Transportation problem into initial simplex tableau












0














enter image description here



I did (b) .



For (a), I got this



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2le 5\ x_3+x_4le4\ x_1+x_3=3\ x_2+x_4ge4\ x_ige0$$



The standard form is



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2+x_5= 5\ x_3+x_4+x_6=4\ x_1+x_3=3\ x_2+x_4-x_7=4\ x_ige0$$



When writting the initial simplex tableau I found a problem because there are only 3 basic variables and 4 constraints, what do I do to solve the problem?










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  • use the two-phase simplex method
    – LinAlg
    Nov 30 '18 at 0:16










  • @LinAlg How? ${}$
    – Al t.
    Nov 30 '18 at 5:04










  • @LinAlg Do you know where can I find the matrix table representation of two-phase simplex method? In my book (Hillier) there is only the matrix table representation of the simplex.
    – Al t.
    Nov 30 '18 at 5:04










  • @Alt. Why is the last constraint an inequality?
    – callculus
    Nov 30 '18 at 15:23












  • @callculus Do you mean the $x_ige0$?
    – Al t.
    Nov 30 '18 at 19:25


















0














enter image description here



I did (b) .



For (a), I got this



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2le 5\ x_3+x_4le4\ x_1+x_3=3\ x_2+x_4ge4\ x_ige0$$



The standard form is



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2+x_5= 5\ x_3+x_4+x_6=4\ x_1+x_3=3\ x_2+x_4-x_7=4\ x_ige0$$



When writting the initial simplex tableau I found a problem because there are only 3 basic variables and 4 constraints, what do I do to solve the problem?










share|cite|improve this question






















  • use the two-phase simplex method
    – LinAlg
    Nov 30 '18 at 0:16










  • @LinAlg How? ${}$
    – Al t.
    Nov 30 '18 at 5:04










  • @LinAlg Do you know where can I find the matrix table representation of two-phase simplex method? In my book (Hillier) there is only the matrix table representation of the simplex.
    – Al t.
    Nov 30 '18 at 5:04










  • @Alt. Why is the last constraint an inequality?
    – callculus
    Nov 30 '18 at 15:23












  • @callculus Do you mean the $x_ige0$?
    – Al t.
    Nov 30 '18 at 19:25
















0












0








0







enter image description here



I did (b) .



For (a), I got this



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2le 5\ x_3+x_4le4\ x_1+x_3=3\ x_2+x_4ge4\ x_ige0$$



The standard form is



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2+x_5= 5\ x_3+x_4+x_6=4\ x_1+x_3=3\ x_2+x_4-x_7=4\ x_ige0$$



When writting the initial simplex tableau I found a problem because there are only 3 basic variables and 4 constraints, what do I do to solve the problem?










share|cite|improve this question













enter image description here



I did (b) .



For (a), I got this



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2le 5\ x_3+x_4le4\ x_1+x_3=3\ x_2+x_4ge4\ x_ige0$$



The standard form is



$$min 3x_1+2.7x_2+2.9x_3+2.8x_4\ s.t. x_1+x_2+x_5= 5\ x_3+x_4+x_6=4\ x_1+x_3=3\ x_2+x_4-x_7=4\ x_ige0$$



When writting the initial simplex tableau I found a problem because there are only 3 basic variables and 4 constraints, what do I do to solve the problem?







optimization linear-programming simplex operations-research






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asked Nov 29 '18 at 21:47









Al t.Al t.

4291519




4291519












  • use the two-phase simplex method
    – LinAlg
    Nov 30 '18 at 0:16










  • @LinAlg How? ${}$
    – Al t.
    Nov 30 '18 at 5:04










  • @LinAlg Do you know where can I find the matrix table representation of two-phase simplex method? In my book (Hillier) there is only the matrix table representation of the simplex.
    – Al t.
    Nov 30 '18 at 5:04










  • @Alt. Why is the last constraint an inequality?
    – callculus
    Nov 30 '18 at 15:23












  • @callculus Do you mean the $x_ige0$?
    – Al t.
    Nov 30 '18 at 19:25




















  • use the two-phase simplex method
    – LinAlg
    Nov 30 '18 at 0:16










  • @LinAlg How? ${}$
    – Al t.
    Nov 30 '18 at 5:04










  • @LinAlg Do you know where can I find the matrix table representation of two-phase simplex method? In my book (Hillier) there is only the matrix table representation of the simplex.
    – Al t.
    Nov 30 '18 at 5:04










  • @Alt. Why is the last constraint an inequality?
    – callculus
    Nov 30 '18 at 15:23












  • @callculus Do you mean the $x_ige0$?
    – Al t.
    Nov 30 '18 at 19:25


















use the two-phase simplex method
– LinAlg
Nov 30 '18 at 0:16




use the two-phase simplex method
– LinAlg
Nov 30 '18 at 0:16












@LinAlg How? ${}$
– Al t.
Nov 30 '18 at 5:04




@LinAlg How? ${}$
– Al t.
Nov 30 '18 at 5:04












@LinAlg Do you know where can I find the matrix table representation of two-phase simplex method? In my book (Hillier) there is only the matrix table representation of the simplex.
– Al t.
Nov 30 '18 at 5:04




@LinAlg Do you know where can I find the matrix table representation of two-phase simplex method? In my book (Hillier) there is only the matrix table representation of the simplex.
– Al t.
Nov 30 '18 at 5:04












@Alt. Why is the last constraint an inequality?
– callculus
Nov 30 '18 at 15:23






@Alt. Why is the last constraint an inequality?
– callculus
Nov 30 '18 at 15:23














@callculus Do you mean the $x_ige0$?
– Al t.
Nov 30 '18 at 19:25






@callculus Do you mean the $x_ige0$?
– Al t.
Nov 30 '18 at 19:25












2 Answers
2






active

oldest

votes


















1














I did a part of (a) and (c).



(a) I almost agree with your answer.



Introduce the following variables.



$x_1$ - the number of pints bought from Dick today,



$x_2$ - the number of pints bought from Dick tomorrow,



$x_3$ - the number of pints bought from Harry today,



$x_4$ - the number of pints bought from Harry tomorrow.



Remark that even if Tom can keep brew bought today for tomorrow, this is wrong, because he can buy it cheaper tomorrow. So without loss of generality we can assume that Tom buys $3$ pints of brew today and $4$ pints of brew and tomorrow. Then we have the following linear programming (LP) problem.



$min 3x_1+2.7x_2+2.9x_3+2.8x_4\
s.t. x_1+x_2le 5
\ x_3+x_4le 4
\ x_1+x_3=3
\ x_2+x_4=4
\ x_ige 0.$



(c) Replacing $x_3$ by $3-x_1$ and $x_4$ by $4-x_2$ we obtain the following LP problem



$min f(x_1,x_2)=0.1x_1-0.1x_2+19.9\
s.t. 3le x_1+x_2le 5
\ 0le x_1le 3
\ 0le x_2le 4.$



This LP problem has only two variables and can be solved graphically.



enter image description here



The domain of $(x_1,x_2)$ lies between the black segments.



The lines with $f(x_1, x_2) =operatorname{const}$ are parallel to the straight line $x_1=x_2$, thus the minimum of $f$ on the domain of $(x_1,x_2)$ is achieved when $x_1=0$ and $x_2=4$.



Thus the final answer is $x_1=0$, $x_2=4$, $x_3=3$, $x_4=0$. In this case $f=19.5$.






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  • Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
    – Al t.
    Dec 4 '18 at 20:55










  • How would you construct the initial simplex tableau of the LP in (c) ?
    – Al t.
    Dec 4 '18 at 20:56












  • @Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
    – Alex Ravsky
    Dec 4 '18 at 23:54










  • Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
    – Al t.
    Dec 6 '18 at 2:12





















1





+50









Your initial linear programming problem is equivalent to



$$min 3x_1 + 2.7 x_2 + 2.9x_3 + 2.8 x_4 + Mx_7 + Mx_8$$
subject to $$x_1+x_2 + x_5 = 5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3+x_7 =3$$
$$x_2+x_4+x_8 = 4$$
$$x ge 0$$



The big $M$ is going to force $x_7$ and $x_8$ to be $0$ at optimal value and you can begin the simplex at $x = (0,0,0,0,5,4,3,4)$.



To formulate this as a transportation problem, one possible way is to imagine that on the third day, all the remaining beers are going to be donated for free. Hence the total demand is equal to the total supply and the objective function is preserved. The parameter table can be listed as follows. The last column represents the supply amount and the last row represents the demand amount. The other indicates the corresponding cost.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & 3 & 2.7 & 0 & 5 \ hline
S_2 & 2.9 & 2.8 & 0 & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



To solve the problem, you can reduce it to a $2$ dimensional problem and solve it.



Or just for fun, you can do the following:



In each of the boxes below, I use the convention of $(c_{ij}, y_{ij})$ where $y_{ij}=x_{ij}$ if it is basic and $y_{ij}=c_{ij}-u_i-v_j$ otherwise. I will use purple to indicate the first case.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}3) & (2.7,color{purple}2) & (0,0.1) & 5 \ hline
S_2 & (2.9, - 0.2) & (2.8,color{purple}2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



It is not optimal as we can see a negative reduce cost. That will be the entering variable.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}1) & (2.7,color{purple}4) & (0,-0.1) & 5 \ hline
S_2 & (2.9, color{purple}2) & (2.8,0.2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, 0.1) & (2.7,color{purple}4) & (0,color{purple}1) & 5 \ hline
S_2 & (2.9, color{purple}3) & (2.8,0.1) & (0,color{purple}1) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



Now, all the reduce cost are nonnegative. The solution suggest that on the first day, we should get $3$ pints at $2.9$ and on the second day, we should get $4$ pints at $2.7$.



Edit:



Let me avoid using big $M$ method since you are not familiar with it.



To implement the simplex algorithm, we need to find our basis matrix.



These are our linearity constraints.
$$x_1+x_2+x_5=5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3 =3$$
$$x_2+x_4 = 4$$
$$x ge 0$$



We want to pick $4$ non-basic variables. For example, let's pick $x_1=3, x_2=4$, let $x_3=0$ and $x_4=0$. Now, we can let $x_5=2$ and $x_6=4$.



That is we pick $x_1, x_2, x_5, x_6$ to be non-basic. Then our $$B = begin{bmatrix}1 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0\
0 & 1 & 0 & 0 end{bmatrix}.$$



From there we can easily compute $-c_B'B^{-1}A, c'-C_B'B^{-1}A, B^{-1}b, B^{-1}A$. That is we can construct our simplex tableu and start from there.



Remark:






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  • Thank you for your answer Siong Thye Goh.
    – Al t.
    Dec 5 '18 at 19:38










  • Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
    – Al t.
    Dec 7 '18 at 18:51












  • so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
    – Siong Thye Goh
    Dec 7 '18 at 19:18










  • I don't see how your comment answer my question in the comment
    – Al t.
    Dec 8 '18 at 18:30






  • 1




    I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
    – Siong Thye Goh
    Dec 9 '18 at 7:34











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2 Answers
2






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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














I did a part of (a) and (c).



(a) I almost agree with your answer.



Introduce the following variables.



$x_1$ - the number of pints bought from Dick today,



$x_2$ - the number of pints bought from Dick tomorrow,



$x_3$ - the number of pints bought from Harry today,



$x_4$ - the number of pints bought from Harry tomorrow.



Remark that even if Tom can keep brew bought today for tomorrow, this is wrong, because he can buy it cheaper tomorrow. So without loss of generality we can assume that Tom buys $3$ pints of brew today and $4$ pints of brew and tomorrow. Then we have the following linear programming (LP) problem.



$min 3x_1+2.7x_2+2.9x_3+2.8x_4\
s.t. x_1+x_2le 5
\ x_3+x_4le 4
\ x_1+x_3=3
\ x_2+x_4=4
\ x_ige 0.$



(c) Replacing $x_3$ by $3-x_1$ and $x_4$ by $4-x_2$ we obtain the following LP problem



$min f(x_1,x_2)=0.1x_1-0.1x_2+19.9\
s.t. 3le x_1+x_2le 5
\ 0le x_1le 3
\ 0le x_2le 4.$



This LP problem has only two variables and can be solved graphically.



enter image description here



The domain of $(x_1,x_2)$ lies between the black segments.



The lines with $f(x_1, x_2) =operatorname{const}$ are parallel to the straight line $x_1=x_2$, thus the minimum of $f$ on the domain of $(x_1,x_2)$ is achieved when $x_1=0$ and $x_2=4$.



Thus the final answer is $x_1=0$, $x_2=4$, $x_3=3$, $x_4=0$. In this case $f=19.5$.






share|cite|improve this answer























  • Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
    – Al t.
    Dec 4 '18 at 20:55










  • How would you construct the initial simplex tableau of the LP in (c) ?
    – Al t.
    Dec 4 '18 at 20:56












  • @Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
    – Alex Ravsky
    Dec 4 '18 at 23:54










  • Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
    – Al t.
    Dec 6 '18 at 2:12


















1














I did a part of (a) and (c).



(a) I almost agree with your answer.



Introduce the following variables.



$x_1$ - the number of pints bought from Dick today,



$x_2$ - the number of pints bought from Dick tomorrow,



$x_3$ - the number of pints bought from Harry today,



$x_4$ - the number of pints bought from Harry tomorrow.



Remark that even if Tom can keep brew bought today for tomorrow, this is wrong, because he can buy it cheaper tomorrow. So without loss of generality we can assume that Tom buys $3$ pints of brew today and $4$ pints of brew and tomorrow. Then we have the following linear programming (LP) problem.



$min 3x_1+2.7x_2+2.9x_3+2.8x_4\
s.t. x_1+x_2le 5
\ x_3+x_4le 4
\ x_1+x_3=3
\ x_2+x_4=4
\ x_ige 0.$



(c) Replacing $x_3$ by $3-x_1$ and $x_4$ by $4-x_2$ we obtain the following LP problem



$min f(x_1,x_2)=0.1x_1-0.1x_2+19.9\
s.t. 3le x_1+x_2le 5
\ 0le x_1le 3
\ 0le x_2le 4.$



This LP problem has only two variables and can be solved graphically.



enter image description here



The domain of $(x_1,x_2)$ lies between the black segments.



The lines with $f(x_1, x_2) =operatorname{const}$ are parallel to the straight line $x_1=x_2$, thus the minimum of $f$ on the domain of $(x_1,x_2)$ is achieved when $x_1=0$ and $x_2=4$.



Thus the final answer is $x_1=0$, $x_2=4$, $x_3=3$, $x_4=0$. In this case $f=19.5$.






share|cite|improve this answer























  • Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
    – Al t.
    Dec 4 '18 at 20:55










  • How would you construct the initial simplex tableau of the LP in (c) ?
    – Al t.
    Dec 4 '18 at 20:56












  • @Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
    – Alex Ravsky
    Dec 4 '18 at 23:54










  • Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
    – Al t.
    Dec 6 '18 at 2:12
















1












1








1






I did a part of (a) and (c).



(a) I almost agree with your answer.



Introduce the following variables.



$x_1$ - the number of pints bought from Dick today,



$x_2$ - the number of pints bought from Dick tomorrow,



$x_3$ - the number of pints bought from Harry today,



$x_4$ - the number of pints bought from Harry tomorrow.



Remark that even if Tom can keep brew bought today for tomorrow, this is wrong, because he can buy it cheaper tomorrow. So without loss of generality we can assume that Tom buys $3$ pints of brew today and $4$ pints of brew and tomorrow. Then we have the following linear programming (LP) problem.



$min 3x_1+2.7x_2+2.9x_3+2.8x_4\
s.t. x_1+x_2le 5
\ x_3+x_4le 4
\ x_1+x_3=3
\ x_2+x_4=4
\ x_ige 0.$



(c) Replacing $x_3$ by $3-x_1$ and $x_4$ by $4-x_2$ we obtain the following LP problem



$min f(x_1,x_2)=0.1x_1-0.1x_2+19.9\
s.t. 3le x_1+x_2le 5
\ 0le x_1le 3
\ 0le x_2le 4.$



This LP problem has only two variables and can be solved graphically.



enter image description here



The domain of $(x_1,x_2)$ lies between the black segments.



The lines with $f(x_1, x_2) =operatorname{const}$ are parallel to the straight line $x_1=x_2$, thus the minimum of $f$ on the domain of $(x_1,x_2)$ is achieved when $x_1=0$ and $x_2=4$.



Thus the final answer is $x_1=0$, $x_2=4$, $x_3=3$, $x_4=0$. In this case $f=19.5$.






share|cite|improve this answer














I did a part of (a) and (c).



(a) I almost agree with your answer.



Introduce the following variables.



$x_1$ - the number of pints bought from Dick today,



$x_2$ - the number of pints bought from Dick tomorrow,



$x_3$ - the number of pints bought from Harry today,



$x_4$ - the number of pints bought from Harry tomorrow.



Remark that even if Tom can keep brew bought today for tomorrow, this is wrong, because he can buy it cheaper tomorrow. So without loss of generality we can assume that Tom buys $3$ pints of brew today and $4$ pints of brew and tomorrow. Then we have the following linear programming (LP) problem.



$min 3x_1+2.7x_2+2.9x_3+2.8x_4\
s.t. x_1+x_2le 5
\ x_3+x_4le 4
\ x_1+x_3=3
\ x_2+x_4=4
\ x_ige 0.$



(c) Replacing $x_3$ by $3-x_1$ and $x_4$ by $4-x_2$ we obtain the following LP problem



$min f(x_1,x_2)=0.1x_1-0.1x_2+19.9\
s.t. 3le x_1+x_2le 5
\ 0le x_1le 3
\ 0le x_2le 4.$



This LP problem has only two variables and can be solved graphically.



enter image description here



The domain of $(x_1,x_2)$ lies between the black segments.



The lines with $f(x_1, x_2) =operatorname{const}$ are parallel to the straight line $x_1=x_2$, thus the minimum of $f$ on the domain of $(x_1,x_2)$ is achieved when $x_1=0$ and $x_2=4$.



Thus the final answer is $x_1=0$, $x_2=4$, $x_3=3$, $x_4=0$. In this case $f=19.5$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 17:22

























answered Dec 3 '18 at 17:16









Alex RavskyAlex Ravsky

39.4k32181




39.4k32181












  • Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
    – Al t.
    Dec 4 '18 at 20:55










  • How would you construct the initial simplex tableau of the LP in (c) ?
    – Al t.
    Dec 4 '18 at 20:56












  • @Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
    – Alex Ravsky
    Dec 4 '18 at 23:54










  • Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
    – Al t.
    Dec 6 '18 at 2:12




















  • Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
    – Al t.
    Dec 4 '18 at 20:55










  • How would you construct the initial simplex tableau of the LP in (c) ?
    – Al t.
    Dec 4 '18 at 20:56












  • @Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
    – Alex Ravsky
    Dec 4 '18 at 23:54










  • Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
    – Al t.
    Dec 6 '18 at 2:12


















Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
– Al t.
Dec 4 '18 at 20:55




Thank you for your answer Alex, I agree that it should be $x_2+x_4=4$ in the LP problem.
– Al t.
Dec 4 '18 at 20:55












How would you construct the initial simplex tableau of the LP in (c) ?
– Al t.
Dec 4 '18 at 20:56






How would you construct the initial simplex tableau of the LP in (c) ?
– Al t.
Dec 4 '18 at 20:56














@Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
– Alex Ravsky
Dec 4 '18 at 23:54




@Alt First I would look in Chaps. 3 and 4 for its definition. The same concerns part (b). This should be an easy exercise, and, maybe, the book suggests obtain (c) from (b). I guess the solution of this exersise should show your knowledge of mentioned methods (which may be needed at the exam). Unfortunately, this stuff is covered by a gap in my education, so my answer is partial.
– Alex Ravsky
Dec 4 '18 at 23:54












Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
– Al t.
Dec 6 '18 at 2:12






Right I should reread chaps. 3 and 4. As mentioned before I've solved (b). Btw thank you again:)
– Al t.
Dec 6 '18 at 2:12













1





+50









Your initial linear programming problem is equivalent to



$$min 3x_1 + 2.7 x_2 + 2.9x_3 + 2.8 x_4 + Mx_7 + Mx_8$$
subject to $$x_1+x_2 + x_5 = 5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3+x_7 =3$$
$$x_2+x_4+x_8 = 4$$
$$x ge 0$$



The big $M$ is going to force $x_7$ and $x_8$ to be $0$ at optimal value and you can begin the simplex at $x = (0,0,0,0,5,4,3,4)$.



To formulate this as a transportation problem, one possible way is to imagine that on the third day, all the remaining beers are going to be donated for free. Hence the total demand is equal to the total supply and the objective function is preserved. The parameter table can be listed as follows. The last column represents the supply amount and the last row represents the demand amount. The other indicates the corresponding cost.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & 3 & 2.7 & 0 & 5 \ hline
S_2 & 2.9 & 2.8 & 0 & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



To solve the problem, you can reduce it to a $2$ dimensional problem and solve it.



Or just for fun, you can do the following:



In each of the boxes below, I use the convention of $(c_{ij}, y_{ij})$ where $y_{ij}=x_{ij}$ if it is basic and $y_{ij}=c_{ij}-u_i-v_j$ otherwise. I will use purple to indicate the first case.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}3) & (2.7,color{purple}2) & (0,0.1) & 5 \ hline
S_2 & (2.9, - 0.2) & (2.8,color{purple}2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



It is not optimal as we can see a negative reduce cost. That will be the entering variable.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}1) & (2.7,color{purple}4) & (0,-0.1) & 5 \ hline
S_2 & (2.9, color{purple}2) & (2.8,0.2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, 0.1) & (2.7,color{purple}4) & (0,color{purple}1) & 5 \ hline
S_2 & (2.9, color{purple}3) & (2.8,0.1) & (0,color{purple}1) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



Now, all the reduce cost are nonnegative. The solution suggest that on the first day, we should get $3$ pints at $2.9$ and on the second day, we should get $4$ pints at $2.7$.



Edit:



Let me avoid using big $M$ method since you are not familiar with it.



To implement the simplex algorithm, we need to find our basis matrix.



These are our linearity constraints.
$$x_1+x_2+x_5=5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3 =3$$
$$x_2+x_4 = 4$$
$$x ge 0$$



We want to pick $4$ non-basic variables. For example, let's pick $x_1=3, x_2=4$, let $x_3=0$ and $x_4=0$. Now, we can let $x_5=2$ and $x_6=4$.



That is we pick $x_1, x_2, x_5, x_6$ to be non-basic. Then our $$B = begin{bmatrix}1 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0\
0 & 1 & 0 & 0 end{bmatrix}.$$



From there we can easily compute $-c_B'B^{-1}A, c'-C_B'B^{-1}A, B^{-1}b, B^{-1}A$. That is we can construct our simplex tableu and start from there.



Remark:






share|cite|improve this answer























  • Thank you for your answer Siong Thye Goh.
    – Al t.
    Dec 5 '18 at 19:38










  • Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
    – Al t.
    Dec 7 '18 at 18:51












  • so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
    – Siong Thye Goh
    Dec 7 '18 at 19:18










  • I don't see how your comment answer my question in the comment
    – Al t.
    Dec 8 '18 at 18:30






  • 1




    I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
    – Siong Thye Goh
    Dec 9 '18 at 7:34
















1





+50









Your initial linear programming problem is equivalent to



$$min 3x_1 + 2.7 x_2 + 2.9x_3 + 2.8 x_4 + Mx_7 + Mx_8$$
subject to $$x_1+x_2 + x_5 = 5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3+x_7 =3$$
$$x_2+x_4+x_8 = 4$$
$$x ge 0$$



The big $M$ is going to force $x_7$ and $x_8$ to be $0$ at optimal value and you can begin the simplex at $x = (0,0,0,0,5,4,3,4)$.



To formulate this as a transportation problem, one possible way is to imagine that on the third day, all the remaining beers are going to be donated for free. Hence the total demand is equal to the total supply and the objective function is preserved. The parameter table can be listed as follows. The last column represents the supply amount and the last row represents the demand amount. The other indicates the corresponding cost.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & 3 & 2.7 & 0 & 5 \ hline
S_2 & 2.9 & 2.8 & 0 & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



To solve the problem, you can reduce it to a $2$ dimensional problem and solve it.



Or just for fun, you can do the following:



In each of the boxes below, I use the convention of $(c_{ij}, y_{ij})$ where $y_{ij}=x_{ij}$ if it is basic and $y_{ij}=c_{ij}-u_i-v_j$ otherwise. I will use purple to indicate the first case.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}3) & (2.7,color{purple}2) & (0,0.1) & 5 \ hline
S_2 & (2.9, - 0.2) & (2.8,color{purple}2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



It is not optimal as we can see a negative reduce cost. That will be the entering variable.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}1) & (2.7,color{purple}4) & (0,-0.1) & 5 \ hline
S_2 & (2.9, color{purple}2) & (2.8,0.2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, 0.1) & (2.7,color{purple}4) & (0,color{purple}1) & 5 \ hline
S_2 & (2.9, color{purple}3) & (2.8,0.1) & (0,color{purple}1) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



Now, all the reduce cost are nonnegative. The solution suggest that on the first day, we should get $3$ pints at $2.9$ and on the second day, we should get $4$ pints at $2.7$.



Edit:



Let me avoid using big $M$ method since you are not familiar with it.



To implement the simplex algorithm, we need to find our basis matrix.



These are our linearity constraints.
$$x_1+x_2+x_5=5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3 =3$$
$$x_2+x_4 = 4$$
$$x ge 0$$



We want to pick $4$ non-basic variables. For example, let's pick $x_1=3, x_2=4$, let $x_3=0$ and $x_4=0$. Now, we can let $x_5=2$ and $x_6=4$.



That is we pick $x_1, x_2, x_5, x_6$ to be non-basic. Then our $$B = begin{bmatrix}1 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0\
0 & 1 & 0 & 0 end{bmatrix}.$$



From there we can easily compute $-c_B'B^{-1}A, c'-C_B'B^{-1}A, B^{-1}b, B^{-1}A$. That is we can construct our simplex tableu and start from there.



Remark:






share|cite|improve this answer























  • Thank you for your answer Siong Thye Goh.
    – Al t.
    Dec 5 '18 at 19:38










  • Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
    – Al t.
    Dec 7 '18 at 18:51












  • so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
    – Siong Thye Goh
    Dec 7 '18 at 19:18










  • I don't see how your comment answer my question in the comment
    – Al t.
    Dec 8 '18 at 18:30






  • 1




    I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
    – Siong Thye Goh
    Dec 9 '18 at 7:34














1





+50







1





+50



1




+50




Your initial linear programming problem is equivalent to



$$min 3x_1 + 2.7 x_2 + 2.9x_3 + 2.8 x_4 + Mx_7 + Mx_8$$
subject to $$x_1+x_2 + x_5 = 5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3+x_7 =3$$
$$x_2+x_4+x_8 = 4$$
$$x ge 0$$



The big $M$ is going to force $x_7$ and $x_8$ to be $0$ at optimal value and you can begin the simplex at $x = (0,0,0,0,5,4,3,4)$.



To formulate this as a transportation problem, one possible way is to imagine that on the third day, all the remaining beers are going to be donated for free. Hence the total demand is equal to the total supply and the objective function is preserved. The parameter table can be listed as follows. The last column represents the supply amount and the last row represents the demand amount. The other indicates the corresponding cost.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & 3 & 2.7 & 0 & 5 \ hline
S_2 & 2.9 & 2.8 & 0 & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



To solve the problem, you can reduce it to a $2$ dimensional problem and solve it.



Or just for fun, you can do the following:



In each of the boxes below, I use the convention of $(c_{ij}, y_{ij})$ where $y_{ij}=x_{ij}$ if it is basic and $y_{ij}=c_{ij}-u_i-v_j$ otherwise. I will use purple to indicate the first case.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}3) & (2.7,color{purple}2) & (0,0.1) & 5 \ hline
S_2 & (2.9, - 0.2) & (2.8,color{purple}2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



It is not optimal as we can see a negative reduce cost. That will be the entering variable.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}1) & (2.7,color{purple}4) & (0,-0.1) & 5 \ hline
S_2 & (2.9, color{purple}2) & (2.8,0.2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, 0.1) & (2.7,color{purple}4) & (0,color{purple}1) & 5 \ hline
S_2 & (2.9, color{purple}3) & (2.8,0.1) & (0,color{purple}1) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



Now, all the reduce cost are nonnegative. The solution suggest that on the first day, we should get $3$ pints at $2.9$ and on the second day, we should get $4$ pints at $2.7$.



Edit:



Let me avoid using big $M$ method since you are not familiar with it.



To implement the simplex algorithm, we need to find our basis matrix.



These are our linearity constraints.
$$x_1+x_2+x_5=5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3 =3$$
$$x_2+x_4 = 4$$
$$x ge 0$$



We want to pick $4$ non-basic variables. For example, let's pick $x_1=3, x_2=4$, let $x_3=0$ and $x_4=0$. Now, we can let $x_5=2$ and $x_6=4$.



That is we pick $x_1, x_2, x_5, x_6$ to be non-basic. Then our $$B = begin{bmatrix}1 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0\
0 & 1 & 0 & 0 end{bmatrix}.$$



From there we can easily compute $-c_B'B^{-1}A, c'-C_B'B^{-1}A, B^{-1}b, B^{-1}A$. That is we can construct our simplex tableu and start from there.



Remark:






share|cite|improve this answer














Your initial linear programming problem is equivalent to



$$min 3x_1 + 2.7 x_2 + 2.9x_3 + 2.8 x_4 + Mx_7 + Mx_8$$
subject to $$x_1+x_2 + x_5 = 5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3+x_7 =3$$
$$x_2+x_4+x_8 = 4$$
$$x ge 0$$



The big $M$ is going to force $x_7$ and $x_8$ to be $0$ at optimal value and you can begin the simplex at $x = (0,0,0,0,5,4,3,4)$.



To formulate this as a transportation problem, one possible way is to imagine that on the third day, all the remaining beers are going to be donated for free. Hence the total demand is equal to the total supply and the objective function is preserved. The parameter table can be listed as follows. The last column represents the supply amount and the last row represents the demand amount. The other indicates the corresponding cost.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & 3 & 2.7 & 0 & 5 \ hline
S_2 & 2.9 & 2.8 & 0 & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



To solve the problem, you can reduce it to a $2$ dimensional problem and solve it.



Or just for fun, you can do the following:



In each of the boxes below, I use the convention of $(c_{ij}, y_{ij})$ where $y_{ij}=x_{ij}$ if it is basic and $y_{ij}=c_{ij}-u_i-v_j$ otherwise. I will use purple to indicate the first case.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}3) & (2.7,color{purple}2) & (0,0.1) & 5 \ hline
S_2 & (2.9, - 0.2) & (2.8,color{purple}2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



It is not optimal as we can see a negative reduce cost. That will be the entering variable.



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, color{purple}1) & (2.7,color{purple}4) & (0,-0.1) & 5 \ hline
S_2 & (2.9, color{purple}2) & (2.8,0.2) & (0,color{purple}2) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



begin{array}{|c|c|c|c|c|}hline
& D_1 & D_2 & D_3 & text{Supply} \ hline
S_1 & (3, 0.1) & (2.7,color{purple}4) & (0,color{purple}1) & 5 \ hline
S_2 & (2.9, color{purple}3) & (2.8,0.1) & (0,color{purple}1) & 4 \ hline
text{Demand} & 3 & 4 & 2 & \ hline
end{array}



Now, all the reduce cost are nonnegative. The solution suggest that on the first day, we should get $3$ pints at $2.9$ and on the second day, we should get $4$ pints at $2.7$.



Edit:



Let me avoid using big $M$ method since you are not familiar with it.



To implement the simplex algorithm, we need to find our basis matrix.



These are our linearity constraints.
$$x_1+x_2+x_5=5$$
$$x_3+x_4+x_6=4$$
$$x_1+x_3 =3$$
$$x_2+x_4 = 4$$
$$x ge 0$$



We want to pick $4$ non-basic variables. For example, let's pick $x_1=3, x_2=4$, let $x_3=0$ and $x_4=0$. Now, we can let $x_5=2$ and $x_6=4$.



That is we pick $x_1, x_2, x_5, x_6$ to be non-basic. Then our $$B = begin{bmatrix}1 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0\
0 & 1 & 0 & 0 end{bmatrix}.$$



From there we can easily compute $-c_B'B^{-1}A, c'-C_B'B^{-1}A, B^{-1}b, B^{-1}A$. That is we can construct our simplex tableu and start from there.



Remark:







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 7:31

























answered Dec 5 '18 at 18:27









Siong Thye GohSiong Thye Goh

100k1465117




100k1465117












  • Thank you for your answer Siong Thye Goh.
    – Al t.
    Dec 5 '18 at 19:38










  • Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
    – Al t.
    Dec 7 '18 at 18:51












  • so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
    – Siong Thye Goh
    Dec 7 '18 at 19:18










  • I don't see how your comment answer my question in the comment
    – Al t.
    Dec 8 '18 at 18:30






  • 1




    I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
    – Siong Thye Goh
    Dec 9 '18 at 7:34


















  • Thank you for your answer Siong Thye Goh.
    – Al t.
    Dec 5 '18 at 19:38










  • Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
    – Al t.
    Dec 7 '18 at 18:51












  • so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
    – Siong Thye Goh
    Dec 7 '18 at 19:18










  • I don't see how your comment answer my question in the comment
    – Al t.
    Dec 8 '18 at 18:30






  • 1




    I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
    – Siong Thye Goh
    Dec 9 '18 at 7:34
















Thank you for your answer Siong Thye Goh.
– Al t.
Dec 5 '18 at 19:38




Thank you for your answer Siong Thye Goh.
– Al t.
Dec 5 '18 at 19:38












Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
– Al t.
Dec 7 '18 at 18:51






Why did you add $x_7$ and $x_8$ in the constraints of the linear program?
– Al t.
Dec 7 '18 at 18:51














so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
– Siong Thye Goh
Dec 7 '18 at 19:18




so that if i want to start my simplex algorithm, there is an obvious starting point, $x_5$ to $x_8$ can be my basic variables and we set the rest to be zero. However, at the optimal value, $x_7$ and $x_8$ will be forced to be zero as the coefficient is $M$ in the objective value.
– Siong Thye Goh
Dec 7 '18 at 19:18












I don't see how your comment answer my question in the comment
– Al t.
Dec 8 '18 at 18:30




I don't see how your comment answer my question in the comment
– Al t.
Dec 8 '18 at 18:30




1




1




I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
– Siong Thye Goh
Dec 9 '18 at 7:34




I believe the question is asked to test whether you are able to use big M method but of course, you don't have to use it. As long as you can pick your non-basic columns/ variables, you can do the simplex algorithm. I have edited to show how to do so. (remark: Depending on how simplex is taught to you, i am not sure if what i say makes sense to you, if it doesn't, you might want to tell me more about what you really know and what you do not know).
– Siong Thye Goh
Dec 9 '18 at 7:34


















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