Industrial Engineering Test 1

Features of product or line layout:

• Semi-skilled workers operate two or more machines.
• Highly specialized machines, jigs and fixtures are used.
• Conveyorized movement of inventories
• Small in-process inventory

Concept:

A basic feasible solution can be obtained by:-

1. North - west corner method

2. Least cost cell method

3. Row minima method

4. Column minima method

5. Vogel’s approximation method

Methods of giving optimality test:

1. Stepping stone method.

2. Modified distribution method (MODI Method)

Given:

D = 1800, C = 40, C₀ = Rs. 20, C_h = 0.2 × 40 = Rs. 8

Now,

Optimum order quantity,

\({Q^*} = \sqrt {\frac{{2D{C_0}}}{{{C_h}}}} = \sqrt {\frac{{2\; \times \;1800\; \times \;20}}{8}} \)

Q* = 94.86

∴ Q* ≃ 95

Concept:

Average time in system (W_s) = Waiting time + Service time (Vulcanizing time)

\({W_s} = {W_q} + \frac{1}{\mu } = \frac{\lambda }{{\mu \left( {\mu - \lambda } \right)}} + \frac{1}{\mu } = \frac{1}{{\mu - \lambda }}\)

λ = 2 belts per shift

μ = 5 belts per shift

\({W_s} = \frac{1}{{5 - 2}} = \frac{1}{3}shift\)

Concept:

\({\rm{Average\;inprocess\;inventory}} = \frac{{\sum jobs \times time}}{{\sum time\;\left( {total\;time} \right)}}\)

Calculation:

\({\rm{Average\;inprocess\;inventory}} = \frac{{\left( {5 \times 1} \right) + \left( {4 \times 2} \right) + \left( {3 \times 3} \right) + \left( {2 \times 4} \right) + \left( {1 \times 5} \right)}}{{1 + 2 + 3 + 4 + 5}}\)

\(\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{5{\rm{\;Jobs}} \to 1{\rm{\;hr}}}\\{4{\rm{\;Jobs}} \to 2{\rm{\;hr}}}\\{3{\rm{\;Jobs}} \to 3{\rm{\;hr}}}\end{array},}&{\begin{array}{*{20}{c}}{2{\rm{\;Jobs}} \to 4{\rm{\;hr}}}\\{1{\rm{\;Job}} \to 5{\rm{\;hr}}}\end{array}}\end{array}\) 

\(\therefore \;Average\;in\;process\;inventory\;\left( {AIP} \right) = \frac{{\left( {5 + 8 + 9 + 8 + 5} \right)}}{{15}}\;\)

Concept:

\({\rm{Time\;between\;orders\;}} = \frac{Q}{A} \times {\rm{Time\;period}}.{\rm{\;}}\left( {365{\rm{\;days}}} \right)\)

\(Q = \sqrt {\frac{{2D{C_0}}}{{{C_H}}}} \)

Where D = A = annual demand, C₀ = Ordinary cost, C_h = Holding cost

Calculation:

\(Q = \sqrt {\frac{{2\; \times \;120000\; \times \;100}}{5}}\)

∴ Q = 2190.89∼2191

Now,

\(\therefore T = \frac{Q}{A} \times 365 = \frac{{2191}}{{120000}} \times 365\)

∴ T = 6.664 days

Concept:

Poisson formula,

\(P\left( n \right) = \frac{{{{\left( {\lambda T} \right)}^n}{e^{ - \lambda T}}}}{{n!}}\)

λ = arrival rate

n = number of customers

T = Time period

Calculation:

\(P\left( s \right) = \frac{{{{\left( {3\; \times \;1} \right)}^5}\;{e^{ - 3 \times 1}}}}{{5!}}\)

\(P\left( s \right) = \frac{{{3^5}}}{{120}} \times {e^{ - 3}}\)

∴ P(s) = 0.101

Concept:

F_{t + 1} = F_t + α (D_t - F_t)

Calculation:

F_Feb = 76 + 0.4 (72 - 76).

FFeb= 76 - 1.6

∴ FFeb= 74.4

Now,

F_March = 74.4 + 0.4 (80 - 74.4)

∴ FMarch = 76.64

Now,

F_April = F_March + α (D_March – F_March)

FApril = 76.64 + 0.4 (75 - 76.64)

∴ F_April = 75.984

Concept:

\({\rm{Service\;level}} = \frac{{Gain}}{{Gain + loss}}\)

Calculation:

Gain = Selling price – cost price

Gain = 10 – 6

∴ Gain = Rs. 4

Now,

Loss = cost price – Rebate (scrap value)

Loss = 6 – 2

∴ Loss = Rs. 4

\(\therefore {\rm{Service\;level}} = \frac{{Gain}}{{Gain + loss}}\)

\(\therefore {\rm{Service\;level}} = \frac{4}{{4 + 4}}\)

∴ Service level = 0.5 or 50%

Explanation:

LPP is max z = 5x₁ + 3x₂

3x₁ + 5x₂ ≤ 15

5x₁ + 2x₂ ≤ 10

Primal is maximization type, so dual will be minimization type and ‘≤’ will change into ‘≥’

The column coefficient will become row coefficients for dual

Let w₁ and w₂ be dual variables, then

Min z* = 15w₁ + 10w₂

3w­₁ + 5w₂ ≥ 5

5w₁ + 2w₂ ≥ 3  

Critical ratio (C.R) Rule: - A rule for sequencing, that is an index number computed by dividing the time remaining until due date by the work time remaining.

\(CR = \frac{{Time\;remianing\;}}{{workdays\;remianing}} = \frac{{Due\;date - Today's\;date}}{{\begin{array}{*{20}{c}} {work\;\left( {lead} \right)time}\\ {\;remining} \end{array}}}\)

• The critical ratio gives priority to jobs that must be done to keep shipping on schedule.
• A job with a low CR (less than 1.0) is one that is falling behind schedule.
• If CR = 1.0 (exactly), job is on schedule.
• If CR > 1.0, means the job is ahead of schedule & has some slack.

Solution:

Job B has the lowest CR ratio, therefore, Job B must be on highest priority.

Explanation:

First, calculating forecast for 5^th year using weighted moving average.

∴ Forecast = (105 × 0.4) + (125 × 0.3) + (110 × 0.2) + (120 × 0.1)

∴ Forecast = 113.5

Now,

Using exponential smoothening method

Given:

α = 0.4, F₁ = 110, D₁ = 120

Now,

F₂ = F₁ + α (D₁ – F₁)

F₂ = 110 + 0.4 (120 - 110)

∴ F₂ = 114

Now,

F₃ = F₂ + α (D₂ – F₂)

F₃ = 114 + 0.4 (110 - 114)

∴ F₃ = 112.4

Now,

F₄ = F₃ + α (D₃ – F₃)

F₄ = 112.4 + 0.4 (125 – 112.4)

∴ F₄ = 117.44

Now,

F₅ = 117.44 + 0.4 (105 – 117.44)

∴ F₅ = 112.464

Now,

Required difference = 113.5 – 112.464

Concept:

\({\rm{Critical\;ratio\;}}\left( {{\rm{CR}}} \right) = \frac{{\left( {Due\;data - current\;data} \right)}}{{Work\;remaining}}\)

Calculation:

(least CR first)

\(A = \frac{{28 - 20}}{{10}} = 0.8\)

\(B = \frac{{26 - 20}}{8} = 0.75\)

\(C = \frac{{24 - 20}}{7} = 0.57\)

\(D = \frac{{32 - 20}}{7} = 1.71\)

\(E = \frac{{30 - 20}}{{12}} = 0.83\)

∴ The order is C, B, A, D, E

Average job completion time \( = \frac{{59}}{5}\) days.

Total lateness = 6 days

Required ratio \( = \frac{{\left( {\frac{{59}}{5}} \right)}}{6} = 1.97\)

Concept:

\(EOQ = \sqrt {\frac{{2D{C_0}}}{{{C_h}}}}\)

D = annual demand, C₀ = ordering cost, C_h = holding cost

Total cost = ordering + carrying + purchasing

\(TC = {C_0}\frac{D}{Q} + {C_c}\frac{Q}{2} + D\left( P \right)\)

Calculation cost corresponding to discount given.

Calculation:

\(EOQ = \sqrt {\frac{{2D{C_0}}}{{{C_H}}}} = \sqrt {\frac{{2\; \times \;125\; \times \;18}}{{20}}} \)

EOQ = 15 parts

\(TC = 18\;\left( {\frac{{125}}{{15}}} \right) + 20 \times \left( {\frac{{15}}{2}} \right) + 125 \times 100 = Rs.\;12800\)

For 50 units.

\(TC = 18 \times \frac{{125}}{{50}} + 20\left( {\frac{{50}}{2}} \right) + 125\left( {100 - 6} \right) = Rs.\;12,295\)

For 100 units.

\(TC = 18 \times \frac{{125}}{{100}} + 20\left( {\frac{{100}}{2}} \right) + 125\;\left( {100 - 8} \right) = Rs.\;12522.5\) 

∴ 50 unit is the most feasible (lowest cost)

Explanation:

\({\rm{Arrival\;rate\;}}\left( \lambda \right) = \frac{1}{9}\)

\({\rm{Service\;rate\;}}\left( \mu \right) = \frac{1}{5}\)

\(\rho = \frac{\lambda }{\mu } = \frac{5}{9}\)

Now,

\({W_s} = \frac{1}{{\mu - \lambda }} = \frac{1}{{\frac{1}{5} - \frac{1}{9}}} = \frac{{45}}{4}\)

(W_s → waiting time in system)

→ Probability of waiting for more than 10 minutes,

\(P\left( {T > 12\;min} \right) = \rho \cdot {e^{ - T/{W_s}}}\)

\(P\left( {T > 12\;min} \right) = \frac{5}{9} \cdot {e^{ - 12/\left( {45/4} \right)}}\)

\(P\left( {T > 12\;min} \right) = \frac{5}{9} \cdot {e^{\frac{{ - 48}}{{45}}}}\)

P(T > 12 min) = 0.1911

∴ P(T > 12 min) = 19.11%

Explanation:

Mean square error is given by:

\(MSE = \frac{{\mathop \sum \nolimits_{i = 1}^n {{\left( {{D_i} - {F_i}} \right)}^2}}}{n}\)

\(\mathop \sum \limits_{i = 1}^5 {\left( {{D_i} - {F_i}} \right)^2} = 64 + 400 + 25 + 25 + 225\)

= 739

Mean square error (MSE) \( = \frac{{739}}{5}\)

∴ Mean square error (MSE) = 147.8