D2 January 2006 Q4

EdexcelOld spec16 marksTransportation Problems

4. The following minimising transportation problem is to be solved.

	J	K	Supply
A	12	15	9
B	8	17	13
C	4	9	12
Demand	9	11

(a) Complete the table below.

	J	K	Supply
A	12	15	9
B	8	17	13
C	4	9	12
Demand	9	11	34

(2)

(b) Explain why an extra demand column was added to the table above. (2)

A possible north-west corner solution is:

	J	K	L
A	9	0
B		11	2
C			12

After three iterations of the stepping-stone method the table becomes:

	J	K	L
A		8	1
B			13
C	9	3

(d) Taking the most negative improvement index as the entering square for the stepping stone method, solve the transportation problem. You must make your shadow costs and improvement indices clear and demonstrate that your solution is optimal. (11)

Scheme	Marks
Adds zero for costs in third column	B1
Adds 14 as the demand value	B1
	(2)

Scheme	Marks
The total supply is greater than the total demand	B2, 1, 0
	(2)

Scheme	Marks
The solution would otherwise be degenerate	B1
	(1)

Mark scheme (d)

Scheme

Marks

		10	15	0
		J	K	L
0	A		8	1
0	B			13
−6	C	9	3

\(I_{AJ} = 12 - 0 - 10 = 2\)
\(I_{BJ} = 8 - 0 - 10 = -2\)
\(I_{BK} = 17 - 0 - 15 = 2\)
\(I_{CL} = 0 + 6 - 0 = 6\)

M1 A1
A1
A1 (4)

	J	K	L
A		\(8 - \theta\)	\(1 + \theta\)
B	\(\theta\)		\(13 - \theta\)
C	\(9 - \theta\)	\(3 + \theta\)

\(\theta = 8\)
Entering square BJ
Exiting square AK

M1
A1ft (2)

		8	13	0
		J	K	L
0	A			9
0	B	8		5
−4	C	1	11

\(I_{AJ} = 12 - 0 - 8 = 4\)
\(I_{AK} = 15 - 0 - 13 = 2\)
\(I_{BK} = 17 - 0 - 13 = 4\)
\(I_{CL} = 0 + 4 - 0 = 4\)

M1 A1ft
A1ft
A1ft
A1 (5)

No negatives, so optimal

(11)

(16 marks)

Notes