notes for industrial engineering for gate

1. Voddepalli Soumith
INDUSTRIAL ENGINEERING

2. Break-even Analysis...........................................................................................................................................................................................................2
Forecasting.............................................................................................................................................................................................................................2
Inventory.................................................................................................................................................................................................................................4
Linear programming ..........................................................................................................................................................................................................9
Transportation Method..................................................................................................................................................................................................13
Project Management........................................................................................................................................................................................................19
Queuing Theory.................................................................................................................................................................................................................23

3. Break-even Analysis
In Break-even analysis, graph is drawn between the cost and No.
of units, in which total cost, variable cost, fixed cost, total
revenues will be drawn and from them various factors can be
analysed. There is point where total cost and total revenues
meet, that point is called Break-even Point.
Break-even analysis shows no. of units to be sold to pay off the
cost of doing business. The Quantity at which there is no profit,
no loss this is called Break-even point. After break-even point
sold units will generate profit.
Profit Ratio is defined as the ratio of the profit area to the sum of
the profit and loss areas in a break-even chart.
Profit Ratio =
Area of profit region
Area of profit and loss region
Margin of Safety = Actual sales − BEP
Forecasting
Forecasting is process of predicting or estimating the future value of a variable.
There are 2 types of forecasting, intrinsic and extrinsic forecasting.
With Intrinsic forecasting, forecasting models based on historical data use extrapolation to generate estimates for the
future.
Extrinsic forecasting looks outward to external factors and assumes that internal forecasts can be correlated to
external factors.
Delphi method
The Delphi method is a structured communication technique or method, originally developed as a systematic,
interactive forecasting method which relies on a panel of experts.
The experts answer questionnaires in two or more rounds. After each round, a facilitator(leader) analyses the
questionnaires and removes irrelevant information and provides an anonymised summary of the experts' forecasts
from the previous round as well as the reasons they provided for their judgments.
Experts are encouraged to revise their earlier answers considering replies of other members.
By this process, the range of answers decreases and this process continues till the group converges to few relevant
solutions.
Regression analysis
Regression analysis is a set of statistical processes for estimating the relationships among variables. It includes many
techniques for modelling and analysing several variables, when the focus is on the relationship between a dependent
variable and one or more independent variables.
Regression analysis helps one understand how the typical value of the dependent variable changes when any one of the
independent variables is varied, while the other independent variables are held fixed.
In this method usually, the data is fitted in a trend line and later the demand is for the future years can be predicted by
this trend line.
Time-series analysis is a technique that attempts to predict the future by using historical data.

4. Trend line
Assume trend line is a straight line of equation 𝒚 = 𝒎𝒙 + 𝒄
∑𝑦 = 𝑚∑𝑥 + 𝑐∑1
∑ 𝑦
𝑛
1
= 𝑚 ∑ 𝑥
𝑛
1
+ 𝑐 ∑ 1
𝑛
1
∑𝑦 = 𝑚∑𝑥 + 𝑐𝑛
From 1 & 2,
𝑚 =
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑥2 − (∑ 𝑥)2
𝑐 =
1
𝑛
(∑𝑦 − 𝑚∑𝑥)
Coefficient of correlation(r) =
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√𝑛 ∑ 𝑥2 − (∑ 𝑥)2√𝑛 ∑ 𝑦2 − (∑ 𝑦)2
Coefficient of correlation tells us how many % of data points are either on the trend line or in the close proximity.
If r=0.8, it means 80% of data points are in close proximity with the trend line and the line is upwards sloping.
Forecasting errors
Mean Absolute Deviation (𝐌𝐀𝐃) =
∑ |yi − Fi|n
i=1
n
Mean Sum of Square Errors (𝐌𝐒𝐄) =
∑ (yi − Fi)2n
i=1
n
Average Error (𝐁𝐢𝐚𝐬) =
∑ (yi − Fi)n
i=1
n
Tracking Signal (𝐓𝐒) =
∑ (yi − Fi)n
i=1
𝐌𝐀𝐃
Moving average method
In statistics, a moving average is a calculation to analyse data points by creating series of averages of different subsets
of the full data set.
It is used for medium range forecast and it can be used for maximum forecast up to next quarter.
A moving average is commonly used with time series data to smooth out (to reduce the changes in a process) short-
term fluctuations and highlight longer-term trends or cycles.
In this forecast is based on most recent values.
Simple Moving Average (SMA)
It is the unweighted mean of the previous n data.
An example of a simple equally weighted running mean for a n-day sample of closing price is the mean of the previous
n days' closing prices. If those prices are pn, pn-1, …. p1 then the formula is
p̅sm =
pn + pn−1 + ⋯ + p1
n
=
1
n
∑ pi
n
i=1
Weighted Moving Average (WMA)
It assumes that some data points are significant than others in generating future forecasts. A weighted average is an
average that has multiplying factors to give different weights to data at different positions in the sample window.
WMA =
n × pn + (n − 1) × pn−1 + ⋯ + 2 × p2 + p1
n + (n − 1) + ⋯ + 2 + 1
( 𝑛, (𝑛 − 1), … ,2,1 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑤𝑒𝑖𝑔ℎ𝑡𝑠 𝑜𝑓 pn, pn−1, … , p1)
1
2

5. Exponential Weighted Moving Average (EWMA)
This is a special case of weighted moving average.
The forecast for next period is computed as the weighted average of the immediate past data point and the forecast of
the previous period.
In other words, the previous forecast is adjusted based on deviation (forecast error) of that forecast from the actual
data.
Forecast of nth period can be calculated,
𝐅𝐧 = 𝐅𝐧−𝟏 + α(𝐲 𝐧−𝟏 − 𝐅𝐧−𝟏)
Forecast for nth period =
Forecast for (n-1)th period+Smoothening coefficient × (Actual demand for (n-1)th period ― Forecast for (n-1)th period)
𝐅𝐧 = α𝐲 𝐧−𝟏 + (1 − α)𝐅𝐧−𝟏 → 𝟏 𝐬𝐭
𝐨𝐫𝐝𝐞𝐫 𝐬𝐦𝐨𝐨𝐭𝐡𝐞𝐧𝐢𝐧𝐠 𝐜𝐮𝐫𝐯𝐞
𝐅𝐧−𝟏 = α𝐲 𝐧−𝟐 + (1 − α)𝐅𝐧−𝟐
𝐅𝐧 = α𝐲 𝐧−𝟏 + α(1 − α)𝐲 𝐧−𝟐 + (1 − α)2
𝐅𝐧−𝟐 → 𝟐 𝐧𝐝
𝐨𝐫𝐝𝐞𝐫 𝐬𝐦𝐨𝐨𝐭𝐡𝐞𝐧𝐢𝐧𝐠 𝐜𝐮𝐫𝐯𝐞
𝐅𝐧 = α𝐲 𝐧−𝟏 + α(1 − α)𝐲 𝐧−𝟐 + α(1 − α)2
𝐲 𝐧−𝟑 + ⋯ ⋯ ⋯ ⋯ ⋯ + (1 − α) 𝑛−1
𝐲 𝟏
Values of α closer to 1 have less amount of smoothening effect and give greater weight to recent changes in the
data(unstable), while values of α closer to zero have greater smoothening effect and are less responsive to recent
changes (stable).
Inventory
Types of Inventory costs
• Set-up cost
• Ordering cost
• Carrying cost
• Total inventory cost
• Shortage cost
• Purchase or production cost
• Selling cost
Total Inventory cost = Ordering Cost + Carrying Cost
When a small order is placed, ordering cost will be very high
because more number of orders must be placed per unit time but carrying cost will be very low. When the order
quantity is high no. of orders placed per unit time will decrease but carrying cost will be high.
There exists an optimum order level at which the total cost of inventory will be minimum, this is called Economic Order
Quantity.
Types of Inventories
• Transportation Inventories
• Buffer Inventories
• Anticipation Inventories
• Decoupling Inventories
• Lot―Size Inventories

6. Inventory models
Uncertain demand
The moment we place an order (for inventory), the time-period after which the order reaches to our hand is called lead
time. In this time-period, amount of inventory consumed by the workshop is called lead time consumption or reorder
point.
The level between maximum and minimum stock, which purchasing activities must start replenishing is Re-order level.
There are two types of uncertain models
Periodic Review (P system): Inventory is reviewed at (prefixed) periodic intervals irrespective of the levels to
which inventory drops; an order is placed to bring up the inventory to the maximum level. This is used for moderate
volume items.
Continuous Review (Fixed Quantity or Q system): Inventory is reviewed continuously and when
inventory drops to a certain (prefixed) reorder level, a fixed quantity is ordered. This model is generally used for high
volume, valuable, or important items.
Deterministic Model
Wilson Harris Model
In this inventory model orders of equal size are placed at periodical intervals. The items against an order are replenished
instantaneously and the items are consumed at a constant rate.
𝐂𝐚𝐫𝐫𝐲𝐢𝐧𝐠 𝐂𝐨𝐬𝐭 = Average inventory(Q) × Carrying Cost per unit item = [
Total Area
Time
] × Cc =
0.5QT
T
× Cc =
Q
2
× Cc
𝐎𝐫𝐝𝐞𝐫𝐢𝐧𝐠 𝐂𝐨𝐬𝐭 = No. of orders per unit time × Ordering Cost per unit Time (T) = N × CO
𝐓𝐨𝐭𝐚𝐥 𝐈𝐧𝐯𝐞𝐧𝐭𝐨𝐫𝐲 𝐂𝐨𝐬𝐭 (𝐓𝐈𝐂) = Carrying Cost + Ordering Cost =
Q
2
× Cc + N × CO
𝐓𝐈𝐂 =
𝟏
𝟐
𝐐𝐂 𝐂 + 𝐍𝐂 𝐎 =
𝟏
𝟐
𝐐𝐂 𝐂 +
𝐑
𝐐
𝐂 𝐎
d(TIC)
dQ
=
1
2
Cc −
R
Q2
CO = 0
Q = √
2RCO
Cc
Q is the optimum quantity at which total cost of inventory (TIC) will be minimal which is at Economic Order Quantity.
At EOQ, TIC =
1
2
QCC +
R
Q
CO = √
RCOCC
𝟐
+ √
RCOCC
𝟐
= √𝟐𝐑𝐂 𝐎 𝐂 𝐂
𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐓𝐨𝐭𝐚𝐥 𝐈𝐧𝐯𝐞𝐧𝐭𝐨𝐫𝐲 𝐂𝐨𝐬𝐭 = √𝟐𝐑𝐂 𝐎 𝐂 𝐂
(N =
1
T
, R =
Q
T
⇒
1
T
=
R
Q
, N =
1
T
=
R
Q
)

7. Production Model
In this model we assume an item that is produced in the factory and it is consumed either by assembly or by directly
giving it to the customer.
𝐊 = rate of production
𝐑 = rate of consumption
𝐊 − 𝐑 = rate of production
− consumption
K > R, so we don’t produce all the time,
we produce a time t1, and then only
consumption takes place for time t2.
So, for t1, the items are produced as well
as consumed. It is produced at the rate of
K consumed, at the rate of R. So,
inventory is built up at the rate of K―R.
𝐐
𝐭 𝟏
= 𝐊 &
𝐐′
𝐭 𝟏
= 𝐊 − 𝐑 &
𝐐
𝐓
= 𝐑 &
𝐐′
𝐭 𝟐
= 𝐑
Total Inventory Cost = Carrying Cost + Ordering Cost
𝐓𝐈𝐂 =
Q′ × Cc
2
+ N × CO
From previous equations,
Q
Q′
=
K
K − R
𝐓𝐈𝐂 =
1
2
(
K − R
K
) Q × Cc + N × CO
d(TIC)
dQ
=
1
2
(
K − R
K
) Cc −
R
Q2
CO = 0
Q = √
2KRCO
(K − R)Cc
At EOQ, TIC =
1
2
QCC +
R
Q
CO = √
RCOCC
𝟐
(
K − R
K
) + √
RCOCC
𝟐
(
K − R
K
) = √ 𝟐𝐑𝐂 𝐎 𝐂 𝐂 (
𝐊 − 𝐑
𝐊
)
𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐓𝐨𝐭𝐚𝐥 𝐈𝐧𝐯𝐞𝐧𝐭𝐨𝐫𝐲 𝐂𝐨𝐬𝐭 = √ 𝟐𝐑𝐂 𝐎 𝐂 𝐂 (
𝐊 − 𝐑
𝐊
)
Safety Stock
During lead times, when there are uncertainties in the demand
and the order may not be able to deliver within the lead time.
Safety stock is being provided as a backup.
( N =
1
T
=
R
Q
)

8. Probabilistic model
In probabilistic models, demands are described by probability distributions based on which inventory is decided.
Single-period inventory model with probabilistic demand
This term single period refers to the situation where the inventory is perishable (will get used) and demand for that
inventory exists only for the period at which it is ordered. Newspaper when ordered today has demand only today,
tomorrow it has less demand.
Increment analysis is used to determine the optimal order quantity for a single-period inventory model with
probabilistic demand. The increment analysis addresses the how-much-to order question by comparing the cost or loss
of ordering one additional unit with the cost or loss of not ordering one additional unit.
Co: Cost per unit of overestimating demand; represents the loss of ordering one additional unit that may not sell.
Cu: Cost per unit of underestimating demand; represents the loss of not ordering one additional unit or profit lose for
which demand existed otherwise.
Let the probability of the demand of inventory being
more than a certain level y is P (D > y), and the
probability of the demand of inventory being less
than or equal to this level y is P (D ≤ y).
Expected loss (EL) is given by either of the two
conditions
Overestimation ⇒ EL (y + 1) = Co × P (D ≤ y)
Underestimation ⇒ EL (y) = Cu × P (D > y)
Following which the optimal order quantity (y’) can be found as follows:
EL (y’ + 1) = EL (y’)
P(D ≤ y′) = Cu × P(D > y′)
It is known that
P(D > y′) = 1 − P(D ≤ y′)
Substituting it, we have
Co × P(D ≤ y′) = Cu × [1 − P(D ≤ y′)]
Solving for P (D ≤ y’), we have
Inventory Control
ABC Analysis [Pareto (80-20) rule]
In this analysis, the inventory items are classified based on their
usage in monetary terms. It is very common to observe that
usually a small number of items account for a large share of total
cost of materials and comparatively large number of items
involve an insignificant share.
Based on this criterion, items are divided into 3 categories,
A → High consumption value items
B → Moderate consumption value items
C → Low consumption value item
𝐏(𝐃 ≤ 𝐲′) =
𝐂 𝐮
𝐂 𝐮 + 𝐂 𝐨

9. VED Analysis [V-Vital, E-Essential, D-Desirable]
Inventories are classified according to the criticality of item and it is used in controlling the level of inventory of spare
parts.
XYZ Analysis
X―Highest inventory
Y―Medium inventory
Z―Lowest inventory
In this the inventories are classified according to its value in stores and it is primarily used in reviewing the inventory.
FNSD [F-Fast, N-Normal, S-Slow, D-Dead]
In this the inventories are classified according to the movement of items from the stores.
SDE [S-Scarce, D-Difficult, E-Easy]
In this the inventories are classified according to the availability of items.
HML [H-High, M-Medium, L-Low]
In this the inventories are classified according to the cost of items.

10. Linear programming
Optimization is a process of either maximizing or minimizing a specific quantity called objective which depends upon a
finite number of variables.
Graphical method
Graphical solution consists two steps:
1. Determination of the feasible solution space.
2. Determination of the optimum solution from among all the points in the solution space.
Example
𝐌𝐚𝐱𝐢𝐦𝐢𝐳𝐞 𝐙 = 𝟒𝟓𝐱 + 𝟒𝟎𝐲, 𝐒𝐮𝐛𝐣𝐞𝐜𝐭𝐞𝐝 𝐭𝐨 𝟐𝐱 + 𝐲 ≤ 𝟗𝟎, 𝐱 + 𝟐𝐲 ≤ 𝟖𝟎, 𝐱 + 𝐲 ≤ 𝟓𝟎, 𝐱, 𝐲 ≥ 𝟎
1. Plotting 2x + y ≤ 90, x + 2y ≤ 80, x + y ≤ 50, x, y ≥ 0 on the graph
2. Find the feasible solution (Solution set) in the graph
3. Mark the end points O, A, B, C, D.
4. Substitute those points (x, y) in Z=45x+40y.
5. The maximum value of Z gives the required (x, y).
Point Z=45x+40y
O (0,0) 0
A (0,40) 1600
B (20,30) 2100
C (40,10) 2200
D (45,0) 2025
‘C’ is the optimum point.
Direct method
To find the point of maxima directly we need to understand the
following theory
1. Initially the objective function line is drawn passing through the
origin, the moment the objective function line touches the feasible region
or the feasible point that is the point of minima.
2. The farthest point at which objective function line touching the
feasible region is the point of maxima.
Slope method
Slope of objective function(m) = −
45
40
= −1.125
Slope of first constraint(m1) = −2
Slope of second constraint(m2) = −
1
2
= −0.5
Slope of third constraint(m3) = −1
As m lies between m1 & m3, the optimum point will be the intersection of the line 1 & 3.

11. Simplex method
When here are more than 2 variables, graphical method cannot be used to solve optimization problems. As it was
explained in the previously, optimal solution exists always at the corner point of feasible region, the simplex method is a
systematic procedure at finding corner point solution and taking them for optimality.
Simplex procedure is for profit maximization and if our objective is loss minimization then the problems have to be
converted into profit maximization by multiplying the objective function by ‘-’ sign before starting simplex procedures.
Example
𝐌𝐚𝐱𝐢𝐦𝐢𝐳𝐞 𝐳 = 𝟏𝟐𝐱 𝟏 + 𝟏𝟔𝐱 𝟐 𝐬𝐮𝐛𝐣𝐞𝐜𝐭𝐞𝐝 𝐭𝐨 𝟏𝟎𝐱 𝟏 + 𝟐𝟎𝐱 𝟐 ≤ 𝟏𝟐𝟎, 𝟖𝐱 𝟏 + 𝟖𝐱 𝟐 ≤ 𝟖𝟎, 𝐱 𝟏, 𝐱 𝟐 ≥ 𝟎
We need to add slack variables to subjective functions and objective function,
Maximize z = 12x1 + 16x2 + 0 × s1 + 0 × s2
10x1 + 20x2 + s1 = 120
8x1 + 8x2 + s2 = 80
x1, x2, s1, s2 ≥ 0
Initial Simplex table
CBi
Cj 12 16 0 0
Solution RatioBasic
Variable
x1 x2 s1 s2
0 s1 10 20 1 0 120 120/20 6 least value
0 s2 8 8 0 1 80 80/8 10
Zj 0 0 0 0 0
Cj– Zj 12 16 0 0
Highest
value
Zj = ∑(CBi)
2
i=1
(aij) → Z1 = CB1 × x11 + CB2 × x12 = 0 × 10 + 0 × 8 = 0
Optimality Condition
For Maximization Problem → all Cj – Zj ≤ 0.
For Minimization Problem → all Cj – Zj ≥ 0.
1. Calculate Zj & Cj– Zj.
2. If Cj– Zj≤0, table is over but here all are positive. So, table should be modified.
3. Find out the key row & key column & key element.
4. Least value in ratio forms key row, Highest value in Cj– Zj forms key column, their intersection forms key
element (20).
5. Here s1 is the leaving variable and x2 is the entering variable.
6. In Modified table, divide all elements in key row with key element.
7. In modified table,
New value in non key row = Old Value −
Corresponding Key column value × Corresponding Key row value
Key Element
Modified Simplex table
CBi
Cj 12 16 0 0
Solution Ratio
B. V x1 x2 s1 s2
16 x2 1/2 1 1/20 0 6 6/0.5 12
0 s2 4 0 - 2/5 1 32 32/4 8 Least value
Zj 8 16 4/5 0 96
Cj– Zj 4 0 - 4/5 0
Highest
value

12. 1. Calculate Zj & Cj– Zj.
2. If Cj– Zj≤0, table is over but here all are positive. So, table should be modified.
3. Find out the key row & key column & key element (4).
4. Here s2 is the leaving variable and x1 is the entering variable.
5. In Re-Modified table, divide all elements with key element in key row.
6. In re-modified table,
New value in non key row = Old Value −
Corresponding Key column value × Corresponding Key row value
Key Element
Re-Modified Table
CBi
Cj 12 16 0 0
Solution
B. V x1 x2 s1 s2
16 x2 0 1 1/10 - 1/8 2
12 x1 1 0 - 1/10 1/4 8
Zj 12 16 2/5 1 128
Cj– Zj 0 0 - 2/5 -1 OPTIMUM
1. Calculate Zj & Cj– Zj.
2. Here Cj– Zj≤0. So, we get the final optimized solution from this table.
3. x1=8, x2=2, Z=128 is the final solution.
BIG M Method
It is a modified version of simplex method.
Example
𝐌𝐢𝐧𝐢𝐦𝐢𝐳𝐞 𝐳 = 𝟕𝒙 𝟏 + 𝟏𝟓𝒙 𝟐 + 𝟐𝟎𝒙 𝟑 𝐬𝐮𝐛𝐣𝐞𝐜𝐭𝐞𝐝 𝒕𝒐 𝟐𝒙 𝟏 + 𝟒𝒙 𝟐 + 𝟔𝒙 𝟑 ≥ 𝟐𝟒, 𝟑𝒙 𝟏 + 𝟗𝒙 𝟐 + 𝟔𝒙 𝟑 ≥ 𝟑𝟎, 𝒙 𝟏, 𝒙 𝟐, 𝒙 𝟑 ≥ 𝟎
Add slack variables to subjective and objective functions
Min z = 7x1 + 15x2 + 20x3 + 0 × 𝐬 𝟏 + 0 × 𝐬 𝟐 + M × 𝐀 𝟏 + M × 𝐀 𝟐
Subjected to,
2x1 + 4x2 + 6x3 − 𝐬 𝟏 + 𝐀 𝟏 = 24
3x1 + 9x2 + 6x3 − 𝐬 𝟐 + 𝐀 𝟐 = 30
𝐱 𝟏, 𝐱 𝟐, 𝐱 𝟑, 𝐬 𝟏, 𝐬 𝟐, 𝐀 𝟏, 𝐀 𝟐 ≥ 0
Initial Big-M Table
CBi
Cj 7 15 20 0 0 M M
Sol.n Ratio
B. V x1 x2 x3 s1 s2 A1 A2
R1 M A1 2 4 6 -1 0 1 0 24 24/4 6
R2 M A2 3 9 6 0 -1 0 1 30 30/9 3.33
Least
positive
Zj 5M 13M 12M –M –M M M 54M
Cj– Zj 7-5M 15-13M
20-
12M
–M –M 0 0
Most
Negative
Zj = ∑(CBi)
2
i=1
(aj) → Z1 = CB1 × x1 = M × 5 = 5M
1. Calculate Zj & Cj– Zj.
2. Optimal Condition for Minimization is Cj– Zj≥0.
3. Most Negative value in (Cj– Zj) forms Key column (x2).
4. Divide the solution values with key column value to get ratio.
5. Least value in ratio forms key row(A2).
6. Their intersection forms key element, 9 is the key element.
7. Here A2 is the leaving variable and x2 is the entering variable.
8. Now Calculate R3 & R4.

13. R4 =
Old value
key element
R3 = Old Value −
Corresponding Key column value × Corresponding Key row value
Key Element
R3 = Old Value − (Corresponding Key column value × R4)
Modified Big-M table
CBi
Cj 7 15 20 0 0 M M
Sol.n RatioB. V
x1 x2 x3 s1 s2 A1 A2
R3 M A1 2/3 0 10/3 -1 4/9 1 ― 32/3 32/10
Least
positive
R4 15 x2 1/3 1 2/3 0 -1/9 0 ― 10/3 5
Zj (2/3) M+5 15 (10/3) M+10 –M (4/9) M-(5/3) M ― (32/3) M+50
Cj– Zj (2/3) M+2 0 – (10/3) M+10 M – (4/9) M+(5/3) 0 ―
Most
Negative
1. Calculate Zj & Cj– Zj.
2. Most Negative value in (Cj– Zj) forms Key column(x3).
3. Divide the solution values with key column values to get ratio.
4. Least value in ratio, forms key row(A1).
5. Here 9 is the key element.
6. Here A1 is the leaving variable and x3 is the entering variable.
7. Now Calculate R5 & R6.
R6 =
Old value
key element
R5 = Old Value − (Corresponding Key column value × R6)
Re-Modified Big-M table
CBi
Cj 7 15 20 0 0 M M
Sol.n
B. V x1 x2 x3 s1 s2 A1 A2
R5 20 x3 1/5 0 1 -3/10 2/15 ― ― 16/5
R6 15 x2 1/5 1 0 1/5 -1/5 ― ― 6/5
Zj 7 15 20 -3 -1/3 ― ― 82
Cj– Zj 0 0 0 3 1/3 ― ―
1. Calculate Zj & Cj– Zj.
2. Cj– Zj≥0, So it is the optimal table.
𝐱 𝟏 = 𝟎, 𝐱 𝟐 =
𝟔
𝟓
, 𝐱 𝟑 =
𝟏𝟔
𝟓
& 𝐙 = 𝟖𝟐

14. Transportation Method
Transportation problems are meant to optimize loss or cost minimization. It is a special kind of LPP in which goods are
transported from a set of sources to a set of destinations subject to supply and demand of the sources and destination
respectively, such that the total cost of transportation is minimized.
Type 1 – Balanced Transportation Problem S=D.
Type 2 – Unbalanced Transportation Problem S≠D.
Methods: -
1. Finding the initial feasible solution
2. Finding optimization.
Example
Destination
A B C Supply
Source
1
2 7 5 200
2
3 4 2 300
3 5 4 7 500
Demand 200 400 400 1000
Supply = Demand
North West Corner Cell Method
Destination
A B C D Supply
Source
1
250
3
50
1 7 4 300 50 0
2 2
300
6
100
5 9 400 100 0
3 8 3
300
3
200
2 500 200 0
Demand 250 350 400 200 1200
0 300 300 0
0 0
1. In Northwest corner cell method, first we look for the North-west corner in the table.
2. We see whether supply or demand is greater than each other and fill the least value in northwest corner.
A B C D Supply
1
250
3
1
7 4 300 50
2 2 6 5 9 400
3 8 3 3 2 500
Demand 250 350 400 200 1200
0
3. Supply from 1 is reduced, as some of it is allocated to A.
4. We close the column or row, which is exhausted. Here it’s Demand for ‘A’.
5. We check for the new northwest corner in the new table and repeat the process.
NEW TABLE

15. A B C D Supply
1
250
3
50
1 7 4 300 50 0
2 2
300
6 5 9 400 100
3 8 3 3 2 500
Demand 250 350 400 200 1200
0 300
0
6. Allot the remaining Supply from 1 to Northwest corner (B).
7. Fill the remaining demand of B and close the column ‘B’ and close the row ‘1’ as supply is completed.
8. Check the northwest corner in new table.
A B C D Supply
1
250
3
50
1 7 4 300 50 0
2 2
300
6
100
5 9 400 100 0
3 8 3
300
3
200
2 500 200 0
Demand 250 350 400 200 1200
0 300 300 0
0 0
9. Fill the remaining supply from ‘2’ in northwest corner and close the row as supply from 2 is exhausted.
10. Fill the remaining demand from 3 in C.
11. Fill the final demand in D and complete the table.
𝑭𝒊𝒏𝒂𝒍 𝒄𝒐𝒔𝒕 𝒂𝒇𝒕𝒆𝒓 𝒂𝒍𝒍𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒖𝒓𝒄𝒆 𝒕𝒐 𝒅𝒆𝒔𝒕𝒊𝒏𝒂𝒕𝒊𝒐𝒏
= 250 × 3 + 50 × 1 + 300 × 6 + 100 × 5 + 300 × 3 + 200 × 2 = 4400
Least cost Method
In this method we will be first selecting the least subjective variable and allocating the supply to it completely and
repeating the process, till all the supply is allocated to meet the demand.
A B C D Supply
1 3
300
1 7 4 300 0
2 2 6 5 9 400
3 8 3 3 2 500
Demand 250 350 400 200 1200
50
1. Least value ‘1’ is selected and supply is allocated and the row is closed as supply from ‘1’ is exhausted.
2. From the remaining again select the least value.
3. Now allocate the supply to A until its demand.
4. Close the column as demand is meet.
5. Find the new least value in remaining table and repeat the process till supply meets the demand.
NEW TABLE
Least
value

16. A B C D Supply
1 3
300
1 7 4 300 0
2
250
2 6 5 9 400 150
3 8 3 3 2 500
250 350 400 200 1200
Demand 0 50
A B C D Supply
1 3
300
1 7 4 300 0
2
250
2 6 5 9 400 150
3 8 3 3
200
2 500
Demand 250 350 400 200 1200
0 50 0
A B C D Supply
1 3
300
1 7 4 300 0
2
250
2
50
6
100
5 9 400 150 0
3 8 3
300
3
200
2 500 300 0
250 350 400 200 1200
0 50 100 0
0 0
𝑭𝒊𝒏𝒂𝒍 𝒄𝒐𝒔𝒕 𝒂𝒇𝒕𝒆𝒓 𝒂𝒍𝒍𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒖𝒓𝒄𝒆 𝒕𝒐 𝒅𝒆𝒔𝒕𝒊𝒏𝒂𝒕𝒊𝒐𝒏
= 300 × 1 + 250 × 2 + 50 × 6 + 100 × 5 + 300 × 3 + 200 × 2 = 2900
Vogel’s Approximation
1. Find out the difference between the least value and its next least value in the row and column and write separately.
A B C D Supply
1
3 1 7 4 300 2
2
250
2 6 5 9 400 3
3
8 3 3 2 500 1
Demand 250 350 400 200 1200
1 2 2 2

17. 2. Find the greatest value and allocate to its least value in table completely until total supply or demand is
exhausted.
3. Close the column as demand is exhausted and continue the process.
A B C D Supply
1 3
300
1 7 4 300 2 3
2
250
2 6 5 5 150 3 0
3 8 3 3 2 500 1 1
Demand 250 350 400 200 1200
1 2 2 2
― 2 2 2
A B C D Supply
1 3
300
1 7 4 300 2 3 ― ―
2
250
2 6
150
5 5 150 3 0 0 1
3 8
50
3
250
3
200
2 500 1 1 1 0
Demand 250 50 400 200 1200
1 2 2 2
― 2 2 2
― 3 2 7
― 3 2 ―
Continue the process till all the demand is supplied.
𝑭𝒊𝒏𝒂𝒍 𝒄𝒐𝒔𝒕 𝒂𝒇𝒕𝒆𝒓 𝒂𝒍𝒍𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒖𝒓𝒄𝒆 𝒕𝒐 𝒅𝒆𝒔𝒕𝒊𝒏𝒂𝒕𝒊𝒐𝒏
= 250 × 2 + 300 × 1 + 150 × 5 + 250 × 3 + 200 × 3 + 50 × 3 = 2850
𝑃𝑖𝑗 = 𝑢𝑖 + 𝑣𝑗 − 𝐶𝑖𝑗
MODI METHOD
The solution we get from previous methods is not optimal. So, we need to optimize it. We will be optimizing it by modi
(u-v method).
1. First, we get the solution by Northwest corner method.
A B C D Supply
1
250
3
50
1
7
4
300
2
2
300
6
100
5
5
400
3
8 3
250
3
150
2 500
Demand 250 350 400 200 1200
2. After Solving we need to find u1 & v1.

18. 3. Keep u1 as zero (0) and find remaining u & v by the formula ui + vi=Cij.
v1 v2 v3 v4
3 1 0 -1
u1 0 200
3 50
1 7 4
u2 5 2 250
6 100
5 5
u3 3 8 3
250
3
150
2
4. Then find Penalties(Pij) for un-allocated cells, 𝐏𝐢𝐣 = 𝐮𝐢 + 𝐯𝐣 – 𝐂𝐢𝐣
P13 0+0-7 -7
P14 0-1-4 -5
P21 5+3-2 6
P24 5-1-9 -5
P31 3+3-8 -8
P32 3+1-3 1
5. If all the penalties are less than zero, then table is optimized.
6. Out of penalties, find out the highest positive one and check out the box in table. Here it is 2×1.
7. Now form a closed loop, loop from checked box (2×1) consisting only allocated cells.
8. Give the selected box a positive value and alternatively give negative and positive value.
v1 v2 v3 v4
3 1 0 -1
u1 0
200
3
50 1
7 4(―) (+)
u2 5 2
300
6
100
5 9(+) (―)
u3 3 8 3
250
3
150
2
9. After closing the loop, check out the negative cells.
10. Select the least value and add it to positive cells and subtract from negative cells.
11. Here it is 200(C11), add 200 to C12 & C21 and subtract it from C22.
12. We will get a new table, check whether No. of rows + No. of columns –1 = No. of allocated cells.
13. If condition is satisfied repeat the process.
14. We will get new ui & vj & Cij.
v1 v2 v3 v4
-3 1 0 -1
u1 0 3
250
1 7 4
u2 5
200
2
100
6
100
5 9
u3 3 8 3
250
3
150
2
15. Again, find Penalties(Pij) for un-allocated cells, 𝐏𝐢𝐣 = 𝐮𝐢 + 𝐯𝐣 – 𝐂𝐢𝐣.
16. Out of penalties, find out the highest positive one and check out the box in table. Here it is 3×2.

19. P11 0-3-3 -6
P13 0+0-7 -7
P14 0-1-4 -5
P24 5-1-9 -5
P31 3+3-8 -8
P32 3+1-3 1
17. Now form a closed loop, loop from checked box (3×2) consisting only allocated cells.
18. Select the least value in negative boxes and add it to positive cells and subtract from negative cells.
19. Here it is 50(C22), add 50 to C23 & C32 and subtract it from C33.
v1 v2 v3 v4
-3 1 0 -1
u1 0 3
250
1 7 4
u2 5
200
2
50
6
100
5 9(―) (+)
u3 3 8 3
250
3
150
2(+) (―)
20. We will get a new table
v1 v2 v3 v4
-3 1 0 -1
u1 0 3
250
1 7 4
u2 5
200
2 6
150
5 9
u3 3 8
50
3
200
3
150
2
21. Check whether No. of rows + No. of columns –1 = No. of allocated cells, (3+4-1=7).
22. If condition is satisfied repeat the process.
23. We will get new ui & vj & Cij.
v1 v2 v3 v4
-2 1 1 0
u1 0 3
250
1 7 4
u2 4
200
2 6
150
5 9
u3 2
8
50
3
200
3
150
2
24. Then find Penalties(Pij) for un-allocated cells, 𝐏𝐢𝐣 = 𝐮𝐢 + 𝐯𝐣 – 𝐂𝐢𝐣
P11 0-2-3 -5
P13 0+1-7 -6
P14 0+0-4 -4
P22 4+1-6 -1
P24 4+0-9 -5
P31 2-2-8 -8
25. Here all the penalties are negative, so table is optimized.
𝑭𝒊𝒏𝒂𝒍 𝑶𝒑𝒕𝒊𝒎𝒂𝒍 𝒄𝒐𝒔𝒕 𝒂𝒇𝒕𝒆𝒓 𝒂𝒍𝒍𝒐𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒖𝒓𝒄𝒆 𝒕𝒐 𝒅𝒆𝒔𝒕𝒊𝒏𝒂𝒕𝒊𝒐𝒏
= 250 × 1 + 200 × 2 + 150 × 5 + 50 × 3 + 200 × 3 + 150 × 2 = 2450

20. Project Management
 A project consists of interrelated activities which are to be executed in a certain order before the entire task is
completed.
 The activities are interrelated in a logical sequence, which is known as precedence relationship.
 Project is represented in the form of a network for the purpose of analytical treatment to get solutions for
scheduling and controlling its activities.
Phases of Project Management
Planning
• Dividing the project into distinct activities.
• Estimating time requirement for each activity.
• Establishing precedence relationship among the activities.
• Construction of the arrow diagram (network).
Scheduling
• Determines the start and end time of each and every activity.
Controlling
• Uses the arrow diagram and time chart for continuous monitoring and progress reporting.
Guidelines for network construction
1. The starting event and ending event of an activity are called tail event and head event, respectively.
2. The network should have unique starting node (Tail event).
3. The network should have a unique completion node (Head event).
4. No activity should be represented by more than one arc.
5. No two activities should have the same starting node and the same ending node.
6. Dummy activity is an imaginary activity indicating precedence relationship only. Duration of a dummy activity is
zero.

21. Critical Path method
Activity Immediate Predecessor(s) Duration (months)
A ― 2
B ― 5
C ― 4
D B 5
E A 7
F A 3
G B 3
H C, D 6
I C, D 2
J E 5
K F, G, H 4
L F, G, H 3
M I 12
N J, K 8
CPM Network
Finding out critical path
1. Determine the earliest start times (ES) of all the nodes. This is called forward pass.
2. Determine latest completion times (LC) of various nodes. This is called backward pass.
𝐂𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧 𝐟𝐨𝐫 𝐜𝐫𝐢𝐭𝐢𝐜𝐚𝐥 𝐩𝐚𝐭𝐡 → 𝐄𝐒𝐢(𝐬𝐭𝐚𝐫𝐭) = 𝐋𝐂𝐢(𝐬𝐭𝐚𝐫𝐭) & 𝐄𝐒𝐣(𝐞𝐧𝐝) = 𝐋𝐂𝐣(𝐞𝐧𝐝)
𝐄𝐒𝐣(𝐞𝐧𝐝) − 𝐄𝐒𝐢(𝐬𝐭𝐚𝐫𝐭) = 𝐋𝐂𝐣(𝐞𝐧𝐝) − 𝐋𝐂𝐢(𝐬𝐭𝐚𝐫𝐭) = 𝐃𝐢𝐣 (𝐃𝐮𝐫𝐚𝐭𝐢𝐨𝐧)

22. Total Floats
It is amount of time that the completion time of an activity can be delayed without affecting the project completion time.
𝐓𝐅𝐢𝐣 = 𝐋𝐂𝐣(𝐞𝐧𝐝) − 𝐄𝐒𝐢(𝐬𝐭𝐚𝐫𝐭) − 𝐃𝐢𝐣(𝐃𝐮𝐫𝐚𝐭𝐢𝐨𝐧)
Free Floats
It is the amount of time that the activity completion time can be delayed without affecting the earliest start time of
immediate successor activities in the network.
𝐅𝐅𝐢𝐣 = 𝐄𝐒𝐣(𝐞𝐧𝐝) − 𝐄𝐒𝐢(𝐬𝐭𝐚𝐫𝐭) − 𝐃𝐢𝐣(𝐃𝐮𝐫𝐚𝐭𝐢𝐨𝐧)
Summary of Total Floats and free Floats
Activity Duration (months) Total Float (TFij) Free Floats (FFij)
A 2 6 0
B 5 0 0
C 4 6 6
D 5 0 0
E 7 6 0
F 3 11 11
G 3 8 8
H 6 0 0
I 2 4 0
J 5 6 6
K 4 0 0
L 3 9 9
M 12 12 4
N 8 0 0
Any critical activity will have zero total float & zero free float, based on this property we can determine critical activities.
𝐂𝐫𝐢𝐭𝐢𝐜𝐚𝐥 𝐩𝐚𝐭𝐡 → 𝐁 − 𝐃 − 𝐇 − 𝐊 − 𝐍
Project Evaluation and Review Technique (PERT)
In CPM, activities are deterministic in nature (activities have fixed duration).
In PERT, each activity will have 3-time estimates.
1. Optimistic time (o)
2. Most likely time (m)
3. Pessimistic time(p)
Activity Immediate Predecessor(s)
Duration (weeks)
o m p
A ― 5 6 7
B ― 1 3 5
C ― 1 4 7
D A 1 2 3
E B 1 2 9
F C 1 5 9
G C 2 2 8
H E, F 4 4 10
I D 2 5 8
J H, G 2 2 8

23. Project Network
Expected Duration and Variance of each activity
Mean duration = Expected Duration (te) =
to + 4tm + tp
6
Variance = σ2
= [
tp − to
6
]
2
Activity
Duration (weeks)
Expected Duration Variance
o (to) m (tm) p (tp)
A 5 6 7 6 0.11
B 1 3 5 3 0.44
C 1 4 7 4 1.00
D 1 2 3 2 0.11
E 1 2 9 3 1.78
F 1 5 9 5 1.78
G 2 2 8 3 1.00
H 4 4 10 5 1.00
I 2 5 8 5 1.00
J 2 2 8 3 1.00
Critical path
We take the expected duration as duration of each activity

24. Probability of Completion
Probability of completing the project on or before 22 weeks
The PERT follows normal distribution.
Activity Mean Duration Variance(σ2)
C 4 1.00
F 5 1.78
H 5 1.00
J 3 1.00
TOTAL (C+F+H+J) 17 4.78
𝜎 = √4.78 = 2.19 Weeks
𝑃(𝑥 ≤ 22) = 𝑃 [
𝑥 − 𝜇
𝜎
≤
22 − 17
2.19
] = 𝑃(𝑧 ≤ 2.28)
𝑃(𝑥 ≤ 22) = 0.9887
This value is obtained from standard normal distribution table. Therefore, the probability of completing the project on or
before 22 weeks is 0.9887.
Queuing Theory
A queuing system is a set of customers, set of servers, and an order whereby customers arrive and are processed.
Queue Length (Lq) No. of customers waiting in the line at any moment.
Lq = ∑(n − c)Pn
∞
n=c
System Length (Ls) The average number of customers in the system.
Ls = ∑ nPn
∞
n=0
Ls − Lq ≠ 1, as there is idle of time in service
Idle Period When all the units in the queue are served, the idle period of the server begins and it continues up to the
arrival of unit. This period of a server is the time during which he remains free because there is no customer present in
the system.
Average length of line It is defined by the no. of customers in the queue per unit time.
Waiting time It is the time up to which a unit has to wait in the queue before it is taken into service.
Service time The time taken for servicing of a unit is called as service time.
Busy period Busy period of a server is the time during which he remains busy in servicing. Thus it is the time between
the start of a service of the first unit to the end of service of the last unit in the queue.
Mean Arrival rate (λ) The mean arrival rate in a waiting line is defined as the expected no. of arrivals/unit time. Arrival
rate follows Possion distribution.
Mean Service rate (μ)The mean servicing rate for a particular servicing station is defined as the expected no. of
services completed in a time of length unity, given that servicing is going in throughout the entire time limit. It follows
exponential distribution.
c=no. of customers
Pn=Probability that there are n
people in the system

25. Percentage utilization or Traffic intensity (ρ) it is the ratio of its men arrival rate and mean servicing rate.
ρ =
Mean arrival rate
Mean service rate
=
λ
μ
The state of a system
It involves the study of a system’s behavior over time.
Transient state
A system is said to be in transient state when its operating characteristics are dependent on time. Thus, a Queuing
system is said to be in transient state when the probability distribution of arrivals, waiting time and servicing time of the
customers are dependent on time. This state occurs at the beginning of operations of the system.
Steady state
A system is said to be in steady state when its operating characteristics become independent of time. This state occurs in
the long run of the system.
Let Pn (t) denote the probability that there are n units in the system at time t then the system acquires steady state as
t→∞ if lim
n→∞
Pn(t) = Pn (independent of t).
Explosive state
If the arrival rate of the system is more than its servicing rate, the length of the queue will go increasing with time and
will tend to ∞ at t→∞. This state of the system is said to be explosive state.
Formulae
→ Pn(Probability that n customers will arrive) = ρn(Service utilization) 𝑛
× Po(Probability that system is idle)
→ Pn = ρn
Po = (
λ
μ
)
n
Po
→ ∑ Pn
∞
n=0
= 1 ⇒ 𝐏𝐨 = 𝟏 − 𝛒
→ (Length Of System) 𝐋 𝐬 = ∑ nPn
∞
n=0
=
𝛒
𝟏 − 𝛒
→ (Length Of Queue) 𝐋 𝐪 = ∑(n − c)Pn
∞
n=c
=
𝛒 𝟐
𝟏 − 𝛒
→ 𝐋 𝐬 − 𝐋 𝐪 = 𝛒
→ (Waiting time in system)𝐖𝐬 =
𝐋 𝐬
𝛌
=
𝛒
(𝟏 − 𝛒)𝝀
=
𝟏
𝝁 − 𝝀
→ (Waiting time in Queue)𝐖𝐪 =
𝐋 𝐪
𝛌
=
𝛒 𝟐
(𝟏 − 𝛒)𝝀
=
𝝀
(𝝁 − 𝝀)𝝁
→ Probability that the arrival has to wait in the system for service, more than unit time (𝐭) = 𝑒
−
t
ws
→ Probabilty that arrival has to wait in the queue for more than time (𝐭) = ρ𝑒
−
t
ws
→ Average length of non– empty queues
⇒ It is the system length by neglecting the probability when there is nobody in the system =
𝐋 𝐬
𝟏 − 𝐏𝐨
Ls =
∑ n × Pn
∞
n=0
∑ Pn
∞
n=0 (= 𝟏)
= ∑ n × Pn
∞
n=0
= 0. P0 + 1. P1 + 2. P3 + ⋯ + ∞

26. Multiple Services Single queue
→ 𝑃𝑛 = {
𝜌 𝑛
𝑛!
𝑃𝑜 … … … 𝑖𝑓 (0 ≤ 𝑛 ≤ 𝑐)
𝜌 𝑛
𝑐 𝑛−𝑐 × 𝑐!
𝑃𝑜 … 𝑖𝑓 (𝑛 ≥ 𝑐)
→ 𝐿 𝑠 = ∑ 𝑛𝑃𝑛
∞
𝑛=0
→ 𝐿 𝑞 = ∑(𝑛 − 𝑐)𝑃𝑛
∞
𝑛=0
→ 𝑊𝑠 =
𝐿 𝑠
𝜆
→ 𝑊𝑞 =
𝐿 𝑞
𝜆