Industrial Engineering Test 2

Concept:

P = Potential profit loss per unit for not meeting the demand

L = Unsold item loss per unit

In these models in order to maximise profit we select ordering quantity in such a manner that:

\(P\left( {s - 1} \right) < \frac{P}{{P + L}} \le P\left( s \right)\)

We calculate the service level using the given data:

Service level \(S = \frac{P}{{P + L}}\)

This service level is then compared with the cumulative probability table and the demand rate is found.

Calculation:

P = 100 – 50 = Rs. 50

L = 50 – 20 = Rs. 30

\(S = \frac{P}{{P + L}} = \frac{{50}}{{30 + 50}} = 0.625\)

Concept:

As per the exponential smoothening method:

Ft = Ft-1 + α [Dt-1 – Ft-1]

Calculation:

Given: Dt-1 = 450, Ft-1 = 470, Ft = 455

Ft = Ft-1 + α [Dt-1 – Ft-1]

455 = 470 + α [450 - 470]

Concept:

In the weighted moving average method, more weight is added to the recent period demand

Here (n = 4), number of periods

So, weights are

\(\frac{n}{{\sum n }},\frac{{n - 1}}{{\sum n }},\frac{{n - 2}}{{\sum n }},\frac{{n - 3}}{{\sum n }}\)

Here, 

\(\sum {n = \frac{{n\left( {n + 1} \right)}}{2}} = \frac{{4\left( {4 + 1} \right)}}{2} = 10\)

Weights = 0.4, 0.3, 0.2, 0.1

By Johnson Rule

D → B → C → A

⇒ MST = 23 min

idle time M₁ = 23 – 17 = 6

idle time M₂ = 2 

Total idle time = 6 + 2 = 8

Total available time per week = 40 hours = 2400 minutes

Now time required for one component = 30 minutes

∴ Ideally no. of components produced \(= \frac{{2400}}{{30}} = 80\)

But efficiency of machine = 80%

Scrap = 25%

∴ Actual no. of components produced per machine in one week = 80 × 0.8 × 0.75 = 48

Desired output = 1200 pieces per week

Explanation:

Balance Delay(BD) – It is the ratio of total idle time of the job on the assembly line to the total time spent by the job on the assembly line.

\(BD = \frac{{(n \times {T_c}) - {T_{wc}}}}{{n \times {T_c}}}\)

where

T_C = Cycle time (time between two successive products coming out of the production line, it maximum of the work stations time)

T_{WC =} Total work content (Equal to the sum of the processing time of each operation)

n = number of work stations

Calculation:

Given:

T_C = 27, n = 6

T_WC = 22 + 24 + 26 +27 + 25 + 26  = 150

Balance delay is:

\(BD= \frac{{(n \times {T_c}) - {T_{wc}}}}{{n \times {T_c}}}= \frac{{(6 \times 27) - 150}}{{6 \times 27}} \times 100 = 7.4\% \)

TQM or Total Quality Management is the process of reducing or eliminating errors in manufacturing, streamlining supply chain management, improving customer experience, and ensuring that employees are up to speed with their training. TQM is a process-oriented management system and considers the complete organization as integrated, is customer focussed, demands continuous improvement, and fact-based decision making to meet the quality standards as planned and desired during the planning stage. TQM involves the integrated approach where a bottom level employee can interact with top management and a have involvement in quality-related parameters. It is a part of ISO 9000 certification.

Hence the most basic attribute is the direct involvement of top management 

Simplex method is a step by step procedure in which we proceed in a systematic manner from an initial feasible solution with an improvement in every iteration until we reach optimum solution.

i) All the resource value or constraints should be non negative.

ii) All the inequalities of the constraint should be converted to equalities with the help of slack or surplus variables.

Arrival rate (λ) = 24 customer/hr

Service rate (μ) = 32 customer per hour

\(\rho = \frac{\lambda }{\mu } = \frac{{24}}{{32}} = 0.75\)

Probability of having n customer in the system is given as P_n = ρⁿ⋅P_o

Where P_o = - 1 - ρ

So probability of 5 customer in the queue implies 6 customers in the system

⇒ P₆ = (0.75)⁶ (1 - 0.75) = 0.045

In % terms, probability = 4.5%

Concept:

D = Annual demand of inventory (units/week)

Q = Quantity to be ordered at each order point (units/order)

C = Costs of purchasing one unit of inventory (Rs/unit)

C_o = Costs of placing one order (Rs/order)

C_h = Costs of holding one unit in inventory for one complete year (Rs/unit/week)

Total annual cost:

\(TAC = D.C + \frac{D}{Q}{C_o} + \frac{Q}{2}{C_h}\)

At EOQ: \(TA{C^*} = D.C + \sqrt {2D{C_O}{C_h}}\)

∴ The total optimist cost of the product is:

\(TA{C^*} = RC + \sqrt {2R{C_O}{C_h}} \)

The selling price will be:

SP = D × unit selling price

And the difference will give the profit.

Calculation:

Given, C_o = Ordering cost = Rs 80, D = Demand = 300 units/week = 15600 units/year, C = Rs 50/unit, C_o = Rs 80/unit

Holding cost is C_h = 8% per year cost of product

\({C_h} = \frac{8}{{100}} \times 50\;per\;unit\;per\;year = \frac{8}{{100}} \times \frac{{50}}{{52}}\;per\;unit\;per\;week = {\rm{Rs\;}}0.0769\)

\(TA{C^*} = 300 \times 50 + \sqrt {2 \times 300 \times 80 \times 0.0769} = Rs\;15060.75\)

Selling price is \(SP = 300 \times 60 = Rs\;18000\)

Thus, profit is:

\(P = 18000 - 15060.75 = Rs\;2939.24\)

Since F_t+1 = F_t + α (D_t – F_t).

F_February = F_January + α (D_January – F_January).

Since for the forecast of February, January month is not known. Thus, for period of January both demand & forecast will be taken as same.

F_January = D_January = 48

F_February = 48 + 0.32 (48 – 48)

F_February = 48

F_March = F_February + α (D_February - F_February)

= 48 + 0.32 (60 – 48).

= 51.84

F_April = 51.84 + 0.32 (56 – 51.84).

= 53.1712

F_May = 53.1712 + 0.32 (53 – 53.1712).

F_May = 53.116

Alternate solution:

By basic definition:

F_t+1= α D_t + α (1-α) D_t-1 + α (1-α)² D_t-2 + α (1- α)³ D_t-3 + …..… + α (1-α)^t-2D₁+ (1-α)^t-1 F₁

F_May = α D_April + (1-α) F_April

F_April = α D_March + (1-α) F_March

Thus F_May = α D_April + (1-α) {α D_March + (1-α) F_March}

F_May = α D_April + α (1-α) D_March + (1-α )² F_March

F_May = α D_April + α (1-α) D_March +(1-α )² {α D_February + (1-α) F_February}

F_May = α D_April + α (1-α) D_March + α (1-α )² D_February + (1-α)³ F_February

F_May = α D_April + α (1-α) D_March + α (1-α )² D_February + (1-α)³ {α D_January + (1-α) F_January}

F_May = α D_April + α (1-α) D_March + α (1-α )² D_February + α (1-α)³ D_January + (1-α)⁴ F_January

F_May = 0.32× 53 + 0.32 × 0.68 × 56 + 0.32 × 0.68² × 60 + 0.32 × 0.68³ × 48 + 0.68⁴ × 48

F_May = 53.1164

\(MAD = \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {{D_i} - {f_i}} \right|}}{n} = \frac{{\left| {160 - 170} \right| + \left| {180 - 190} \right| + \left| {165 - 190} \right| + \left| {175 - 200} \right| + \left| {200 - 180} \right|}}{5}\)

\( = \frac{{10 + 10 + 25 + 25 + 20}}{5}\)

\(= \frac{{90}}{5} = 18\)

\(MAPE = \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {\frac{{{D_i} - {f_i}}}{{{D_i}}}} \right|}}{n} \times 100\% \)

\( = \frac{{\left| {\frac{{\left( {160 - 170} \right)}}{{160}}} \right| + \left| {\frac{{\left( {180 - 190} \right)}}{{180}}} \right| + \left| {\frac{{\left( {165 - 190} \right)}}{{165}}} \right| + \left| {\frac{{\left( {175 - 200} \right)}}{{175}}} \right| + \left| {\frac{{\left( {200 - 180} \right)}}{{200}}} \right|}}{5} \times 100\% \)

MAPE = 10.25%

Concept:

In early due date policy, the jobs are arranged in such a way that the job with less due date comes first.

Where as in Short processing time technique, the jobs are arranged in such a way that jobs with less processing time comes first.

Calculation:

After arranging the data as per EDD technique,

Total Flow time = 5 + 9 + 15 + 18 + 21 = 68 days

Mean flow time = 68/5 = 13.6 days (Option 1)

Total tardiness = 1 day (Option 2)

Average tardiness = 1/5 = 0.2 day

If shortest processing time technique is used,

Mean flow time will be 11 days and

The total tardiness is 10 days

Concept: - Smallest time machine A_r (on Plane turning machine)

Smallest time on machine C_S (On step turning machine)

If  max B_i ≤ smallest A_r

     max B_i ≤ smallest C_s

If these conditions satisfy, then three machines A, B, C can be hypothesized into two machines namely G & H to find optimal sequence.

The time elements for machine G & H are: -

G_i = A_i + B_i

 H_i = C_i + B_i

Now by applying Johnson’s & Bellman method, the optimal sequence is: -

Concept:

Duality in Linear programming problem (LPP): It means a linear programming problem has another LPP which is derived from it.  

Original LPP is known as primal and derived LPP is known as Dual.

Explanation:

Some points about dual LPP:

1) The final simplex table giving optimal solution of the primal also contains optimal solution of its dual in itself.

2) If either the primal or the dual problem has a finite optimal solution, then the other problem also has a finite optimal solution.

3) If either problem has an unbounded optimum solution, then the other problem has no feasible solution at all

Calculation:

Let x₁, x₂ be the variables and P be the function in primal linear program. For dual linear program, let y₁, y₂ be the variables and Q be the function.

From the given data, the original linear program will be

Primal:

Maximize P = 2x₁ + 4x₂ (Option 1)

Subject to

x₁ + 3x₂ ≤ 96

2x₁ + 5x₂ ≤ 100

x₁ ≥ 0; x₂ ≥ 100

Using graphical method, the boundary points are (0, 20) and (50, 0)

P will be maximum at (50, 0)

∴ optimal solution will be x₁ = 50; x₂ = 0 (Option 3)

Dual:

Dual can be found by using the below formula,

If primal is given as

Maximize C^T x, subject to A x ≤ b

Then dual will be

Minimize b^T y, subject to Ay ≥ c

Therefore, the dual will be

Minimize Q = 96y₁ + 100y₂ (Option2 is wrong)

Subject to

y₁ + 2y₂ ≥ 2

3y₁ + 5y₂ ≥ 4

y₁ ≥ 0, y₂ ≥ 0

After solving, the boundary points will be (2, 0) and (0, 1)

Q will be minimum at (0, 1) (Option 4)

Solution can be verified by checking the values of P and Q which should be equal.

\(\lambda = \frac{{30}}{{60 \times 24}} = \frac{1}{{48}}\) locomotives per minutes

\(\mu = \frac{1}{{36}}\) Locomotives per minute

\(\rho = \frac{\lambda }{\mu } = \frac{{36}}{{48}} = 0.75\) 

Length of system (L_s)₁ = \(\frac{\rho }{{1 - \rho }} = \frac{{0.75}}{{1 - 0.75}} = 3\;locomotives\)

Now, input changed to 35 per day,

So \(\lambda = \frac{{35}}{{60 \times 24}} = \frac{7}{{288}}\)

\(\begin{array}{l}\mu = \frac{1}{{36}}\\\therefore \rho = \frac{\lambda }{\mu } = \frac{7}{{288}} \times 36 = \frac{7}{8}\end{array}\)  

Now, length of system, (L_s)₂ = \(\frac{{\frac{7}{8}}}{{1 - \frac{7}{8}}} = 7\) locomotives

Concept:

The PERT (Project Evaluation and Review Technique) technique is used, when activity time estimates are stochastic in nature. For each activity, three values of time (optimistic, most likely, pessimistic) are estimated.

• Optimistic time (to) estimate is the shortest possible time required for the completion of an activity
• Most likely time (tm) estimate is the time required for the completion of activity under normal circumstances
• Pessimistic time (tp) estimate is the longest possible time required for the completion of an activity

In PERT expected time of an activity is determined by using the below-given formula:

\({t_e} = \frac{{\left( {{t_o} + 4{t_m} + {t_p}} \right)}}{6}\)

Calculation:

\({t_e} = \frac{{\left( {{2} + 4\times {5} + {8}} \right)}}{6} =5\)

te = 5 minutes