 60 /15  4 units per hour. Given the ratio of units produced
must be 4 to 5 to 6, the process can produce 4 units/ hr of A, 5 units / hr of B and
6 units / hr of C.

Q3.3. Cranberry
Cranberries arrive at a rate of 150 barrels per hour. They get processed at a rate of 100 barrels
per hour. Thus, inventory accumulates at a rate of 150-100 = 50 barrels per hour. This happens
while trucks arrive, i.e. from 6am to 2pm. The highest inventory level thereby is 8h*50 barrels
per hour = 400 barrels. From these 400 barrels, 200 barrels are in the bins, the other 200 barrels
are in trucks.
(a) 200 barrels
(b) From 2pm onwards, no additional cranberries are received. Inventory gets depleted at a rate
of 100 barrels per hour. Thus, it will take 2h until the inventory level has dropped to 200
barrels, at which time all waiting cranberries can be stored in the bins (no more truck
waiting)
(c) It will take another 2 hours until all the bins are empty
(d) Since the seasonal workers only start at 10:00am, the first 4 hours of the day we accumulate
4hours * 50barrels per hour = 200 barrels. For the remaining time that we receive incoming
cranberries, our processing rate is higher (125 barrels per hour). Thus, inventory only
accumulates at a rate of 25 (150-125 barrels per hour). Given that this happens over 4 hours,
we get another 100 barrels in inventory. At 2pm, we thereby have 300 barrels in inventory.
After 2pm, we receive no further cranberries, yet we initially process cranberries at a rate of
125 barrels per hour. Thus, it only takes 100 barrels /125 barrels/hour  0.8 hours  48 minutes
until all bins are empty. From then, we need another 2h until the bins are empty.

Q3.4. Western Pennsylvania Milk
We start the day with 25,000 gallons of milk in inventory. From 8am onwards, we produce 5,000
gallons, yet we ship 10,000 gallons. Thus inventory is depleted at a rate of 5000 gallons per hour,
which leaves us without milk after 5 hours (at 1pm). From then onwards, clients will have to
wait. This situation gets worse and worse and by 6pm (last client arrives), we are short 25,000
gallons.
(a) 1pm
(b) Clients will stop waiting when we have worked off our 25,000 gallon backlog that we are
facing at 6pm. Since we are doing this at a rate of 5,000 gallon per hour, clients will stop
waiting at 11pm (after 5 more hours).
(c) At 6pm, we have a backlog of 25,000 gallons, which is equivalent to 20 trucks
(d) The waiting time is the area in the triangle

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

, - width: beginning of waiting (1pm) to end of waiting (11pm) = 10 hours
- height: maximum number of trucks waiting: 20 (see part c above)
Hence, we can compute the area in the triangle as: 0.5*10hours*20trucks = 100 truck* hours
The cost for this waiting is 50$ / truck  hour 100 truck hours  5000$

Q3.5. Bagel Store
(a) The bottleneck is “Veggies on Bagel”, as it has the highest implied utilization.

Resource Available Grilled Veggie Cream Total Implied
minutes per veggie bagel cheese utilization
hour bagel bagel
Cut 60 3*3 3*11 3*4 54
Grilled 60 3*10 0 0 30 30/60
stuff
Veggies on 60 3*5 11*5 0 70
Bagel
Cream 60 0 0 4*4 16
cheese
Wrap 60 3*2 11*2 4*2 36

(b) If we want to keep the product mix constant (i.e. keep the ration between the various bagel
types at 3:11:4), we need to scale down demand by . This leads to the following flow
rates:
- Grilled veggie: 3  bagels / hour
- Veggie: 11 bagels / hour
- Cream cheese: 4 bagels / hour
If we try to fulfill as much demand as possible, we encounter a problem with Veggie bagels and
Grilled + Veggie bagels. Given that we have an implied utilization of % at the bottleneck,
we can only fulfill % of demand for these two bagel types. We can meet all of demand for
the cream cheese bagels.

Q3.6. Valley Forge Income Tax
Recall that the demand for the 4 groups is:
1: 15%
2: 5%
3: 50%
4:30%

There are three resources, the admin, the junior person, and the senior person. We compute the
minutes of capacity they have available every month as well as their work-load, given that there
are 50 incoming returns to be processed. This leads to the following computations.

Resource Minutes 1 2 3 4 Total Implied
per month work- utilization

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

, load
Admin 9600 0.15*50 0.05*50 0.5*50 0.3*50 3575 0.372
*(20+50) *(40+80) *(20+30) *(40+60)
Senior 9600 0.15*50 0.05*50 0.5*50 0.3*50 1325 0.138
*(30+20) *(90+60) *(0+5) *(0+30)
Junior 9600 0.15*50 0.05*50 0.5*50 0.3*50 6650 0.693
*120 *300 *80 *200

(a) The junior person is the bottleneck, her implied utilization is the highest

(b) Implied utilizations (see table)

(c) One should consider: revenue, future revenue (lifetime value of customer), to what extent the
request draws on bottleneck capacity

(d) not at all, since improvements at non-bottleneck steps don’t increase capacity

Q3.7. Car Wash Supply Process
a. The implied utilization at wheel cleaning is: 42 minutes of work from package 3 a day,
84 minutes of work from package 4 per day, giving a total of 126 minutes per day.
Relative to 720 minutes of available time, that gives an implied utilization of 17.5%.
b. The highest implied utilization is at step 1 (automated washing machine) where the
implied utilization is 55.55%.
c. The new bottleneck will be the interior cleaning employee with an implied utilization of
111%
d. Every day, we are 80 minutes of work short at step “interior cleaning”. That corresponds
to four customers with package 4.

Q3.8. Starbucks
a. Time required of the frozen drink maker = 20*2 = 40 minutes. Each worker has 60
minutes available, so the utilization for the frozen drink maker:
b. Highest implied utilization is for the Espresso Drink Maker with 116.66%
c. The workload on the cashier is as follows. 25 Drip coffee customer per hour at
1/ 3 min per customer  min / hr . 5 Ground customers per hour at 1 min per
customer  5 min / hr . 120 customers per hour at 1/ 3 min to pay per
customer  40 min / hr . 30% of 120 customers per hour buying food, which is 36
customers per hour and they require 1/ 3 min per customer, for a total of 12 min / hr . In
total the workload on the cashier is  5  40  12  65.33 min / hr . The implied
utilization is 65.  108.9% . The implied utilization of the other persons don’t
change and they are 66.66% and 116.66%.

Q3.9. Paris Airport
a. The implied utilization levels of the resources are computed as follows

Servers Avail. Min/Hr Requested Imp. Ulll.

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

, Security 4 240 150 62.50%
Agents 6 360 440 122.22%
Kiosk 3 180 160 88.89%

b. The backlog at the bottleneck accumulates at a rate of 80 “requested minutes” per hour.
After 4 hours this is 4*80 = 320 “requested minutes” of work from the agents. This takes
6 agents   0.89 hours to complete, or 53.4 minutes after the last arrival
c. When Kim arrives, the backlog is half as long as the final backlog in question 2, e.g.
there are 160 “requested minutes” of work in front of Kim. This will take the agents
 0.45 hours  26.6 minutes to complete. Kim takes 3 minutes to complete
service with the agent and 30 seconds to pass through security (there is no line at
security) so the answer is 30 minutes and 10 seconds
d. We compute the new utilizations as follows:

Servers Avail Min/Hr Requested Imp. Ulll.
Security 4 240 150 62.50%
Agents 6 360 376 104.44%
Kiosk 3 180 32 17.78%

Extra work accumulates at a rate of 16 minutes per hour for the agents, for a total of 64
minutes after the last arrival. This takes the 6 agents  0.178 hours  10.7 minutes to
complete, so 8:10 PM is the closest answer.
Matching Supply with Demand: An Introduction to Operations Management
5e
Solutions to Chapter Problems

Chapter 4
Estimating and Reducing Labor Costs

Q4.1. Empty System Labor Utilization
(a) Time to complete 100 units:

#1 The process will take 10 + 6 + 16 minutes = 32 minutes to produce the first unit.

#2 We know from problem xyz that resource 2 is the bottleneck and the process capacity is
0.1666 units per minute

#3 Time to finish 100 units

99 units
 32 minutes   626 minutes
0.1666 units / min

(b) + (c) + (d) Use Exhibit for Labor computations

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

,#1 Capacities are:
Resource 1: 2 /10 units/ minute  0.2 units / minute
Resource 2: 1/ 6 units/ minute  0.1666 units / minute
Resource 3: 3 /16 units/ minute  0.1875 units / minute
Resource 2 is the bottleneck and the process capacity is 0.1666 units/minute

#2 Since there is unlimited demand, the flow rate is determined by the capacity and thereby
0.1666 units/minute ; this corresponds to a cycle time of 6 minutes/unit

6 10$ / h
#3 Cost of direct labor   6$ / unit
60  0.1666 units / h

#4 Compute the idle time of each worker for each unit:
Idle time for workers at resource 1  6 min / unit  2  10min / unit  2 min / unit
Idle time for worker at resource 2  6 min / unit 1  6 min / unit  0 min / unit
Idle time for workers at resource 3  6 min / unit  3  16 min / unit  2 min / unit

#5 Labor content  10  6  16 min / unit  32 min / unit

32
#6 Average labor utilization   0.8888
32  4

Q4.2. Assign tasks to workers
(a)
Worker Task(s) Processing Time (sec) Capacity (units per hour)
1 1 30 120
2 2 25 144
3 3,4 75 48
4 5,6 45 80
The capacity of the current line is restricted by the capacity of the step with the longest
processing time. Therefore, capacity  1/ 75 sec  48 units per hour .

(b)
Worker Task(s) Processing Time (sec) Capacity (units per hour)
1 1,2 55 65.45
2 3 35 102.86
3 4 40 90
4 5,6 45 80
Therefore, capacity of the revised line  1/ 55 sec  65.45 units per hour .

(a) No matter how you organize the tasks, maximum capacity of the line is 65.45 units per hour,
i.e. at a cycle time of 55 seconds.

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

,Q4.3. Power Toys
(a) Since every resource has exactly one worker assigned to it, the bottleneck is the assembly
station with the highest processing time (#3)

(b) Capacity  1/ 90 sec  40 units per hour

(c) Direct labor cost  Labor cost per hour / flow rate  9 15$ / h / 40 trucks per hour  3.38 $ / truck

(d) Direct labor cost in work cell  (75  85  90  65  70  55  80  65  80)sec / truck  $15 / hr
 2.77$ / truck

(e) Utilization  flow rate / capacity 85 sec / 90 sec  94.4%

(f)
Worker Station(s) Processing Time (sec) Capacity (units per hour)
1 1 75 48
2 2 85 42.35
3 3 90 40
4 4,5 135 26.67
5 6,7 135 26.67
6 8,9 145 24.83

(g) Capacity  1/145 units / second  24.83 toy - trucks per hour

Q4.4. 12 tasks to 4 workers
(a)
Worker Task(s) Processing Time (sec) Capacity (units per hour)
1 1,2,3 70 51.43
2 4,5,6 55 65.45
3 7,8,9 85 42.35
4 10,11,12 60 60

(a) Capacity  1/ 85 sec  42.35 units per hour

(b) Direct labor content  (70  55  85  60) sec  270 sec/ unit or 4.5 min / unit

(c) Labor utilization  labor content / ( labor content  total idle time )  270sec / (270  15 
30  0  25 sec)  79.41%

(d) Note that we are facing a machine paced line, thus the first unit will take 4*85 seconds top go
through the empty system. Flow Time  4  85 sec  99 /(1/ 85 sec)  8755 sec or 145.92 min
or 2.43 hrs

(e) There are multiple ways to achieve this capacity. This table shows only one example.

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

, Worker Task(s) Processing Time (sec) Capacity (units per hour)
1 1,2 55 65.45
2 3,4,5 50 51.43
3 6,7 70 51.43
4 8,9 35 102.86
5 10,11,12 60 60
Capacity  1/ 70 units / sec  51.43 units per hour

(f) There are multiple ways to achieve this capacity. This table shows only one example.
Worker Task(s) Processing Time (sec) Capacity (units per hour)
1 1,2 55 65.45
2 3,4,6 55 65.45
3 5,8,10 55 65.45
4 7 50 72
5 9,11,12 55 65.45
Capacity  1/ 55 units / sec  65.45 units per hour

(g) We have to achieve a cycle time of  50 seconds/unit. The following task allocation
includes a lot of idle time, but is the only way to achieve the cycle time, given the constraints
we face.

Worker Task(s) Processing Time (sec)
1 1 30
2 2,3 40
3 4,5 35
4 6 20
5 7 50
6 8,9 40
7 10,11 40
8 12 20
Therefore, a minimum of 8 workers are required to achieve a capacity of 72 units per hour.

Q4.5. Geneva Watch
(a) Station E is the bottleneck with a process capacity of 1 unit every 75 seconds.

(b) Capacity  1/ 75 sec  48 watches per hour

(c) Direct labor content = 68 + 60 + 70 + 58 + 75 + 64 = 395 sec

(d) Utilization  60 sec / 75 sec  80%

(e) Idle time  (75  70)sec/ 75 sec  60 min per hour  4 min per hour ; as an alternative
computation, we can observe that the worker has 5 seconds idle time per cycle (i.e. per unit)
and that there are 48 cycles (units) per hour. Thus, the idle time over the course of an hour is
240 seconds = 4 minutes.

© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.

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