PRODUCTION AND OPERATIONS MANAGEMENT

PRACTICALS

Chapter- Inventory management

1. Condition--[When lead time is not given]

A product demand that is normally distributed during the lead time, with a mean of 350
units and a standard deviation of 10. Company wants to follow a policy that result in
stockouts occurring only 5 % of the time on an order. How much safety stock should be
maintain and what will be the Reordering point? (Assume Z value as 1.64).
2. Condition—[ demand is variable but lead time is constant]
Average Daily demand for certain product is 60 units per day ,with a standard deviation of 7
units . Lead time is 6 days , Cost of placing order is Rs. 10/- per order and annual holding cost
is Rs. 0.50 per unit, Assume sales occur entire 365 days. Calculate Reorder quantity and
reorder point to satisfy 95% probability of no stock out during lead time. (Assume Z value as
1.64).
3. For a company average daily demand is normally distributed, with a mean of 15 units and a
standard deviation of 3. Lead time is exactly 4 days. Company wants to maintain a 97%
service level. What is the reorder point, and how much safety stock should be carried?
(Assume Z value as 1.88).
NOTE: SERVICE LEVEL = 1 – Probability of stockout
4. Condition—[ demand is constant but lead time is variable]
The daily demand is constant at 25 units per day, and the lead time is normally distributed,
with a mean of 6 days and standard deviation of 3. Company wants to maintain a 98%
service level on this product. What is the reorder point?

1. A company uses 1200 unit per month of an electric component each costing Rs.2. placing each
order cost Rs. 50 and carrying cost is 6% per year of the average inventory. 1. Find EOQ. 2. If the
company gets 5% discount if it place single order, should they accept the discount order?

2. An auto industry purchases spark plugs at the rate of Rs. 25 per piece. The annual consumption of
spark plug is 18,000 nos. if the ordering cost is Rs. 250 per order and carrying cost 25% pa. what
would be EOQ? If the supplier offers a discount of 5% for order quantity of 3000 no per order, do you
accept the offer?

3. For a given item, there is constant demand rate. Annual demand is 60,000 nos. the price per item is
Rs. 30. The ordering cost is estimated as Rs. 300 per order and carrying cost is 30% per annum. What
should be the optimal ordering quantity? If 2000 unit is purchased at time, a discount of 5% on unit
price is given , is offer accepted?

4. A company uses every month 1500 units of a component costing Rs. 1.20 each. Each purchase
order cost Rs. 50 and inventory carrying cost is 6% per yr of the average inventory. A] Find EOQ. B]
if 5% discount is given for a single order placement, is it worth?

EOQ METHOD:

(a) ECONOMIC ORDER QUANTITY (EOQ)

The optimal size of an order for replenishment of inventory is called economic
order quantity. Economic order quantity (EOQ) or optimum order quantity is that
size of the order where total inventory costs (ordering costs + carrying costs) are minimized.
Economic order quantity can be calculated from any of the following
two methods:

Where, EOQ = Economic Order Quantity

R = Annual Requirement or consumption in units
O = Ordering Cost per order
C = Carrying Cost per unit per year
No. of orders = R/EOQ
Time gap between two orders = No. of days in a year/No. of orders
Total Cost = Purchase Cost + Carrying Cost + Order Cost
= (R x Unit Price) + (EOQ/2 x C) + (R/EOQ x O)
(b) RE-ORDER POINT
After determining the optimum quantity of purchase order, the next problem is to specify the
point of time when the order should be placed. Re-order level is that level of inventory at
which an order should be placed for replenishing the current stock of inventory. The
determination of re-order point depends upon the lead time, usage rate and safety stock.
These terms are explained below:
1. Lead Time: Lead time refers to the time gap between placing the order and actually
receiving the items ordered.
2. Usage Rate: It refers to the rate of consumption of raw material per day.
Usage Rate = Total annual consumption / No. of days in a year
3. Safety Stock: It is the minimum quantity of inventory which a firm decides to maintain
always to protect itself against the risk and losses likely to occur due to stoppage in
Production and loss of sale, due to non- availability of inventory.
Formulae:
Re Order Point = (Lead Time x Usage Rate) + Safety Stock or
Re Order Point = Maximum usage x Maximum Re Order Period
Safety Stock = Usage Rate x Days of safety

Formulae
Maximum Level = (ROL + ROQ) – (Minimum Usage x Minimum Re Order
Period) or Maximum Level = Safety Stock + EOQ

Minimum Level = Re-order level – (Normal Usage Rate x Normal Re-order

period) = Re Order Level – (Normal Usage x Average Re Order Period)

Average stock level = (Maximum level + Minimum level) / 2 or

Average stock level = (Minimum level + 1/2 Re- order Quantity)

Q.4. Two components P and Q are used as follow:

Normal usage 600 units per week each.
Maximum usage 900 units per week each
Minimum usage 300 units per week each
Re-order quantity- P = 4,000 units; Q = 7,000 units
Re-order period- P = 4 to 6 weeks; Q = 2 to 4 weeks
a. Re-order level
b. Minimum Level
c. Maximum Level
d. Average stock level

Q.5. 1. Annual demand of EMR- 1,000 units

2. Cost of placing an order Rs. 100
3. Annual carrying cost per units Rs. 10
4. Normal usages 100 units per week
5. Minimum usages 50 units per week
6. Maximum usages 150 units per week
7. Re-order period 2 to 6 weeks
Calculate the following:
a. Re-order Quantity
b. Re-order level
c. Minimum Level
d. Maximum Level
e. Average stock level

MODULE 3 OPERATIONS SCHEDULING (SUMS)

Q.1. Calculate Idle time for each machine. Also find Total Elapsed time.

Book A B C D E F
Printing 30 120 50 20 90 110
time (hrs.)
Binding 80 100 90 60 30 10
time (hrs.)

Q.2. Find the sequence that minimize the total Elapsed time required to complete the following jobs
on 2 machines:

Job 1 2 3 4 5 6 7 8 9
Machine A 12 15 14 19 16 18 17 15 14
Machine B 16 18 17 14 13 19 13 18 21

Q.3. Processing times of 7 jobs on three machines I,II,III are given. The order of processing is I-II-III.
Determine the sequence that minimizes the total elapsed time required to complete the jobs. Also
evaluate the idle time of machines.

Jobs I II III
A 30 40 60
B 80 30 70
 C 70 20 50
D 40 50 110
E 90 10 50
F 80 40 60
G 70 30 120

Q.4. Six task must go through 3 machines A,B,C in the order ABC process time in days are given in the
following table:

Determine the sequence for the five jobs that minimize the elapsed time.

Job J1 J2 J3 J4 J5 J6
Machine A 8 3 7 2 5 1
Machine B 3 4 5 2 1 6
Machine C 8 7 6 9 10 9

Chapter PERT & CPM:

1. Prepare a PERT CHART from following data:

Activity 1-2 1-6 2-3 2-4 3-5 4-5 6-7 5-8 7-8

Time 8 4 14 7 13 8 10 5 20

2. Draw pert chart and find expected time.

activity sequence Optimistic time Most likely Pessimistic time

time

A 1-2 7 12 13

B 1-3 7 10 12

C 2-5 8 13 15

D 3-5 10 12 22

E 5-6 10 14 18

3. Prepare a network and determined critical path:

Activity Preceding activity Estimated time

A - 2

B A 3

C B 4

D - 5
 E D 2

F C,E 5

4. Determined critical path for the following project. Find EFT,LFT and FLOAT TIME

Job Time

1-2 2

1-3 5

2-4 4

3-4 3

3-5 5

4-6 6

5-7 2

6-7 4

5. Time estimates and predecessors of each activity in a project are given below
⮚ Draw a CPM network diagram

⮚ Find out critical path & duration of CP

⮚ Compute EST,EFT,LST & LFT

⮚ Compute the free float of each activity

Activity A B C D E F G H

Predecessors - A A B B D,E D C,F,G

Time 2 4 8 3 2 3 4 8

Project crashing/ Crashing CPM network

6. Construct the project network and find optimal project duration by crashing activities.
Consider Indirect cost = Rs. 200 per week.

Activity Time (weeks) NormalCost (RS) Crash Time(week) Crash cost (Rs.)

1-2 7 700 4 850

1-3 5 500 3 700

1-4 8 600 5 1200

2-5 9 800 7 1250

3-5 5 700 3 1000

3-6 6 1100 5 1300

4-6 7 1200 5 1450

5-7 2 400 1 500

6-7 3 500 2 850

 i Crashingtdot) Gash
Cest- nornad
Cas
Actushes Time N. Coat Crashfie Crash last Slope
TATme CiThe
J-2 850 50
|-3 500 700
l-4 600 5 1200 200
2-5 9 800 1250 925

3-5 5 100 1000

3-6 6 1300 200

4-6 l200 5 1450 |25

5-4 400 500
6-4 3 500 850 350
Step Find Sope = Cesh Cost - Normal Cost
Nor mal ime - Crh Tine
9 Draw Netoork nd ckeal PaH
Stap |Tme Taken
Path
|-2-54 18CP
S-3-5-7 12
6 -3-6t
-4-6-4 J18 CP

* Note : Crashing is oone ou Cribcal Path

Tojeet Compleion TmeCostiu of Pojeol with 18 Weeleu.
Step 3 Findl totl
Cost of Projeel (Total e? N. Cost) - 6500
Norma Inalieal 3,600
+ Tota Tndieal Cost (20o oele x 18 welula |10,1eD R.

1
step 4 Deeide Chashing Aeffoiy find Crash linait slope
minimum
Gashing linit Sopen
CHkeal Pth (Cikeal Aefoiliy LCost
|-2-5-4 I-2 4-4 =(3) 2 50
9-5 225
5-4
|-4 8-5- 3 200
4-6 7-5-(2)1 125

350
Iterahion Repea all
Stp 1 Draw olagam Airad Criheal Palk
Path Tie Taleen
|-2-5-+
(6) 12
2
5 5 |-3-6 -+ 14
I 4-6-+

Total Cost + Direet lost hDirest Cosl

Totl Cost = Pevious -10075 R
= 10, 100 +50+|25)- 200
Step 3 Chashing aehivity ) find Cresh linit 2slope
Actuty Chash So
ing Gmit| slope
Cikeal Path GiBeal
I-2 (2)1il 50
|- 2-5-+w1
2-5 225

I-4-6-+ -4 200

4-6 125

3SO
2
Tterabion 3
Arnd cp
Step 1 Drao Nelawok Fathhe laln
2-2-54 CP
|-3-5-+
5
l3- 6-7 14
6
1-4-6- I6 cP
CP

Step 2 Final Tötal Cot wih l6 weles complekin TF

Total Cost - Ruvions total Cost + Diretlost -Tlivetlost
= lo, 0+s0+ 125-200
=lo, o25 Rs
Crashing Akvily Find Cradh tine gslope.
Step 3
Citeal Path cical Aehuil Crash i slope
|- 2-5-+
2-5 225

5-4

-4-6 + |4 3 200

4-6
350
I teration 4

Path Tine Tahen

|-2-5-+ 15
5
|-3-5-+ 12
6 1-3-6 - 14
| 4-6-7

3
 stp 2 Find Total ost of Pojet oith t6 days.
lost
10tal Cost = FewiousTotal Cost + Diret ost - Ihdiret
025 + (50+ 20D)+ 200
= l0,
l00 R.

KStop Crashing because total Cost iu hgh

than 3rdl leralon.