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Notes blog

Damped free Vibration-Numerical 1

A mass of 85 kg is supported on a spring which deflects 18 mm under the weight of the mass. The vibrations of the mass are constrained to be linear and vertical. A dashpot is provided which reduces the amplitude to one-quarter of its initial value in two complete oscillations. Calculate magnitude of the damping force at unit velocity and periodic time of damped vibration.


fig_damped-free-vibration-numerical-1 Fig-1

Given data:

mass  m=85\ m = 85 kg
Static deflection,  x=18 mm=0.018 m\ x = 18\ \text{mm} = 0.018\ \text{m}

x0=x_0 = initial amplitude
x2=x_2= final amplitude after two complete cycle =14= {1\over4}

Taking ratio,

x0x2=x014x0=4{x_0 \over x_2}={x_0 \over {1 \over 4} x_0}=4

Damping force:

F=c.x˚=c\text F = c.\mathring x= c


ξ=cccORξ=c2mω0(ascc=2mωn) c=(2mωn)ξ  (1)\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad &\xi= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore \ &c=(2m\omega_n)\xi \ \dots\dots\dots\dots \ (1) \end{aligned}

1. The Logarithmic decrement:

δ=1nloge(x0xn)=12loge(x0x2)=12loge(4)\begin{aligned} \delta &= {1 \over n} log_e\bigg ({x_0 \over x_n}\bigg)\\ &= {1 \over 2} log_e\bigg ({x_0 \over x_2}\bigg)\\ &= {1 \over 2} log_e(4) \end{aligned}

2. Damping factor:

ξ=δ4π2+δ2=δ4π2+δ2=0.1091\begin{aligned} \xi &= {\delta \over \sqrt{4 \pi^2 + {\delta}^2} } \\ &={\delta \over \sqrt{4 \pi^2 + {\delta}^2} }\\ &= 0.1091 \end{aligned}

3. Frequency of undamped free vibration:

The natural circular frequency of vibration is,

ωn=k m=m g x m=gx=9.810.018=23.34 rad/s\begin{aligned} \omega_n &= {\sqrt{k \over \ m}} \\ &= {\sqrt{m \ g \over \ x \ m}}\\ &= {\sqrt{g \over x}} \\ &= {\sqrt{9.81 \over 0.018}}\\ &= 23.34 \ \text {rad/s} \end{aligned}

From equation (1); we get,
Damping force at unit velocity;

F=c=(2mωn)ξ=(2 x 85 x 23.34) x (0.1091)=91.67 N\begin{aligned} \text F = c &= (2m\omega_n)\xi \\ &= (2 \ \text x \ 85\ \text x \ 23.34)\ \text x \ (0.1091) \\ &= 91.67 \ \text N \end{aligned}

The time period of damped vibration is,

tp=2πωd=2πωn 1ξ2=2π(23.34) 1(0.1091)2=0.2708 sec\begin{aligned} t_p &= {2\pi \over \omega_d} \\ &= {2\pi \over \omega_n \ \sqrt {1 -\xi^2}} \\ &= {2\pi \over (23.34) \ \sqrt {1 -(0.1091)^2}} \\ &=0.2708 \ \text {sec} \end{aligned}

Logarithmic Decrement (​δ)

fig_logarithmic-decrement fig-1: displacement V/s time curve for under damped system

Logarithmic decrement

Logarithmic decrement is defined as the natural logarithm of the ratio of successive amplitude on the same side of mean position.

The rate of decay in the amplitudes of under-damped system is measured by the parameter known as logarithmic decrement.

Rate of decay in amplitudes depends on the amount of damping present in the system. So if the damping is more, then the rate of decay will also be more.

Let A and B are the two points on the successive cycles which shows maximum deflection as shown in figure.

The periodic time:

tp=t2t1=2πωd=2π(1ξ2) ωn\begin{aligned} t_p &= t_2 − t_1 \\ &= {2 \pi \over \omega_d} \\ &= {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \end{aligned}

The amplitude at time t1t_1 and t2t_2 are:

x1=Xeξωnt1[sin(ωdt1+)]x_1 = Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)]


x2=Xeξωnt2[sin(ωdt2+)]x2=Xeξωn(t1+tp)[sin{ωd(t1+tp)+}]x2=Xeξωn(t1+tp)[sin(ωdt1+ωdtp+)]x2=Xeξωn(t1+tp)[sin(ωdt1+ωd(2πωd)+)]x2=Xeξωn(t1+tp)[sin(ωdt1+2π+)]x2=Xeξωn(t1+tp)[sin{2π+(ωdt1+)}]x2=Xeξωn(t1+tp)[sin(ωdt1+)]\begin{aligned} x_2 &= Xe^{-\xi \omega_n t_2} [sin(\omega_d t_2 + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ \omega_d (t_1 + t_p) + \varnothing \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d t_p + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d \bigg({2 \pi \over \omega_d}\bigg) + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + 2 \pi + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ 2 \pi + (\omega_d t_1 + \varnothing) \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] \\ \end{aligned}

Taking ratio, we get;

x1x2=Xeξωnt1[sin(ωdt1+)]Xeξωn(t1+tp)[sin(ωdt1+)]x1x2=eξωn(t1t1tp)x1x2=eξωntp\begin{aligned} {x_1 \over x_2} &= {Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)] \over Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] }\\ \therefore \quad \quad {x_1 \over x_2} &= e^{-\xi \omega_n (t_1-t_1-t_p)} \\ \therefore \quad \quad {x_1 \over x_2} &= e^{\xi \omega_n t_p} \end{aligned}

The logarithmic decrement is given by;

δ=loge(x1x2)δ=loge(eξωntp)δ=ξωntpδ=ξωn2π(1ξ2) ωnδ=2πξ(1ξ2)\begin{aligned} \delta &= log_e \bigg({x_1 \over x_2}\bigg) \\ \therefore \quad \delta &= log_e (e^{\xi \omega_n t_p}) \\ \therefore \quad \delta &= \xi \omega_n t_p \\ \therefore \quad \delta &= \xi \omega_n {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \\ \therefore \quad \delta &= {2 \pi \xi\over \big(\sqrt {1-\xi^2}\big)} \end{aligned}

The logarithmic decrement can also be determined as follows;

δ=loge(x0x1)=loge(x1x2)=loge(x2x3)==loge(xn1xn)Adding upto ’n’ termsnδ=loge(x0x1)+loge(x1x2)+loge(x2x3)++loge(xn1xn)  nδ=loge(x0x1 .x1x2 .x2x3 . .xn1xn)Ornδ=loge(x0xn)δ=1nloge(x0xn)\begin{aligned} \delta &= log_e \bigg({x_0 \over x_1}\bigg)= log_e \bigg({x_1 \over x_2}\bigg)= log_e \bigg({x_2 \over x_3}\bigg) = \dotso = log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \text {Adding upto 'n' terms}\\ n\delta &= log_e \bigg({x_0 \over x_1} \bigg) + log_e \bigg({x_1 \over x_2}\bigg) + log_e \bigg({x_2 \over x_3}\bigg) + \dotso + log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \therefore \ \ n\delta &= log_e \bigg({x_0 \over x_1} \ . {x_1 \over x_2} \ . {x_2 \over x_3} \ . \dots \ . {x_{n-1} \over x_n}\bigg) \\ \text {Or} \qquad \\ n\delta &= log_e \bigg({x_0 \over x_n}\bigg) \\ \therefore \quad \delta &= {1 \over n} log_e \bigg({x_0 \over x_n}\bigg) \end{aligned}

x0x_0 = amplitude at the starting position
xnx_n = amplitude after ‘n’ cycles

Under-Damped System (​ξ < 1)

Fig_1_under-damped-system Fig-1

Underdamped system  (ξ<1)\ (\xi < 1)

If the damping factor  ξ\ \xi is less than one or the damping coefficient cc is less than critical damping coefficient ccc_c, then the system is said to be an under-damped system.

ξ<1ORccc<1    c<cc\xi < 1 \quad \text OR \quad {c \over c_c} < 1\implies c < c_c
  • We know that roots of differential equations are:
S1=[ξ+ξ21]ωnS2=[ξξ21]ωnS_1 = \big [-\xi + \sqrt{\xi^2 -1} \big] \omega_n \\ S_2 = \big [-\xi - \sqrt{\xi^2 -1} \big] \omega_n
  • But for  ξ<1\ \xi < 1; the roots for under-damped system are given by S1S_1 and S2S_2 as below:
S1=[ξ+i(1ξ2)]ωnS2=[ξi(1ξ2)]ωnWhere i=1 is the imaginary unit of complex root\begin{aligned} S_1 = & \big [-\xi + i \sqrt{(1 - \xi^2)}] \omega_n \\ S_2 = & \big [-\xi - i \sqrt{(1 - \xi^2)}] \omega_n \\ &\text Where \ i = \sqrt{-1} \ \text {is the imaginary unit of complex root} \end{aligned}
  • The roots are complex and negative, so the solution of differential equation is given by
x=AeS1t+BeS2tx=Ae[ξ+i(1ξ2)]ωnt+Be[ξi(1ξ2)]ωntx=A.eξωnt.ei(1ξ2)ωnt+B.eξωnt.ei(1ξ2)ωntx=eξωnt[A.ei(1ξ2)ωnt+B.ei(1ξ2)ωnt]x=eξωnt[A.eiωdt+B.eiωdt][(1ξ2)ωn=ωd]\begin{aligned} x &= Ae^{S_1t} + Be^{S_2t} \\ \therefore x &= Ae^{[-\xi + i \sqrt{(1 - \xi^2)}] \omega_nt} + Be^{[-\xi - i \sqrt{(1 - \xi^2)}] \omega_nt} \\ \therefore x &= A.e^{-\xi\omega_nt}.e^{i \sqrt{(1 - \xi^2)} \omega_nt} + B.e^{-\xi\omega_nt}.e^{- i \sqrt{(1 - \xi^2)} \omega_nt} \\ \therefore x &= e^{-\xi\omega_nt} \big [A.e^{i \sqrt{(1 - \xi^2)} \omega_nt} + B.e^{- i \sqrt{(1 - \xi^2)} \omega_nt}] \\ \therefore x &= e^{-\xi\omega_nt} \big [A.e^{i\omega_dt} + B.e^{- i\omega_dt}] \quad \big [ \because \sqrt {(1 - \xi^2) \omega_n} = \omega_d] \end{aligned}
  • According to Euler’s theorem, above equation can be written as:
x=Xeξωnt[sin(ωdt+)]Where  Xand   are constantsx = Xe^{-\xi \omega_nt} \big [\sin (\omega_dt + \varnothing)] \\ \text Where \ \ X \text and \ \ \varnothing \text { are constants}
  • Above equation shows the equation of motion for an underdamped system, and the amplitude reduces gradually and finally becomes zero after some time.

  • Amplitude decreases by  X.eξωnt\ X.e^{-\xi \omega_nt}

  • The natural angular frequency of damped free vibrations is given by:

ωd=(1ξ2)ωn\omega_d = \sqrt {(1 - \xi^2)}\omega_n
  • Time period for under-damped vibration is given by:
tp=2πωd=2π(1ξ2)ωnsec.t_p = {2\pi \over \omega_d} = {2\pi \over \sqrt{(1 - \xi^2)}\omega_n} sec.

Critically Damped System (ξ = 1)

critically-damped-system-graph Fig.1(Critically Damped System)

  • Critically damped system(ξ=1): If the damping factor ξ is equal to one, or the damping coefficient c is equal to critical damping coefficient "cc", then the system is said to be a critically damped system.
ξ=1ORccc=1    c=cc\xi=1 \quad \text OR \quad {c \over c_c} = 1\implies c = c_c
  • Two roots for critically damped system are given by S1 and S2 as below:
S1=[ξ+ξ21]ωnS2=[ξξ21]ωnS_1 = \big [-\xi + \sqrt{\xi^2 -1} \big] \omega_n \\ S_2 = \big [-\xi - \sqrt{\xi^2 -1} \big] \omega_n
  • For ξ=1ξ=1; S1=S2=ωnS_1 = S_2 = -\omega_n
    Here both the roots are real and equal, so the solution to the differential equation can be given by
x=(A+Bt)eωnt...(1)x = (A + Bt)e^{-\omega_n t} \quad \quad ...(1)
  • Now differentiating equation (1) with respect to ‘t’, we get:
x˚=Beωntωn(A+Bt)eωnt...(2)\mathring x = Be^{-\omega_nt} - \omega_n(A + Bt)e^{-\omega_nt} \quad \quad ...(2)
  • Now, let at
    t=0t = 0 : x=X0x = X_0
    t=0t = 0 : x˚=0\mathring x =0

  • Substituting these values in equation (1):

X0=A...(3)X_0 = A \quad ...(3)
  • Same way, from equation (2), we get
0=Bωn(A+0)0=BωnAB=ωnAB=ωnX0...(4)\begin{aligned} 0&= B - \omega_n(A + 0) \\ 0&= B - \omega_nA \\ B&= \omega_nA \\ B&= \omega_nX_0 \qquad ...(4) \end{aligned}
  • Now putting the values of A and B in equation (1), we get:
x=(X0+ωnX0t)eωntx=X0(1+ωnt)eωnt...(5)\begin{aligned} x&= (X_0 + \omega_nX_0t)e^{-\omega_nt} \\ x&= X_0(1 + \omega_nt)e^{-\omega_nt} \qquad ...(5) \end{aligned}


  • From above equation (5), it is seen that as time t increases, the displacement x decreases exponentially.

  • The motion of a critically damped system is aperiodic (aperiodic motion motions are those motions in which the motion does not repeat after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.

  • For critically damped systems, if a system is displaced from its initial position, it will try to reach its mean position in a very short time.

  • Critically damped systems are generally seen in hydraulic doors closer as it is necessary for the door to come to its initial position in a very short time.

Basics of Program Evaluation and Review Technique (PERT)

PERT is the technique used to find project completion time of “variable activities”. In PERT, the time is combination of three different time estimations. Following are the three different time estimation:

  1. The optimistic time estimate (to): The minimum time required for the completion of the activity as per the predetermined condition.
  2. The pessimistic time estimate (tp): The maximum time that activity will take under worst condition.
  3. The most likely time estimate (tm): The time an activity will take if executed under normal condition.

Important terms in PERT analysis:

  1. Expected time or average time (te): Since there are three time values available in PERT, average time is to be calculate by following formula:
te=to+4tm+tp6t_{e} = {t_o + 4t_m + t_p \over 6}
  1. Variance (V): Variance is given by following formula:
V=(tpto6)2V = \Big({t_p - t_o \over 6}\Big)^2
  1. Standard Deviation (σ\sigma): Standard deviation is square root of summation of variance of critical activities. It is given by following formula:
σ=V1+V2+V3\sigma = \sqrt{V_1 + V_2 + V_3}
  1. Probability of completion of project (z): It is calculated in order to estimate that how many percentages are the chances of completion of project in certain time or given time (t). Let, tcp ​be the time of completion of project on critical path and t is any certain time or given time, then probability of completion of project in that given time t, is given by:
Z=ttcpσZ = {t - t_{cp} \over \sigma}

Numerical on PERT (Program Evaluation and Review Technique)

For the given activities determine:

  1. Critical path using PERT.
  2. Calculate variance and standard deviation for each activity.
  3. Calculate the probability of completing the project in 26 days.
Activity t o t m t p
1-2 6 9 12
1-3 3 4 11
2-4 2 5 14
3-4 4 6 8
3-5 1 1.5 5
2-6 5 6 7
4-6 7 8 15
5-6 1 2 3

Solution: First of all draw the network diagram for given data as shown below:


Here the time for completion of activities are probabilistic. So, using given values of time we will find the expected time to completion the activities and variance.

Expected time te=to+4tm+tp6\text {Expected time } \quad t_e = {t_o + 4t_m + t_p \over 6} Variance V=(tpto6)2\text {Variance } V = \Big({t_p - t_o \over 6}\Big)^2

For each given activity we will calculate the expected time as follows:

Activity t o t m t p te=to+4tm+tp6t_e = {t_o + 4t_m + t_p \over 6} Variance (V)
1-2 6 9 12 6+4×9+126=9{6+4 \times 9+12 \over 6} = 9 (1266)2=1.000({12-6 \over 6})^2 = 1.000
1-3 3 4 11 5 1.778
2-4 2 5 14 6 4.000
3-4 4 6 8 6 0.444
3-5 1 1.5 5 2 0.444
2-6 5 6 7 6 0.111
4-6 7 8 15 9 1.778
5-6 1 2 3 2 0.111

Now based on estimate time, we calculate the EST, EFT, LST and LFT for each activity to find out critical path of project as shown below. (Click here to know about calculation of EST, EFT, LST and LFT from CPM numerical)

Activity Duration EST EFT LST LFT Total Float
1-2 9 0 9 0 9 0
1-3 5 0 5 4 9 4
2-4 6 9 15 9 15 0
3-4 6 5 11 9 15 4
3-5 2 5 7 20 22 15
2-6 6 9 15 18 24 9
4-6 9 15 24 15 24 0
5-6 2 7 9 22 24 15

Here the critical path is along the activities 1-2, 2-4, 4-6. So the critical path is 1-2-4-6. Following diagram is prepared to show critical path along with EST and LFT.


\therefore The critical path = 1-2-4-6 with time duration (tcp)(t_{cp}) of 24 days.

Here standard deviation is calculated for activities of critical path. So we get

σ=V1+V2+V3σ=1+4+1.778=2.6034\begin{aligned} \sigma&= \sqrt{V_1 + V_2 + V_3} \\ \sigma&= \sqrt{1 + 4 + 1.778} \\ &= 2.6034 \end{aligned}

Now the probability of completion of project in that given time (t) of 26 days, can be calculate by below formula,

Z=ttcpσ=26242.6034=0.7682Z = {t - t_{cp} \over \sigma} = {26 - 24 \over 2.6034} = 0.7682

Using table in Appendix-B, we get probability =77.8%= 77.8 \%

As you can see below in Appendix-B the first column Z that is the probability we find out in the example through formula. The second column ψ(z)\psi(z) represent the probability in percentage (%)(\%).




As we have value of Z=0.7682Z=0.7682 now you can see in table we have Z=0.76Z=0.76 which has ψ(z)=0.7764\psi(z)=0.7764 and Z=0.77Z=0.77 has ψ(z)=0.7794\psi(z)=0.7794 we take average of both we get ψ(z)=0.7779=77.8%\psi(z)=0.7779=77.8\%




As you can see for Z=0  Z=0 \space \space ψ(z)=0.5000  \psi(z)=0.5000 \space \space which is 50%50\% and for Z=1  Z=1 \space \space ψ(z)=0.8413  \psi(z)=0.8413 \space \space which is 84.13%84.13\%


Static and Dynamic Balancing

When machine is in working condition different forces are acting on it which may cause machine to vibrate and cause damage to machine parts. The different forces acting on machine parts are static forces and dynamic forces.

Static force

The force which depends on weight of a body, is known as static force. (Generally, a static force acts when vibrations occurs in same plane.)

Dynamic or Inertia force

The force which depends on acceleration of a body, is known as dynamic force. (Generally, dynamic force acts when vibrations occurs in different planes.)

Due to these forces, the efficiency of the system decreases and life span of the system also decreases.
Due to these forces, the machine starts vibrating and sometimes when the vibrations increases, the machine would lift from it’s position and cause damage to other machine or human. So to avoid this, foundation is made below the machine (as you can see in the Fig-1 below), which absorbs the vibration and protects the machine against causing damage. Thus, balancing of the machine is required.

Fig_1_static-and-dynamic-balancing Fig-1


Balancing is the process of eliminating, the effect of static forces and dynamic forces acting on machine components.

When is system said to be unbalanced?

In any system with one or more rotating masses, if the centre of mass of the system does not lie on the axis of rotation, then the system is said to be unbalanced.

As you can see in the Fig-2 we have a rotor which is mounted on a shaft and the shaft has its own axis of rotation. Now you can see that the C.G(centre of gravity) of the rotor is at a distance r from the axis of rotation of the shaft, so when the rotor will start to rotate, a centrifugal force will act on it which is in the outward direction as you can see in Fig-3. Due to this force our system will become unbalanced and it will start to vibrate.

Fig_2_static-and-dynamic-balancing Fig-2

Fig_3_static-and-dynamic-balancing Fig-3

Static balancing:-

A system is said to be statically balanced, if the centre of masses(C.G) of the system lies on the axis of rotation.

Condition for Static Balancing

The resultant of all the centrifugal forces (dynamic forces) acting on the system during rotation must be zero.

Centrifugal forces acting on the system=zeroi.e. +mrω2mrω2=0\begin{aligned} \small ∑ \text {Centrifugal forces acting on the system} &= \text {zero}\\ i.e. \ +mrω^2 – mrω^2 &= 0 \end{aligned}

Dynamic balancing:-

A system is said to be dynamically balanced, if it satisfies following two conditions:-

  1. The resultant of all the dynamic forces acting on the system during rotation must be zero.
Dynamic forces acting on the system=zero∑ \text {Dynamic forces acting on the system} = \text {zero}
  1. The resultant couple due to all the dynamic forces acting on the system during rotation, about any plane, must be zero.
couple=zero∑ \text {couple} = \text {zero}

Transportation Model - Introduction

What is the Transportation Model?

  • Transportation Model is a special case of LPP(Linear Programming Problem) in which the main objective is to transport a product from various sources to various destinations at total minimum cost.

  • In Transportation Models, the sources and destinations are known, the supply and demand at each source and destinations are also known.

  • It is designed to find the best arrangement for transportation such that the transportation cost is minimum.

For example:

  • Consider three companies (Company1, Company2 and Company3) which produce mobile phones and are located in different regions.

  • Similarly, consider three cities (namely CityA, CityB & CityC) where the mobile phones are transported.

  • The companies where mobile phones are available are known as sources and the cities where mobile phones are transported are called destinations.

  • Let,
      Company1 produces a1 units,
      Company2 produces a2 units,
      Company3 produces a3 units.

  • Let,
      demand in CityA is b1 units,
      demand in CityB is b2 units,
      demand in CityC is b3 units.

  • The cost of transportation from each source to destination is given in table


  • The transportation of mobile phones should be done in such a way that the total transportation cost is minimum.

Types of transportation problems:

  • There are two types of transportation problems:

   i) Balanced transportation problem: The sum of supply and sum of demand are same.

Σ Supply=Σ Demand\Sigma \text { Supply} = \Sigma \text { Demand}

    ii) Unbalanced transportation problem: The sum of supply and sum of demand are different.

Σ SupplyΣ Demand\Sigma \text { Supply} \ne \Sigma \text { Demand}

Methods to solve Transportation Model

  1. North-West corner method
  2. Least cost method
  3. Vogel's Approximation Method (VAM)

Industrial applications of Transportation Model

  1. Minimize the transportation cost from source to destination.
  2. Determine lowest cost location for new industries, offices, warehouse, etc.
  3. Determine the number of products to be manufactured according to demand.
  4. Courier Services: Helps in taking proper decisions to find the best route for transportation.

Basics of Linear Programming


At the time of World War- II, G.B. Dantzing was working with US Air force and he was facing many problems such as allocation of weapons to different war locations, military logistics, etc. Since available resources are limited (restricted), it was very crucial to allocate optimum number of weapons or logistics in order to fulfil the objective of winning the war. At that time, he developed the technique to solve such problems, and that technique is known as Linear Programming (LP).

Linear Programming

LP consists of two words – Linear and Programming. Linear word represents that relationship between variables is linear. Programming represents mathematical modelling and development of algorithms to solve a problem that involves optimum allocation of limited or restricted resources.


  • Linear Programming is a mathematical modelling for optimization of a function (Objective Function), subject to restricted resources of variables (represented ‘n’ form of linear equations and/or inequalities).

  • Objective function may be to maximize profit or to minimize loss or any other measures which to be obtained in the best optimal manner. Constraints are restrictions on available man hours, materials, money, machines, etc.

In general, if the resources are unlimited, then there will not need to define any kind of strategy. Since, resources are limited, there should be a well-defined strategy.

Pre-requirements of LP

  1. Decision variables should be interrelated and non-negative. Decision variables are those variables which are decided at the starting of solving the sum.
  2. There must be a well-defined objective function, which can be represented as a linear function of decision variables.
  3. There must be constraints on amount of resources which can be expressed as linear inequalities in terms of decision variables.

Properties/Assumptions of LP

Tip: As suggested in video, properties of LP can be easily remembered as keyword PANC2{PANC}^2.

The amount of each resource consumed and its contribution to profit in objective function must be proportional to the value of each decision variable. For eg: If one egg can provide 8 g of protein, then 10 eggs can provide 80 g (10X8=80 g) of protein.

The total value of the objective function equals the sum of the contribution of each variable to the objective function. Similarly, total resource use is the sum of the resource use of each variable. For eg: Total profit by selling ‘n’ products must be equal to the profit earned by selling the items individually. On the other hand, the total cost of manufacturing ‘n’ products at a time, must be equal to the cost of manufacturing items individually.

It is assumed that values of decision variables are either positive or zero. It cannot be negative.

It is assumed that the solution value of decision variables and the amount of resources used are not be integer values. For eg: It is not possible to have 3.5 as the optimum number of products.

Certainty of coefficients:
In all LP models, the coefficients of objective function, R.H.S. coefficients of constraints and resources values are certainly and precisely known and measurable. Also, it is assumed that those values remain constant irrespective of time.

Advantages of LP:

  1. Optimum allocation of resources: LP shows the allocation of resources to fulfil the objective function and helps in making the optimum use of available resources.
  2. Improvement in quality of decision: LP improves decision quality as it gives accurate values of decision variables to fulfil the objective function.
  3. Exploring bottleneck machines: Bottleneck machines are those machines which cannot meet the demand and because of them other machines are idle. LP model identifies this bottleneck machines and explores the problem of low production capacity of the plant.
  4. Re-evaluation of basic plan: LP re-evaluates the basic plan as per changing conditions. If condition change when the plan is partly carried out, they can be determined so as to adjust the remainder of the plan for best result.

Limitations of LP:

  1. LP treats relationships between variables as Linear. However, always it is not possible to have linearity.
  2. LP assumes that outcome of problem has decision variables as integers.
  3. LP assumes the certainty of the coefficients, but in each case it is not possible to determine the values of coefficients.
  4. LP deals with single objective functions, where as in real life situation business may have multiple objectives.
  5. LP ignores the effect of time.

Applications of LP:

Now-a-days, LP has wide applications in different sectors like industrial, management, farming, financial and many other miscellaneous applications.

1) Product Management:

- Product Mix: Company produces number of products from same limited production resources. In such cases it is essential to determine the quantity of each product to be manufactured knowing the profit induced and amount of materials consumed by each of them. The objective is to maximize profit subject to all constraints. For example, wafer manufacturing companies like Balaji.

Balaji-Wafers Image source: http://www.walkthroughindia.com/

- Blending problems: When a product can be made from a variety of raw materials, each of which has a particular composition and price. The objective here is to determine the minimum cost blend subject to availability of the raw materials and the minimum constraints on certain product constituents.

For example, we are producing some product-X, and for which we have four different raw materials- A, B, C & D. Now, product-X can be made of A blend with B, as well as B with D and A with C. Now, for each particular raw material has different price, so compositions have their different price too. LP helps in selecting optimum composition and its price, subjected to constraints.

blending Image illustrating Blending Problem

- Trim loss: When an item can be made of standard size (like glass, paper,sheet, etc.), the problem that arises is to determine which combination of requirements should be produced from standard materials in order to minimize the trim loss (wastage).

- Production Planning: This deals with the determination of minimum cost production plan over a planning period of an item with a fluctuating demand considering the initial number of units in inventory, production capacity,constraints on production, manpower and all relevant cost factors. The objective here is to minimize total operational costs.

2) Marketing Management:

- Media selection: LP techniques can helps in determining the advertising media mix so as to maximize the effective exposure, subject to limitation of budget, specified exposure rates to different market segments, specified minimum and maximum number of advertisements in various media.

- Travelling salesman problem: The problem of salesman is to find the shortest route starting from a given city, visiting each specified cities and then returning to the original point of departure, provided no city shall be visited twice during the tour.

- Physical distribution: LP determines the most economic and efficient manner of locating manufacturing plants and distribution centers for physical distribution.

3) Agricultural Applications:

- Farm management: LP can be applied in agricultural planning like allocation of limited resources such as available land, labour, water supply and working capital, etc. in such a way so as to maximize net revenue.

4) Financial Management:

- Capital budgeting problems: This deals with the selection of specific investment activity among several other activities.

- Profit planning: This deals with the maximization of profit margin from investment in plant facilities and equipment, cash on hand and inventory.

5) Miscellaneous problems:

- Diet problems: Diet problem includes the optimization of food quantities in order to fulfil the nutrition proportions as well as to minimize the cost of the food. Dieticians can use LP for planning a balance diet for a particular patient.

- Inspection problems: These problems relates with the deciding the optimum number of inspectors in order to inspect the raw materials. Also, the cost of inspectors should be minimum and reliability requires being high. This helps in deciding the optimum time to move on machine inspection based on require output, so as to increase the profit by decreasing the manufacturing time of particular product.

- Military Applications: Military application includes the problem of allocation of limited tools and technologies to various locations.

LPP Formulation with Numericals

To formulate Linear Programming Problem (LPP) for given statement type problems(numerical) is easy if we go through its Mathematical Model.

General Mathematical Modelling of LP

As explained in the video, mathematical model of LP consists of the following:

i) Objective Function (denote with “Z”)
ii) Decision variables (represented in terms of “x”)
iii) Constraints

Note: To know more about above terms you can go through the notes of the Basics of Linear programming, link to the notes- click here

Now, let us generate the mathematical model of the LPP:

Decision variables be x1,x2,x3,x4,,xnx_1, x_2, x_3, x_4,…, x_n with total ‘m’ constraints.
Now, LP can be formulate as:

Optimize (Maximize or Minimize)  Z=c1x1+c2x2+c3x3++cnxnSubject to linear constraints,a11x1+a12x2+a13x3++a1nxn(,=,)b1a21x1+a22x2+a23x3++a2nxn(,=,)b2a31x1+a32x2+a33x3++a3nxn(,=,)b3uptoam1x1+am2x2+am3x3++a1nxn(,=,)bmandx1,x2,x3,x4,,xn0\small \text {Optimize (Maximize or Minimize)} \space \space Z= c_1x_1 + c_2x_2 + c_3x_3 +\dotso+ c_nx_n \\ \begin{aligned} \text {Subject to linear constraints,} \quad a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + \dotso + a_{1n}x_n \quad &(≤ , = , ≥) \quad b_1 \\ \quad a_{21}x_1 + a_{22}x_2 + a_{23}x_3 +\dotso+ a_{2n}x_n\quad &(≤ , = , ≥) \quad b_2 \\ \quad a_{31}x_1 + a_{32}x_2 + a_{33}x_3 +\dotso+ a_{3n}x_n \quad &(≤ , = , ≥) \quad b3\\ \quad \text upto \dotso a_{m1}x_1 + a_{m2}x_2 + a_{m3}x_3 + \dotso+ a_{1n}x_n \quad &(≤ , = , ≥) \quad b_m \\ \text and \quad x_1, x_2, x_3, x_4,\dotso, x_n ≥ 0 \end{aligned}

Thus, we will have any of the three conditions applied on constraints as per the provided problem (,=,)(≤ , = , ≥).

Steps for LPP Formulation

  1. Identify the decision variables:
    It is the most important step on LPP Formulation, because based on the decision variables only, the whole problem governs. We will learn about, how to identify decision variables from given numerical/problem in this note only. (Also you can go through video for the same. Link provided in cover page)

  2. Identify Objective Function (Z) and express it as a linear function of decision variables:
    As we have seen above in mathematical model, objective of any problem can be maximize or minimize. Based on that, we will have to define the objective function in terms of decision variables. (For eg: Maximize Z=2x1+5x2+9x3Z= 2x_1 + 5x_2 + 9x_3)

  3. Identify all constraints and express it as linear inequalities or equalities in terms of decision variables:
    As we know that, in the real world all the resources are limited and so in each particular problem you will find some limitation/constraint on the use of available resources.

  4. Express decision variables as Feasible variables:
    It means that, we have lastly define all the decision variables are greater than or equal to zero (For eg:  x1,x2,x30\ x_1, x_2, x_3 ≥ 0)

Let’s move to numerical now,

Numerical 1: Two products ‘A’ and ‘B’ are to be manufactured. Single unit of ‘A’ requires 2.4 minutes of punch press time and 5 minutes of assembly time, while single unit of ‘B’ requires 3 minutes of punch press time and 2.5 minutes of welding time. The capacity of punch press department, assembly department and welding department are 1200 min/week, 800 min/week and 600 min/week respectively. The profit from ‘A’ is ₹60 and from ‘B’ is ₹70 per unit. Formulate LPP such that, profit is maximized.


Tip: Always relate/treat LPP with real life problems for better understanding and ease of solution

Step 1: Identify the decision variables

Here, the decision has to be taken for how much quantities of product ‘A’ and ‘B’ to be manufactured in order to maximize the profit. (Quantities of product is governing this problem)

Quantity of product ‘A’ manufactured = x1x_1
Quantity of product ‘B’ manufactured = x2x_2

Step 2: Identify Objective Function (Z)

Here, the objective is to maximize the profit from product ‘A’ and ‘B’. Each unit produced product ‘A’ gives profit of ₹60, while ‘B’ gives ₹70. So, the objective function can be defined in terms of decision variables as:

MaximizeZ=60x1+70x2\text Maximize \quad Z = 60x_1 + 70x_2

Step 3: Identify all the constraints

Product ‘A’ manufactured by: Punch press and Assembly
Product ‘B’ manufactured by: Punch press and Welding

So, we have a total of three processes for the manufacturing of two products A and B.
Thus, there are total three constraints.

Now, we will bifurcate this problem based on processes to get a clear idea of constraints on resources, as follows:

Product ‘A’ requires 2.4 min and product ‘B’ requires 3 min of Punch press time.
Also, punch press time is limited to 1200 min/week.

2.4x1+3x21200\therefore 2.4x_1 + 3x_2≤ 1200

Product ‘A’ requires 5 min of Assembly time.
Also, assembly time is limited to 800 min/week.

5x1800\therefore 5x_1≤ 800

Product ‘B’ requires 2.5 min of Welding time.
Also, welding time is limited to 600 min/week.

2.5x1600\therefore 2.5x_1≤ 600

Step 4: Express decision variables as feasible variables

We assume that the decision variables has feasible solution, that implies - decision variables are greater than or equal to zero.

x1,x20x_1, x_2 ≥ 0

So, complete LP can be written as,

Maximize Z=60x1+70x2Subject to 2.4x1+3x212005x18002.5x1600And  x1,x20\text {Maximize Z} = 60x_1 + 70x_2 \\ \begin{aligned} \text {Subject to } 2.4x_1 + 3x_2 &≤ 1200 \\ 5x1 &≤ 800 \\ 2.5x1 &≤ 600 \\ \text And \ \ x_1, x_2 &≥ 0 \end{aligned}

Note that, here the solution provided (in above numerical) to you with steps is just for explanation. You can always skip this step wise solution to LP problems and can use direct method as provided in video of this topic (click here to watch video).

Numerical 2: A tailor prepares two kinds of dresses. First kind of dress is having raw material cost of ₹150 and labour cost of ₹80, while the second kind of dress is raw material cost of ₹250 and labour cost of ₹100. He can sell the dresses of first and second kind at the rate of ₹300 and ₹500 respectively. First kind of dress takes 2 hours and second kind takes 2.5 hours of cutting. The total labour hours are restricted to 84 hours/week. Also, first kind of dress takes 1.5 hours of stitching and second requires 2 hours of stitching. The stitching hours are restricted to 60 hours/week. Formulate LPP, such that maximum profit can be earned by tailor.


The solution of this numerical will be such that, you can present this in your examination. This will save your time and gives complete idea of your explanation too in examination. All the explanation stuff will be provided in {"Explanation"} brackets for your understanding only, in the solution of this problem.

Number of first kind of dress = x1x_1 units
Number of first kind of dress = x2x_2 units

{The decision variables selected here are the units of dresses of both kind so produced, as we are going to maximize the profit of tailor. Also, by understanding the whole problem you can easily make out that, to increase the profit, tailor needs to produce more goods.}

Now, Net Profit earned by selling a unit of first kind of dress =(30015080)=70= ₹(300-150-80) = ₹70

Net Profit earned by selling a unit of first kind of dress =(500250100)=150= ₹(500-250-100) = ₹150

{As you can understand that the problem provided here is based on revenue so generated by selling different kinds of dresses. Now, revenue will be the profit generated after excluding the raw material and manufacturing cost on each of these dresses. So, for example you can see above we have excluded raw material cost of first kind of dress ₹150 and labour cost ₹80 from total selling cost of ₹300.
Thus revenue/net profit generated by selling each unit of first kind of dress is ₹ (300-150-80) = ₹70 and the same for the second kind of dress.}

∴ The objective function can be defined as;

MaximizeZ=70x1+150x2\text {Maximize} \quad Z = 70x_1 + 150x_2

Now, constraints are provided on labour hours for cutting and stitching. Thus, LP is subject to the following constraints;

2x1+2.5x2841.5x1+2x260And  x1,x20\begin{aligned} 2x_1 + 2.5x_2 &≤ 84 \\ 1.5x_1 + 2x_2 &≤ 60 \\ \text {And} \ \ x_1, x_2 &≥ 0 \end{aligned}

{As we can find from problem provided, the cutting and stitching hour requirements for both kinds of dresses. Also total hours are restricted on cutting and stitching is 84 hours/week and 60 hours/week respectively.}

So, complete LP can be written as,

Maximize Z=70x1+150x2Subject to  2x1+2.5x2841.5x1+2x260Andx1,x20\text {Maximize Z}= 70x_1 + 150x_2 \\ \begin{aligned} \text {Subject to} \ \ 2x_1 + 2.5x_2 &≤ 84 \\ 1.5x_1 + 2x_2 &≤ 60 \\ \text {And} \quad x_1, x_2 &≥ 0 \end{aligned}

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