Industrial Engineering

Question 1
A project consists of five activities (A, B, C, D and E). The duration of each activity follows beta distribution. The three time estimates (in weeks) of each activity and immediate predecessor(s) are listed in the table. The expected time of the project completion is __________ weeks (in integer).
\begin{array}{|c|c|c|c|c|} \hline \text{Activity} & \text{Optimistic time (week)} & \text{Most likely time (week)} &\text{Pessimistic time (week)} &\text{Immediatepredecessor(s)} \\ \hline A& 4 & 5 &6 & None\\ \hline B& 1& 3 & 5 & A\\ \hline C& 1& 2 & 3 & A\\ \hline D& 2& 4 & 6 & C\\ \hline E& 3& 4 & 5 &B,D \\ \hline \end{array}
A
12
B
15
C
16
D
18
GATE ME 2022 SET-2      PERT and CPM
Question 1 Explanation: 
\begin{array}{|c|c|} \hline \text{Activity}& t_e=\frac{t_0+4t_m+t_p}{6}\\ \hline A&5\\ \hline B&3\\ \hline C&2\\ \hline D&4\\ \hline E&4 \\ \hline \end{array}

\begin{array}{|c|c|} \hline \text{Path}& \text{Duration}\\ \hline 1-2-4-5(A-B-E)&12\\ \hline 1-2-3-4-5 (A-C-D-E)&15\\ \hline \end{array}
Expected completion time of the project = critical path duration = 15 weeks
Question 2
A manufacturing unit produces two products P1 and P2. For each piece of P1 an P2, the table below provides quantities of materials M1, M2, and M3 required, and also the profit earned. The maximum quantity available per day for M1, M2 and M3 is also provided. The maximum possible profit per day is Rs __________.
\begin{array}{|c|c|c|c|c|} \hline &\text{M1 } &\text{M2}&\text{M3}&\text{Profit per piece (Rs.)} \\ \hline P1 & 2 &2 &0&150\\ \hline P2 & 3& 1&2&100\\ \hline \text{Maximum quantity available per day}& 70& 50&40\\ \hline \end{array}
A
5000
B
4000
C
3000
D
6000
GATE ME 2022 SET-2      Linear Programming
Question 2 Explanation: 
Let,
x_1= No. of units of product P_1
x_2= No. of units of product P_2
Max, Z=150x_1+100x_2
\begin{aligned} s.t.\;\; 2x_1+3x_2 &\leq 70 \\ 2x_1+3x_2&\leq 50 \\ 2x_2 &\leq 40 \\ x_1\geq 0, x_2&\geq 0 \end{aligned}

Point B: Intersection of

\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_2 &\leq 40 \\ x_2=20, x_1&=5 \end{aligned}

Point C: Intersection of
\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_1+x_2 &\leq 50 \\ \text{By sloving }2x_2 &=20 \\ \Rightarrow x_2=10, x_1&=20 \end{aligned}

\begin{aligned} Z_{max}&=150x_1+100x_2\\ Z|_{A(0,20)}&=150(0)+100(20)=2000\\ Z|_{B(5,20)}&=150(5)+100(20)=2750\\ Z|_{C(20,10)}&=150(20)+100(10)=4000\\ Z|_{D(25,0)}&=150(25)+100(0)=3750\\ \Rightarrow x_1&=20,x_2=10\\ Z_{max}&=4000 \end{aligned}
Question 3
Parts P1-P7 are machined first on a milling machine and then polished at a separate machine. Using the information in the following table, the minimum total completion time required for carrying out both the operations for all 7 parts is __________ hours.
\begin{array}{|c|c|c|} \hline \text{Part} & \text{Milling (hours)} &\text{Polishing (hours)} \\ \hline P1 & 8 &6 \\ \hline P2 & 3& 2\\ \hline P3& 3& 4\\ \hline P4& 4& 6\\ \hline P5& 5&7 \\ \hline P6& 6 & 4\\ \hline P7& 2& 1 \\ \hline \end{array}
A
31
B
33
C
30
D
32
GATE ME 2022 SET-2      Production Planning and Control
Question 3 Explanation: 
Sequencing model :
\begin{array}{|c|c|c|}\hline \text{Part}&\text{Milling (hours)}& \text{Polishing (hours)} \\ \hline P_1&8&6\\ \hline P_2&3&2\\ \hline P_3&3&4\\ \hline P_4&4&6\\ \hline P_5&5&7\\ \hline P_6&6&4\\ \hline P_7&2&1\\ \hline \end{array}
Optimum job sequence (According to Johnson's algorithm)
P_3,P_4,P_5,P_1,P_6,P_2,P_7
Optimum Make-span (Minimum total completion time)
\begin{array}{|c|c|c|c|c|c|c|}\hline \text{Part}& \text{Milling}& & &\text{Polishing}&& \\ \hline & \text{In time}& \text{Processing time} &\text{Out time}&\text{In time}& \text{Processing time} &\text{Out time}\\ \hline P_3& 0 & 3 & 3 & 3 &4 &7\\ \hline P_4& 3 & 4 & 7 & 7 &6 &13\\ \hline P_5& 7 & 5 & 12 & 13 &7 &20\\ \hline P_1& 12 & 8 & 20 & 20 &6 &26\\ \hline P_6& 20 & 6 & 26 & 26 &4 &30\\ \hline P_2& 26 & 3 & 29 & 30 &2 &32\\ \hline P_7& 29 & 2 & 31 & 32 &1 &33\\ \hline \end{array}
Minimum total completion time of all parts = 33 hours
Question 4
The demand of a certain part is 1000 parts/year and its cost is Rs. 1000/part. The orders are placed based on the economic order quantity (EOQ). The cost of ordering is Rs. 100/order and the lead time for receiving the orders is 5 days. If the holding cost is Rs. 20/part/year, the inventory level for placing the orders is ________ parts (round off to the nearest integer).
A
14
B
8
C
18
D
22
GATE ME 2022 SET-2      Inventory Control
Question 4 Explanation: 
Inventory control:
Annual demand (D) = 1000 units
Lead time (LT) = 5 days
Inventory level for placing the order = Re-order level (ROL)
Rate of consumption (d) =D/365
ROL=d \times LT=\frac{1000}{365} \times 5=13.69 \approx 14 \; units

Question 5
An electric car manufacturer underestimated the January sales of car by 20 units, while the actual sales was 120 units. If the manufacturer uses exponential smoothing method with a smoothing constant of \alpha = 0.2, then the sales forecast for the month of February of the same year is _______units (in integer).
A
96
B
104
C
120
D
128
GATE ME 2022 SET-2      Production Planning and Control
Question 5 Explanation: 
Actual sales in Jan ( D_{Jan}) = 120 units
Under estimate in Jan = D_{Jan}- F_{Jan}=20
D_{Jan} = 100 units
Smoothing constant ( \alpha = 0.2)
Forecast for Feb ( D_{Jan}) = ?
F_{Feb} = F_{Jan} + \alpha (D_{Jan} - F_{Jan})= 100 + 0.2 (120 - 100) = 104 \;units
Question 6
The product structure diagram shows the number of different components required at each level to produce one unit of the final product P. If there are 50 units of on-hand inventory of component A, the number of additional units of component A needed to produce 10 units of product P is _________ (in integer).

A
160
B
50
C
110
D
170
GATE ME 2022 SET-1      Inventory Control
Question 6 Explanation: 


To produce 10 units of 'P'
No. of units of 'A' required = (4x10)+(2x3x2x10) = 160 units
Net requirement of 'A' = 160 - 50 = 110 units
Question 7
Activities A to K are required to complete a project. The time estimates and the immediate predecessors of these activities are given in the table. If the project is to be completed in the minimum possible time, the latest finish time for the activity G is _____ hours.
\begin{array}{|c|c|c|}\hline \text{Activity} &\text{Time (hours)} & \text{Immediate predecessors}\\ \hline A &2 &- \\ \hline B& 3 &- \\ \hline C& 2 &- \\ \hline D & 4 &A \\ \hline E & 5 & B\\ \hline F & 4 & B\\ \hline G & 3 & C\\ \hline H & 10 &D,E \\ \hline I & 5 & F\\ \hline J & 8 & G\\ \hline K & 3 &H,I,J\\ \hline \end{array}
A
5
B
10
C
8
D
9
GATE ME 2022 SET-1      PERT and CPM
Question 7 Explanation: 


Activity 'G'

LFT of activity 'G' = 10 hours
Question 8
Which one of the following is NOT a form of inventory?
A
Raw materials
B
Work-in-process materials
C
Finished goods
D
CNC Milling Machines
GATE ME 2022 SET-1      Inventory Control
Question 8 Explanation: 


CNC milling machines will not be treated as inventory.
Question 9
In a linear programming problem, if a resource is not fully utilized, the shadow price of that resource is
A
positive
B
negative
C
zero
D
infinity
GATE ME 2022 SET-1      Linear Programming
Question 9 Explanation: 
In a Linear programming problem if a resource is not fully utilized. The shadow price or dual price of the particular resource is zero.
Question 10
The demand and forecast of an item for five months are given in the table.

\begin{array}{|c|c|c|}\hline \textbf{Month} & \textbf{Demand} & \textbf{Forecast} \\ \hline \text{April} & \text{225} & \text{200} \\ \hline \text{May} & \text{220} & \text{240}\\ \hline \text{June} & \text{285} & \text{300} \\ \hline \text{July} & \text{290} & \text{270} \\ \hline \text{August} & \text{250} & \text{230} \\ \hline \end{array}

The Mean Absolute Percent Error (MAPE) in the forecast is _______% (round off to two decimal places)
A
4.45
B
12.25
C
18.42
D
8.08
GATE ME 2021 SET-2      Production Planning and Control
Question 10 Explanation: 
\begin{array}{|c|c|c|c|c|} \hline \text { March } & \mathrm{Di} & \mathrm{Fi} & \mathrm{ei} & \left|\frac{\mathrm{ei}}{\mathrm{Di}} \times 100\right| \\ \hline \text { April } & 225 & 200 & 25 & 11.11 \% \\ \hline \text { May } & 220 & 240 & -20 & 9.09 \% \\ \hline \text { June } & 285 & 300 & -15 & 5.26 \% \\ \hline \text { July } & 290 & 270 & 20 & 6.896 \% \\ \hline \text { August } & 250 & 230 & 20 & 8.0 \% \\ \hline & & & & \sum \frac{e i}{D i} \times 100 \mid=40.356 \\ \hline \end{array}
\text{MAPE}=\frac{\sum\left|\frac{e i}{D i} \times 100\right|}{n}=8.0712 \%


There are 10 questions to complete.

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