Question 1 |
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 1 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 \%
Question 2 |
Daily production capacity of a bearing manufacturing company is 30000 bearings. The daily demand of the bearing is 15000. The holding cost per year of keeping a bearing in the inventory is Rs. 20. The setup cost for the production of a batch is Rs. 1800. Assuming 300 working days in a year, the economic batch quantity in number of bearings is ______ (in integer).
36254 | |
20145 | |
40250 | |
42145 |
Question 2 Explanation:
\begin{aligned} Q^{\star} &=\sqrt{\frac{2 D \times C_{0}}{C_{h}} \times \frac{P}{P-d}} \\ &=\sqrt{\frac{2 \times 15000 \times 300 \times 1800}{20} \times\left(\frac{30000}{30000-15000}\right)} \\ &=40249.2 \simeq 40250 \text { units } \end{aligned}
Question 3 |
A factory produces m(i=1,2,...m) products, each of which requires processing on n(j=1,2,...n) workstations. Let a_{ij} be the amount of processing time that one unit of the i^{th} product requires on the j^{th} workstation. Let the revenue from selling one unit of the i^{th} product be r_i and h_i be the holding cost per unit per time period for the i^{th} product. The planning horizon consists of T \;(t=1,2,...T) time periods. The minimum demand that must be satisfied in time period t is d_{it}, and the capacity of the j^{th} workstation in time period t is c_{jt}. Consider the aggregate planning formulation below, with decision variables S_{it} (amount of product i sold in time period t ), X_{it} (amount of product i manufactured in time period t ) and I_{it} (amount of product i held in inventory at the end of time period t.
max\sum_{t=1}^{T}\sum_{i=1}^{m}(r_iS_{it}-h_iI_{it})
subject to
S_{it}\geq d_{it}\;\;\forall i,t
< capacity constraint >
< inventory balance constraint >
X_{it},S_{it}, I_{it} \geq 0;\; I_{i0}=0
The capacity constraints and inventory balance constraints for this formulation are
max\sum_{t=1}^{T}\sum_{i=1}^{m}(r_iS_{it}-h_iI_{it})
subject to
S_{it}\geq d_{it}\;\;\forall i,t
< capacity constraint >
< inventory balance constraint >
X_{it},S_{it}, I_{it} \geq 0;\; I_{i0}=0
The capacity constraints and inventory balance constraints for this formulation are
\sum_{j}^{m}a_{ij}X_{it}\leq c_{jt}\;\forall \; i,t\text{ and }I_{it}=I_{i,t-1}+X_{it}-d_{it}\; \forall \;i,t | |
\sum_{i}^{m}a_{ij}X_{it}\leq c_{jt}\;\forall \; i,t\text{ and }I_{it}=I_{i,t-1}+X_{it}-d_{it}\; \forall \;i,t | |
\sum_{i}^{m}a_{ij}X_{it}\leq d_{it}\;\forall \; i,t\text{ and }I_{it}=I_{i,t-1}+X_{it}-S_{it}\; \forall \;i,t | |
\sum_{i}^{m}a_{ij}X_{it}\leq d_{it}\;\forall \; i ,t\text{ and }I_{it}=I_{i,t-1}+S_{it}-X_{it}\; \forall \;i,t |
Question 3 Explanation:
\begin{aligned} m & \rightarrow i \ldots m \leftarrow \text { product } \\ n & \rightarrow i \ldots n \leftarrow \text { workstation } \\ a_{\bar{j}} & \rightarrow \text { time } \end{aligned}
r_{i} \rightarrow selling price
h_{i} \rightarrow holding cost
T \rightarrow t=1,2, \ldots T
d_{i t} \rightarrow demand of product in time t
c_{j t} \rightarrow capacity of workstation in time t
S_{i t} \rightarrow Number of product sold in time t
x_{i t} \rightarrow Number of product produced in time t
I_{i t} \rightarrow Number of product i hold in inventory at end of period t
Capacity constraint
a_{i j} x_{i t} \leq c_{j t}
Inventory constraint
l_{i t}=I_{i, t-1}+x_{i t}-S_{i t}
r_{i} \rightarrow selling price
h_{i} \rightarrow holding cost
T \rightarrow t=1,2, \ldots T
d_{i t} \rightarrow demand of product in time t
c_{j t} \rightarrow capacity of workstation in time t
S_{i t} \rightarrow Number of product sold in time t
x_{i t} \rightarrow Number of product produced in time t
I_{i t} \rightarrow Number of product i hold in inventory at end of period t
Capacity constraint
a_{i j} x_{i t} \leq c_{j t}
Inventory constraint
l_{i t}=I_{i, t-1}+x_{i t}-S_{i t}
Question 4 |
A PERT network has 9 activities on its critical path. The standard deviation of each activity on the critical path is 3. The standard deviation of the critical path is
3 | |
9 | |
27 | |
81 |
Question 4 Explanation:
In CPM,
\begin{array}{l} \sigma=\sqrt{\text { sum of variance along critical path }} \\ \sigma=\sqrt{\sigma^{2}+\sigma^{2}+\ldots .+\sigma^{2}} \\ \sigma=\sqrt{9 \sigma^{2}}=\sqrt{9 \times 9}=9 \end{array}
\begin{array}{l} \sigma=\sqrt{\text { sum of variance along critical path }} \\ \sigma=\sqrt{\sigma^{2}+\sigma^{2}+\ldots .+\sigma^{2}} \\ \sigma=\sqrt{9 \sigma^{2}}=\sqrt{9 \times 9}=9 \end{array}
Question 5 |
Activities A, B, C and D form the critical path for a project with a PERT network. The means and variances of the activity duration for each activity are given below. All activity durations follow the Gaussian (normal) distribution, and are independent of each other.
\begin{array}{|l|c|c|c|c|}\hline \text{Activity} & \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \text{Mean (days)} & \text{6} & \text{11} & \text{8} & \text{15} \\ \hline \text{Variance (days$^{2})$} & \text{4} & \text{9} & \text{4} & \text{9}\\ \hline \end{array}
The probability that the project will be completed within 40 days is _____ (round off to two decimal places).
(Note: Probability is a number between 0 and 1).
\begin{array}{|l|c|c|c|c|}\hline \text{Activity} & \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \text{Mean (days)} & \text{6} & \text{11} & \text{8} & \text{15} \\ \hline \text{Variance (days$^{2})$} & \text{4} & \text{9} & \text{4} & \text{9}\\ \hline \end{array}
The probability that the project will be completed within 40 days is _____ (round off to two decimal places).
(Note: Probability is a number between 0 and 1).
0.25 | |
0.5 | |
0.65 | |
0.85 |
Question 5 Explanation:
PERT-CPM
\begin{aligned} T_{S} &=40 \text { days, } T_{E}=6+11+8+15 \\ T_{E} &=40 \text { days, } \\ 2 &=\frac{T_{S}-T_{E}}{\sigma}=0 \rightarrow 50 \% \end{aligned}
Probability of completing project in expected time is always 0.5.
Question 6 |
A set of jobs A, B, C, D, E, F, G, H arrive at time t = 0 for processing on turning and grinding machines. Each job needs to be processed in sequence
- first on the turning machine and second on the grinding machine, and the grinding must occur immediately after turning. The processing times of the jobs are given below.
\begin{array}{|l|l|l|l|l|l|l|l|l|}\hline \text{Job} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{G} & \text{H} \\ \hline \text{Turning (minutes)} & \text{2} & \text{4} &\text{8}&\text{9} &\text{7} &\text{6} &\text{5} &\text{10} \\ \hline \text{Grinding (minutes)} & \text{6} & \text{1} &\text{3} &\text{7} &\text{9}&\text{5} &\text{2} &\text{4} \\ \hline \end{array}
If the makespan is to be minimized, then the optimal sequence in which these jobs must be processed on the turning and grinding machines is
\begin{array}{|l|l|l|l|l|l|l|l|l|}\hline \text{Job} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{G} & \text{H} \\ \hline \text{Turning (minutes)} & \text{2} & \text{4} &\text{8}&\text{9} &\text{7} &\text{6} &\text{5} &\text{10} \\ \hline \text{Grinding (minutes)} & \text{6} & \text{1} &\text{3} &\text{7} &\text{9}&\text{5} &\text{2} &\text{4} \\ \hline \end{array}
If the makespan is to be minimized, then the optimal sequence in which these jobs must be processed on the turning and grinding machines is
A-E-D-F-H-C-G-B | |
A-D-E-F-H-C-G-B | |
G-E-D-F-H-C-A-B | |
B-G-C-H-F-D-E-A |
Question 6 Explanation:
Sequencing,
\begin{array}{|c|c|c|} \hline A & (2) & 6 \\ \hline B & 4 & (1) \\ \hline C & 8 & (3) \\ \hline D & 9 & (7) \\ \hline E & (7) & 9 \\ \hline F & 6 & (5) \\ \hline G & 5 & (2) \\ \hline H & 10 & (4) \\ \hline \end{array}
AEDFHCGB
\begin{array}{|c|c|c|} \hline A & (2) & 6 \\ \hline B & 4 & (1) \\ \hline C & 8 & (3) \\ \hline D & 9 & (7) \\ \hline E & (7) & 9 \\ \hline F & 6 & (5) \\ \hline G & 5 & (2) \\ \hline H & 10 & (4) \\ \hline \end{array}
AEDFHCGB
Question 7 |
For a single item inventory system, the demand is continuous, which is 10000 per year.
The replacement is instantaneous and backorders (S units) per cycle are allowed as
shown in the figure.
As soon as the quantity (Q units) ordered from the supplier is received, the backordered quantity is issued to the customers. The ordering cost is Rs. 300 per order. The carrying cost is Rs. 4 per unit per year. The cost of backordering is Rs. 25 per unit per year. Based on the total cost minimization criteria, the maximum inventory reached in the system is ________ (round off to nearest integer).
As soon as the quantity (Q units) ordered from the supplier is received, the backordered quantity is issued to the customers. The ordering cost is Rs. 300 per order. The carrying cost is Rs. 4 per unit per year. The cost of backordering is Rs. 25 per unit per year. Based on the total cost minimization criteria, the maximum inventory reached in the system is ________ (round off to nearest integer).
181.94 | |
1319.09 | |
1137.14 | |
1802.16 |
Question 7 Explanation:
Given data: D=10000 items/year, C_{o}=Rs. 300/order,
C_{h}=Rs. 4/unit/year, C_{b}=Rs. 25/unit/year,
For minimum tool cost
\begin{aligned} \text { Quantity ordered, } Q=& \sqrt{\frac{2 D C_{o}}{C_{h}} \times\left(\frac{C_{b}+C_{h}}{C_{o}}\right)}\\ &=\sqrt{\frac{2 \times 10000 \times 300}{4} \times\left(\frac{25+4}{25}\right)} \\ Q &=1319.09 \text { unit } \end{aligned}
Now, for minimum cost, optimum units backordered
\begin{aligned} (Q-S) \times C_{h}&=(S) \times C_{b}\\ S\left(C_{b}+C_{h}\right) &=Q \times C_{h} \\ S &=\frac{1319.09 \times 4}{(25+4)}=181.94 \text { units } \end{aligned}
Maximum inventory in the system, Q_{\max }=Q-S
\begin{aligned} &=1319.09-181.94 \\ &=1137.15 \text { unit } \\ &\approx 1137 \text { units } \end{aligned}
Alternate:
Maximum inventory in system =\sqrt{\frac{2 D C_{o}}{C_{h}} \times\left(\frac{C_{b}}{C_{b}+C_{h}}\right)}
Q_{\max }=\sqrt{\frac{2 \times 10000 \times 300}{4} \times\left(\frac{25}{29}\right)}=1137.147 \text { units }
C_{h}=Rs. 4/unit/year, C_{b}=Rs. 25/unit/year,
For minimum tool cost
\begin{aligned} \text { Quantity ordered, } Q=& \sqrt{\frac{2 D C_{o}}{C_{h}} \times\left(\frac{C_{b}+C_{h}}{C_{o}}\right)}\\ &=\sqrt{\frac{2 \times 10000 \times 300}{4} \times\left(\frac{25+4}{25}\right)} \\ Q &=1319.09 \text { unit } \end{aligned}
Now, for minimum cost, optimum units backordered
\begin{aligned} (Q-S) \times C_{h}&=(S) \times C_{b}\\ S\left(C_{b}+C_{h}\right) &=Q \times C_{h} \\ S &=\frac{1319.09 \times 4}{(25+4)}=181.94 \text { units } \end{aligned}
Maximum inventory in the system, Q_{\max }=Q-S
\begin{aligned} &=1319.09-181.94 \\ &=1137.15 \text { unit } \\ &\approx 1137 \text { units } \end{aligned}
Alternate:
Maximum inventory in system =\sqrt{\frac{2 D C_{o}}{C_{h}} \times\left(\frac{C_{b}}{C_{b}+C_{h}}\right)}
Q_{\max }=\sqrt{\frac{2 \times 10000 \times 300}{4} \times\left(\frac{25}{29}\right)}=1137.147 \text { units }
Question 8 |
The forecast for the monthly demand of a product is given in the table below.
The forecast is made by using the exponential smoothing method. The exponential smoothing coefficient used in forecasting the demand is
The forecast is made by using the exponential smoothing method. The exponential smoothing coefficient used in forecasting the demand is
0.1 | |
0.4 | |
0.5 | |
1 |
Question 8 Explanation:
\begin{aligned} F_{t}&=F_{t-1}+\alpha\left(D_{t-1}-F_{t-1}\right)\\ \text{For 2nd month }F_{t}&=31.8, \text{ for }1^{\text {st }} \text{month}\\ F_{t-1} &=32 \text { and } D_{t-1}=30 \\ 31.8 &=32+\alpha(30-32) \\ 2 \alpha &=32-31.8 \\ \alpha &=0.1 \end{aligned}
Question 9 |
Consider the following network of activities, with each activity named A-L, illustrated
in the nodes of the network
The number of hours required for each activity is shown alongside the nodes. The slack on the activity L, is ________ hours.
The number of hours required for each activity is shown alongside the nodes. The slack on the activity L, is ________ hours.
1 | |
2 | |
3 | |
4 |
Question 9 Explanation:
Time along path A-B-C-F-I-J-K = 42 hours
Time along path A-B-C-E-H-L = 31 hours
Time along path A-B-C-D-G-H-L = 40 hours
Slack for L = 42 - 40 = 2 hours
Time along path A-B-C-E-H-L = 31 hours
Time along path A-B-C-D-G-H-L = 40 hours
Slack for L = 42 - 40 = 2 hours
Question 10 |
In Materials Requirement Planning, if the inventory holding cost is very high and the setup cost is zero, which one of the following lot sizing approaches should be used?
Economic Order Quantity | |
Lot-for-Lot | |
Base Stock Level | |
Fixed Period Quantity, for 2 periods |
Question 10 Explanation:
In Lot-for-lot, holding cost is high and ordering
cost is less.
There are 10 questions to complete.