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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Linear Programming
BITSAT 2015 · Mathematics · Linear Programming
Minimise Z=∑_ j=1 ^ n ∑_ i=1 ^ m c_ i j x_ i j Subject to ∑_ i=1 ^...
Minimise
Z
=
∑
j
=
1
n
∑
i
=
1
m
c
i
j
x
i
j
Z=\sum_{j=1}^{n} \sum_{i=1}^{m} c_{i j} x_{i j}
Z
=
∑
j
=
1
n
∑
i
=
1
m
c
ij
x
ij
Subject to
∑
i
=
1
m
x
i
j
=
b
j
,
j
=
1
,
2
,
…
,
n
∑
j
=
1
n
x
i
j
=
b
j
,
i
=
1
,
2
,
…
,
m
\sum_{i=1}^{m} x_{i j}=b_{j}, j=1,2, \ldots, n\sum_{j=1}^{n} x_{i j}=b_{j}, i=1,2, \ldots, m
∑
i
=
1
m
x
ij
=
b
j
,
j
=
1
,
2
,
…
,
n
∑
j
=
1
n
x
ij
=
b
j
,
i
=
1
,
2
,
…
,
m
is a
L
P
P
L P P
L
PP
with number of constr
Options
A
m
−
n
m-n
m
−
n
B
m
m
m
C
m
+
n
\mathrm{m}+\mathrm{n}
m
+
n
D
m
n
\frac{\mathrm{m}}{\mathrm{n}}
n
m
Correct answer
C.
m
+
n
\mathrm{m}+\mathrm{n}
m
+
n
Step-by-step solution
Constraints will be
x
11
+
x
21
+
…
.
+
x
m
l
=
b
1
x
12
+
x
22
+
…
…
x
m
2
=
b
2
x
ln
+
x
2
n
+
…
.
+
x
m
n
=
b
n
x
11
+
x
12
+
…
+
x
1
n
=
b
1
x
21
+
x
22
+
…
+
x
2
n
=
b
2
x
m
l
+
x
m
2
+
…
+
x
m
n
=
b
n
\mathrm{x}_{11}+\mathrm{x}_{21}+\ldots.+\mathrm{x}_{\mathrm{ml}}=\mathrm{b}_{1} \mathrm{x}_{12}+\mathrm{x}_{22}+\ldots \ldots \mathrm{x}_{\mathrm{m} 2}=\mathrm{b}_{2}\mathrm{x}_{\ln }+\mathrm{x}_{2 \mathrm{n}}+\ldots.+\mathrm{x}_{\mathrm{mn}}=\mathrm{b}_{\mathrm{n}}\mathrm{x}_{11}+\mathrm{x}_{12}+\ldots+\mathrm{x}_{1 \mathrm{n}}=\mathrm{b}_{1}\mathrm{x}_{21}+\mathrm{x}_{22}+\ldots+\mathrm{x}_{2 \mathrm{n}}=\mathrm{b}_{2}\mathrm{x}_{\mathrm{ml}}+\mathrm{x}_{\mathrm{m} 2}+\ldots+\mathrm{x}_{\mathrm{mn}}=\mathrm{b}_{\mathrm{n}}
x
11
+
x
21
+
…
.
+
x
ml
=
b
1
x
12
+
x
22
+
……
x
m
2
=
b
2
x
l
n
+
x
2
n
+
…
.
+
x
mn
=
b
n
x
11
+
x
12
+
…
+
x
1
n
=
b
1
x
21
+
x
22
+
…
+
x
2
n
=
b
2
x
ml
+
x
m
2
+
…
+
x
mn
=
b
n
So, total number of constraints
=
m
+
n
=\mathrm{m}+\mathrm{n}
=
m
+
n
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