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}
\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}
12 | |
15 | |
16 | |
18 |
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

\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}
\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}
5000 | |
4000 | |
3000 | |
6000 |
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}
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}
\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}
31 | |
33 | |
30 | |
32 |
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
\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).
14 | |
8 | |
18 | |
22 |
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

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).
96 | |
104 | |
120 | |
128 |
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
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).


160 | |
50 | |
110 | |
170 |
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}
\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}
5 | |
10 | |
8 | |
9 |
Question 7 Explanation:

Activity 'G'

LFT of activity 'G' = 10 hours
Question 8 |
Which one of the following is NOT a form of
inventory?
Raw materials | |
Work-in-process materials | |
Finished goods | |
CNC Milling Machines |
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
positive | |
negative | |
zero | |
infinity |
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)
\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)
4.45 | |
12.25 | |
18.42 | |
8.08 |
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 \%
\text{MAPE}=\frac{\sum\left|\frac{e i}{D i} \times 100\right|}{n}=8.0712 \%
There are 10 questions to complete.
Hello,
I believe the answer to question 8 is option D (CNC Milling Machines). Kindly make the changes accordingly.
Love your effort.
Dear Suraj,
We have updated the answer as D.